Engineering Circuit Analysis Solution Manual From chap 1 to Chap 12

Engineering Circuit Analysis, 7th Edition 1. (a) 12 μs (b) 750 mJ (c) 1.13 kΩ (d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz ...

Author: William Hart Hayt | Jack E. Kemmerly | Steven M. Durbin

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Engineering Circuit Analysis, 7th Edition

1. (a)

12 μs (b) 750 mJ (c) 1.13 kΩ

(d) 3.5 Gbits (e) 6.5 nm (f) 13.56 MHz

Chapter Two Solutions

10 March 2006

(g) 39 pA (h) 49 kΩ (i) 11.73 pA

PROPRIETARY MATERIAL . © 2007 Th e McGraw-Hill Companies, Inc. Lim ited distr ibution perm itted onl y to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


2.

(a) 1 MW (b) 12.35 mm (c) 47. kW (d) 5.46 mA

(e) 33 μJ (f) 5.33 nW (g) 1 ns (h) 5.555 MW


10 March 2006

(i) 32 mm



3. (a)


⎛ 745.7 W ⎞ ⎟ = 298.3 ⎝ 1 hp ⎠

( 400 Hp ) ⎜

10 March 2006

kW

⎛ 12 in ⎞ ⎛ 2.54 cm ⎞ ⎛ 1 m ⎞ (b) 12 ft = (12 ft) ⎜ ⎟⎜ ⎟⎜ ⎟ = 3.658 m ⎝ 1 ft ⎠ ⎝ 1 in ⎠ ⎝ 100 cm ⎠ (c) 2.54 cm =

25.4 mm

⎛ 1055 J ⎞ (d) ( 67 Btu ) ⎜ ⎟ = 70.69 ⎝ 1 Btu ⎠ (e) 285.4´10-15 s =

kJ

285.4 fs



4.


10 March 2006

(15 V)(0.1 A) = 1.5 W = 1.5 J/s. 3 hrs running at this power level equates to a transfer of energy equal to (1.5 J/s)(3 hr)(60 min/ hr)(60 s/ min) = 16.2 kJ



5.


10 March 2006

Motor power = 175 Hp (a) With 100% efficient mechanical to electrical power conversion, (175 Hp)[1 W/ (1/745.7 Hp)] = 130.5 kW (b) Running for 3 hours, Energy = (130.5×103 W)(3 hr)(60 min/hr)(60 s/min) = 1.409 GJ (c) A single battery has 430 kW-hr capacity. We require (130.5 kW)(3 hr) = 391.5 kW-hr therefore one battery is sufficient.



6.


10 March 2006

The 400-mJ pulse lasts 20 ns. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 400

20

Then

t (ns)

P = 400×10-3/20×10-9 = 20 MW.

(b) At 20 pulses per second, the average power is Pavg = (20 pulses)(400 mJ/pulse)/(1 s) = 8 W.



7.


10 March 2006

The 1-mJ pulse lasts 75 fs. (a) To compute the peak power, we assume the pulse shape is square: Energy (mJ) 1

75

Then

t (fs)

P = 1×10-3/75×10-15 = 13.33 GW.

(b) At 100 pulses per second, the average power is Pavg = (100 pulses)(1 mJ/pulse)/(1 s) = 100 mW.



8.


10 March 2006

The power drawn from the battery is (not quite drawn to scale): P (W)

10

6 t (min) 5

7

17

24

(a) Total energy (in J) expended is [6(5) + 0(2) + 0.5(10)(10) + 0.5(10)(7)]60 = 6.9 kJ. (b) The average power in Btu/hr is (6900 J/24 min)(60 min/1 hr)(1 Btu/1055 J) = 16.35 Btu/hr.



9.


10 March 2006

The total energy transferred during the first 8 hr is given by (10 W)(8 hr)(60 min/ hr)(60 s/ min) = 288 kJ The total energy transferred during the last five minutes is given by

∫

300 s

0

10 2 ⎡ 10 ⎤ ⎢⎣ − 300 t + 10 ⎥⎦ dt = − 600 t + 10t

300

= 1.5 kJ 0

(a) The total energy transferred is 288 + 1.5 = 289.5 kJ (b) The energy transferred in the last five minutes is 1.5 kJ




10 March 2006

charge q = 18t2 – 2t4 C.

10. Total

(a) q(2 s) = 40 C. (b) To find the m aximum charge within 0 ≤ t ≤ 3 s, we need to take the second derivitives:

f irst and

dq/dt = 36t – 8t3 = 0, leading to roots at 0, ± 2.121 s d2q/dt2 = 36 – 24t2

substituting t = 2.121 s into the expression for d2q/dt2, we obtain a value of –14.9, so that this root represents a maximum. Thus, we find a maximum charge q = 40.5 C at t = 2.121 s. (c) The rate of charge accumulation at t = 8 s is dq/dt|t = 0.8 = 36(0.8) – 8(0.8)3 = 24.7 C/s. (d) See Fig. (a) and (b). 50 70

(b)

(a)

60 50

0

30

tim e (t)

q (C)

40

20

-50

10 0

-100

-10 -20 0

0.5

1

1.5 tim e (s )

2

2.5

3

-150 0

0.5

1

1.5 i (A )

2

2.5


3


11.


10 March 2006

Referring to Fig. 2.6c,

⎧- 2 + 3e −5t A, i1 (t ) = ⎨ 3t ⎩ - 2 + 3e A,

t0

Thus, (a) i1(-0.2) = 6.155 A (b) i1 (0.2) = 3.466 A (c) To determine the instants at which i1 = 0, we must consider t < 0 and t > 0 separately: for t < 0, - 2 + 3e-5t = 0 leads to t = -0.2 ln (2/3) = +0.0811 s (impossible) for t > 0, -2 + 3e3t = 0 leads to t = (1/3) ln (2/3) = –0.135 s (impossible) Therefore, the current is never negative. (d) The total charge passed left to right in the interval –0. 8 < t < 0.1 s is q(t)

= =

∫

0.1

i (t )dt

−0.8 1

0

−5 t ∫−0.8 ⎡⎣ −2 + 3e ⎤⎦ dt

+

0

3 ⎛ ⎞ = ⎜ −2t − e−5t ⎟ 5 ⎝ ⎠ -0.8

+

∫

0.1

0

⎡⎣ −2 + 3e3t ⎤⎦ dt

( −2t + e ) 3t

0.1 0

= 33.91 C



12.


10 March 2006

Referring to Fig. 2.28,

(a) The average current over one period (10 s) is iavg = [-4(2) + 2(2) + 6(2) + 0(4)]/10

= 800 mA

(b) The total charge transferred over the interval 1 < t < 12 s is qtotal =

∫

12

1

i (t )dt = -4(1) + 2(2) + 6(2) + 0(4) – 4(2) = 4 C

(c) See Fig. below

q (C) 8 -8

16 2

4

10

6 8

12 14

t(s) 16

-16



13. (a)

VBA = – (b) VED =

2 pJ -1.602 × 10-19 C


=

12.48 MV

0 -1.602 × 10-19 C

=

0

3 pJ 1.602 × 10-19 C

=

–18.73 MV

(c) VDC = –

10 March 2006

(d) It takes – 3 pJ to move +1.602x10–19 C from D to C. It takes 2 pJ to move –1.602x10–19 C from B to C, or –2 pJ to move –19 C from B to C, or +2 pJ to move +1.602x10–19 C from C to B. +1.602x10 Thus, it requires –3 pJ + 2 pJ = –1 pJ to m ove +1.602x10 –19 C from D to C to B. Hence,

VDB =

−1 pJ = –6.242 MV. 1.602 × 10-19 C




10 March 2006

14.

+ V1 –

– V2 +

+ –

Voltmeter

+

Voltmeter

From the diagram, we see that V2 = –V1 = +2.86 V.




15. (a)

Pabs = (+3.2 V)(-2 mA) = –6.4 mW

(b)

Pabs = (+6 V)(-20 A) = –120 W

10 March 2006

(or +6.4 mW supplied)

(or +120 W supplied)

(d) Pabs = (+6 V)(2 ix) = (+6 V)[(2)(5 A)] = +60 W (e) Pabs = (4 sin 1000t V)(-8 cos 1000t mA)| t = 2 ms

= +12.11 W



16.


10 March 2006

i = 3te-100t mA and v = [6 – 600t] e-100t mV (a) The power absorbed at t = 5 ms is

[

Pabs = (6 − 600t ) e −100t ⋅ 3te −100t =

]

t = 5 ms

μW

0.01655 μW = 16.55 nW

(b) The energy delivered over the interval 0 < t < ∞ is

∫

∞

0

Pabs dt

=

∫

∞

0

3t (6 − 600t ) e − 200t dt

μJ

Making use of the relationship

∫

∞

0

x n e − ax dx

=

n! a n +1

where n is a positive integer and a > 0,

we find the energy delivered to be = 18/(200)2 - 1800/(200)3 = 0



17.


(a) Pabs = (40i)(3e-100t)| t = 8 ms =

[

di ⎞ ⎛ (b) Pabs = ⎜ 0.2 ⎟ i = - 180 e −100t dt ⎠ ⎝

(

t (c) Pabs = ⎛⎜ 30∫ idt + 20 ⎞⎟ 3e −100t ⎝ 0 ⎠

)

[

360 e −100t

]

2

t = 8 ms

]

2

t = 8 ms

10 March 2006

= 72.68 W

= - 36.34 W

t = 8 ms

t = ⎛⎜ 90e −100t ∫ 3e −100t ′ dt ′ + 60e −100t ⎞⎟ 0 ⎝ ⎠

= 27.63 W t = 8 ms



18.


10 March 2006

(a) The short-circuit current is the value of the current at V = 0. Reading from the graph, this corresponds to approximately 3.0 A. (b) The open-circuit voltage is the value of the voltage at I = 0. Reading from the graph, this corresponds to roughly 0.4875 V, estimating the curve as hitting the x-axis 1 mm behind the 0.5 V mark. (c) We see that th e maximum current corres ponds to zero voltage , and likewise, the maximum voltage occu rs at zero curren t. The m aximum power point, therefore, occurs somewhere between these two points. By trial and error, Pmax is roughly (375 mV)(2.5 A) = 938 mW, or just under 1 W.



19. (a)


10 March 2006

P first 2 hours = ( 5 V )( 0.001 A ) = 5 mW

P next 30 minutes = ( ? V )( 0 A )

= 0 mW

P last 2 hours = ( 2 V )( −0.001 A ) = −2 mW (b) Energy = (5 V)(0.001 A)(2 hr)(60 min/ hr)(60 s/ min) = 36 J (c) 36 – (2)(0.001)(60)(60) = 21.6 J



20.


10 March 2006

Note that in the table below, only th e –4-A sou rce and the –3-A source are actually “absorbing” power; the remaining sources are supplying power to the circuit. Source 2 V source 8 V source -4 A source 10 V source -3 A source

Absorbed Power (2 V)(-2 A) (8 V)(-2 A) (10 V)[-(-4 A)] (10 V)(-5 A) (10 V)[-(-3 A)]

The 5 powe r quantities sum to –4 – 16 + conservation of energy.

Absorbed Power -4W - 16 W 40 W - 50 W 30 W

40 – 50 + 30 = 0, as de

manded from




10 March 2006

21. 20 A

32 V 8V

–16 A

40 V –12 A

40 V P8V supplied

= (8)(8)

=

64 W

(source of energy)

P32V supplied

= (32)(8)

=

256 W

(source of energy)

P–16A supplied

= (40)(–16)

=

–640 W

P40V supplied

= (40)(20)

=

800 W

P–12A supplied

= (40)( –12) =

Check:

∑ supplied power

(source of energy)

–480 W

= 64 + 256 – 640 + 800 – 480 = 0 (ok)



22.


10 March 2006

We are told that V x = 1 V, and from Fig. 2.33 w e see that the current flowing through the dependent source (and hence th rough each element of t he circuit) is 5V x = 5 A. We will co mpute absorbed power by using the current flowing into the po sitive reference terminal of the a ppropriate voltage (p assive sign conven tion), and we will compute supplied power by using the current flowing out of the pos itive reference terminal of the appropriate voltage. (a) The power absorbed by element “A” = (9 V)(5 A) = 45 W (b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and the power supplied by the dependent source = (8 V)(5 A) = 40 W (c) The sum of the supplied power = 5 + 40 = 45 W The sum of the absorbed power is 45 W, so yes, the su m of the power supplied = the sum of the power absorbed, expect from the principle of conservation of energy.

as we



23.


10 March 2006

We are asked to determine the voltage vs, which is identical to the voltage labeled v1. The only remaining reference to v1 is in the expression for the current flowing through the dependent source, 5v1. This current is equal to –i2. Thus, 5 v1 = -i2 = - 5 mA Therefore v1 = -1 mV and so

vs = v1 = -1 mV



24.


10 March 2006

The voltage across the dependent source = v2 = –2ix = –2(–0.001) = 2 mV.



25.


10 March 2006

The battery delivers an energy of 460.8 W-hr over a period of 8 hrs. (a) The power delivered to the headlight is therefore (460.8 W-hr)/(8 hr) = 57.6 W (b) The current through the headlight is equal to the power it absorbs from the battery divided by the voltage at which the power is supplied, or I = (57.6 W)/(12 V) = 4.8 A



26.


10 March 2006

The supply voltage is 110 V, and the m aximum dissipated power is 500 W. The fuse s are specified in te rms of current, so we need to determine the m aximum current th at can flow through the fuse. P=VI

therefore Imax = Pmax/V = (500 W)/(110 V) = 4.545 A

If we choose the 5-A fuse, it will allow up to (110 V)(5 A) = 550 W of power to be delivered to the application (we must assume here that the fuse absorbs zero power, a reasonable assumption in practice). This exceeds the specified maximum power. If we choose the 4.5-A fuse instead, we will hav e a m aximum current of 4.5 A. This leads to a maximum power of (110)(4.5) = 495 W delivered to the application. Although 495 W is less than the m aximum po wer allowed, this fuse will provid e adequate protection for the app lication circuitry. If a faul t occurs and the application circuitry attempts to draw too m uch power, 1000 W for example, the fuse will blow, no current will f low, and the application c ircuitry will be p rotected. However, if the application circuitry tries to draw its maximum rated power (500 W ), the fuse will also blow. In practice, m ost equipm ent will not draw its m aximum rated powe r continuously—although to be safe, we typically assume that it will.



27. (a)

(b)


imax = 5/ 900 =

5.556 mA

imin = 5/ 1100 =

4.545 mA

10 March 2006

p = v2 / R so pmin = 25/ 1100

= 22.73 mW

pmax = 25/ 900

= 27.78 mW



28.


10 March 2006

p = i2 R, so pmin = (0.002)2 (446.5) = 1.786 mW and (more relevant to our discussion) pmax = (0.002)2 (493.5) = 1.974 mW

1.974 mW would be a correct answer, although power ratings are typically expressed as integers, so 2 mW might be more appropriate.



29. (a)


10 March 2006

Pabs = i2R = [20e-12t] 2 (1200) μW

= [20e-1.2] 2 (1200) μW = (b)

43.54 mW Pabs = v2/R = [40 cos 20t] 2 / 1200 W 2

=

[40 cos 2] / 1200 W

=

230.9 mW

(c) =

Pabs = v i=

keep in mind we are using radians

8t 1.5 W 253.0 mW



30.


10 March 2006

It’s probably best to begin this problem by sketching the voltage waveform: v (V) +10 40

60

t (ms)

20 -10

(a) vmax = +10 V (b) vavg = [(+10)(20×10-3) + (-10)(20×10-3)]/(40×10-3)

= 0

(c) iavg = vavg /R = 0 (d) pabs

(e) pabs

max

avg

=

=

2 vmax = (10)2 / 50 = 2 W R

⎤ 1 ⎡ (+10) 2 (−10) 2 ⋅ 20 + ⋅ 20⎥ = 2 W ⎢ R 40 ⎣ R ⎦




10 March 2006

31.

Since we are inform ed that the same current must flow through each com ponent, we begin by defining a current I flowing out of the positive reference term inal of the voltage source. The power supplied by the voltage source is Vs I. The power absorbed by resistor R1 is I2R1. The power absorbed by resistor R2 is I2R2. Since we know that the total power supplied is equal to the total power absorbed, we may write: Vs I = I2R1 + I2R2 or Vs = I R1 + I R2 Vs = I (R1 + R2) By Ohm’s law, I = VR2 / R2 so that

Vs =

VR2 R2

(R1

+ R2 )

Solving for VR2 we find

VR2 = Vs

R2 (R1 + R2 )

Q.E.D.




10 March 2006

32. (a) 6 5 4

c urrent (m A )

3 2 1 0 -1 -2 -3 -4 -1.5

-1

-0.5

0

0.5 voltage (V )

1

1.5

2

2.5

(b) W e see from our answer to part (a ) that this device has a reasonably linear characteristic (a not unreasonable degree of e xperimental error is evident in the data). Thus, we choose to estimate the resistance using the two extreme points:

Reff = [(2.5 – (-1.5)]/[5.23 – (-3.19)] kΩ = 475 Ω Using the last two points instead, we find Reff = 469 Ω, so that we can state with some certainty at least that a reasonable estimate of the resistance is approximately 470 Ω. (c) 2

1.5

c urrent (m A )

1

0.5

0

-0.5

-1

-1.5 -1.5

-1

-0.5

0

0.5 voltage (V )

1

1.5

2

2.5



33.


10 March 2006

Top Left Circuit:

I = (5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW

Top Right Circuit:

I = -(5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW

Bottom Left Circuit: I = (-5/10) mA = -0.5 mA, and P10k = V2/10 mW = 2.5 mW Bottom Right Circuit: I = -(-5/10) mA = 0.5 mA, and P10k = V2/10 mW = 2.5 mW



34. The


10 March 2006

voltage vout is given by

vout

= -10-3 vπ (1000) = - vπ

Since

vπ = vs = 0.01 cos 1000t V, we find that vout

= - vπ =

-0.01 cos 1000t V



35.


10 March 2006

vout = -vπ = -vS = -2sin 5t V vout (t = 0) = 0 V vout (t = 0.324 s) = -2sin (1.57) = -2 V (use care to em ploy r adian m ode on your calculator or convert 1.57 radians to degrees)



36.


10 March 2006

18 AWG wire has a resistance of 6.39 Ω / 1000 ft. Thus, we require 1000 (53) / 6.39 = 8294 ft of wire. (Or 1.57 miles. Or, 2.53 km).



37. tem


10 March 2006

We need to create a 47 0-Ω resistor from 28 AWG wire , knowing that the am bient perature is 108oF, or 42.22oC. Referring to Table 2.3, 28 AWG wire is 65.3 m Ω/ft at 20 oC, and using the equation provided we compute R2/R1 = (234.5 + T2)/(234.5 + T1) = (234.5 + 42.22)/(234.5 + 20) = 1.087 We thus find that 28 AWG wire is (1.087)(65.3) = 71.0 mΩ/ft. Thus, to repair the transmitter we will need (470 Ω)/(71.0 × 10-3 Ω/ft) = 6620 ft (1.25 miles, or 2.02 km). Note: This seems like a lot of wire to be washing up on shore. We may find we don’t have enough. In that case, perhaps we shoul d take our cue from Eq. [6], and try to squash a piece of the wire flat so that it has a very small cross-sectional area…..



38.


10 March 2006

We are given that the conductivity σ of copper is 5.8×107 S/m. (a) 50 ft of #18 (18 AWG) copper wire, which has a diameter of 1.024 mm, will have a resistance of l/(σ A) ohms, where A = the cross-sectional area and l = 50 ft. Converting the dimensional quantities to meters,

l = (50 ft)(12 in/ft)(2.54 cm/in)(1 m/100 cm) = 15.24 m and

r = 0.5(1.024 mm)(1 m/1000 mm) = 5.12×10-4 m so that

A = π r2 = π (5.12×10-4 m)2 = 8.236×10-7 m2 Thus, R = (15.24 m)/[( 5.8×107)( 8.236×10-7)] = 319.0 mΩ (b) We assume that the conductivity value specified also holds true at 50oC. The cross-sectional area of the foil is

A = (33 μm)(500 μm)(1 m/106 μm)( 1 m/106 μm) = 1.65×10-8 m2 So that

R = (15 cm)(1 m/100 cm)/[( 5.8×107)( 1.65×10-8)] = 156.7 mΩ A 3-A current flowing through this copper in the direction specified would lead to the dissipation of

I2R = (3)2 (156.7) mW = 1.410 W



39.


10 March 2006

Since R = ρ l / A, it follows that ρ = R A/ l . From Table 2.4, we see that 28 AWG soft copper wire (cross-sectional area = 0.080 4 mm2) is 65.3 Ω per 1000 ft. Thus, R = 65.3 Ω. l = (1000 ft)(12 in/ft)(2.54 cm/in)(10 mm/cm) = 304,800 mm.

A = 0.0804 mm2. Thus, ρ = (65.3)(0.0804)/304800 = 17.23 μΩ/ mm or

ρ = 1.723 μΩ.cm which is in fact consistent with the representative data for copper in Table 2.3.



40.


10 March 2006

(a) From the text, (1) Zener diodes, (2) Fuses, and (3) Incandescent (as opposed to fluorescent) light bulbs This last on e requires a few facts to be put together. We have st ated that temperature can affect resistance—in other words, if the temperature changes during operation, the resistance will not rem ain constant and hence nonlinear behavior will be observed. Most discrete resistors are rated for up to a sp ecific power in o rder to ensure th at temperature variation during operation will no t significantly change th e res istance value. Light bulbs, however, become rather warm when operating and can experience a significant change in resistance. (b) The energy is dissipated by the resistor, converted to heat which is transferred to the air surrounding the resistor. The resistor is unable to store the energy itself.



41.


10 March 2006

The quoted resistivity ρ of B33 copper is 1.7654 μΩ.cm. A = πr2 = π(10–3)2 = 10–6π m2. l = 100 m.

(1.7654 ×10 Thus, R = ρ l / A =

−6

)

Ωcm (1 m/100 cm )(100 m ) 10−6 π

= 0.5619 Ω

And P = I2R = (1.5)2 (0.5619) = 1.264 W.



42.


10 March 2006

We know that for any wire of cross-sectional area A and length l , the resistan ce is given by R = ρ l / A. If we keep ρ fixed by choosing a material, and A fixed by choosing a wire gauge (e.g. 28 AWG), changing l will change the resistance of our “device.” A simple variable resistor concept, then:

Leads to connect to circuit

Copper wire Rotating sh ort wire de termines length of long wire used in circuit.

But this is s omewhat impractical, as the l eads may turn out to have alm ost the sam e resistance unless we have a very long wi re, which can also be im practical. One improvement would be to replace the coppe r wire shown with a coil of insulated copper wire. A s mall amount of insulation would then need to be removed from where the moveable wire touches the coil so that electrical connection could be made.



43.


10 March 2006

(a) We need to plot the negative and positive voltage ranges separately, as the positive voltage range is, after all, exponential! x 10

-6

10

16

-4

14 12

10

-5

c urrent (A )

c urrent (A)

10 8 6

10

-6

4

10

2

-7

0 -2 -0.7

10 -0.6

-0.5

-0.4

-0.3 -0.2 voltage (V )

-0.1

0

-8

0.1

0

(b) To dete rmine the resist ance of the device at corresponding current:

0.01

0.02

0.03 0.04 voltage (V)

V = 550 mV, we com

0.05

0.06

pute the

I = 10-9 [e39(0.55) – 1] = 2.068 A Thus,

R(0.55 V) = 0.55/2.068 = 266 mΩ (c) R = 1 Ω corresponds to V = I. Thus, we need to solve the transcendental equation

I = 10-9 [e39I – 1] Using a scientific calculator or the tried-and-true trial and error approach, we find that

I = 514.3 mA


0.07


44.


10 March 2006

We require a 10- Ω resistor, and are told it is f or a portable application, implying that size, weight or both would be im portant to consider when selecting a wire gauge. W e have 10,000 ft of each of the gaug es listed in Table 2.3 w ith which to work. Quick inspection o f the values listed e liminates 2, 4 a nd 6 AW G w ire as the ir respec tive resistances are too low for only 10,000 ft of wire. Using 12-AWG wire would require (10 Ω) / (1.59 mΩ/ft) = 6290 ft. Using 28-AWG wire, the narrowest available, would require (10 Ω) / (65.3 mΩ/ft) = 153 ft. Would the 28-AW G wire weight less? Again referring to T able 2.3, we see that the cross-sectional area of 28-AWG wire is 0.0804 mm 2, and that of 12-AWG wire is 3.31 mm2. The volume of 12-AWG wire required is therefore 6345900 mm 3, and that of 28-AWG wire required is only 3750 mm3. The best (but not the only) choice for a portable application is clear: 28-AWG wire!



45. cm


10 March 2006

Our target is a 100-Ω resistor. We see from the plot that at ND = 1015 cm-3, μn ~ 2x103 2 /V-s, yielding a resistivity of 3.121 Ω-cm. At ND = 1018 cm-3, μn ~ 230 cm2/ V-s, yielding a resistivity of 0.02714 Ω-cm. Thus, we s ee that the lower doping level cl early provides m aterial with higher resistivity, requiring less of the available area on the silicon wafer. Since R = ρL/A, where we know R = 100 Ω and ρ = 3.121 Ω-cm for a phosphorus concentration of 1015 cm-3, we need only define the resistor geom etry to complete the design. We choose a geometry as shown in the figure; our contact area is arbitrarily chosen as 100 μm by 250 μm, so that only the length L remains to be specified. Solving, L=

R

ρ

A=

(100 Ω)(100 μ m )(250 μ m ) = 80.1 μ m 4 ( 3.121 Ω cm ) 10 μ m/cm

(

)

Design summary (one possibility): Contact

ND = 1015 cm-3 L = 80.1 μm width = 100 μm

(Note: this is som ewhat atypical; in the semiconductor industry contacts are typically made to the top and/o r bottom surface of a wafer. So, there’s more than one solu tion based on geometry as well as doping level.)

100 μm Wafer surface 250 μm contact 80.1 μm

contact



Chapter Three Solutions

10 March 2006

1.

PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


2.


10 March 2006

(a) six nodes; (b) nine branches.



3.


10 March 2006

(a) Four nodes; (b) five branches; (c) path, yes – loop, no.



4.


10 March 2006

(a) Five nodes; (b) seven branches; (c) path, yes – loop, no.



5.


10 March 2006

(a) The number of nodes remains the same – four (4). (b) The number of nodes is increased by one – to five (5). (c)

i) ii) iii) iv) v)

YES NO YES NO NO

– does not return to starting point – does not return to starting point – point B is crossed twice



6.

(a) By KCL at the bottom node: So iZ = 9 A.


10 March 2006

2 – 3 + iZ – 5 – 3 = 0

(b) If the left-most resistor has a value of 1 Ω, then 3 V appears across the parallel network (the ‘+’ reference terminal being the bottom node) Thus, the value of the other resistor is given by R=

3 = 600 mΩ . −(−5)



7.


10 March 2006

(a) 3 A; (b) –3 A; (c) 0.



8.


10 March 2006

By KCL, we may write: 5 + iy + iz = 3 + ix (a) ix = 2 + iy + iz = 2 + 2 + 0 = 4 A (b) iy = 3 + ix – 5 – iz iy = –2 + 2 – 2 iy Thus, we find that iy = 0. (c) 5 + iy + iz = 3 + ix

→ 5 + ix + ix = 3 + ix so ix = 3 – 5 = -2A.



9.


10 March 2006

Focusing our attention on the bottom left node, we see that ix = 1 A. Focusing our attention next on the top right node, we see that iy = 5 A.



10.


10 March 2006

We obtain the current each bulb draws by dividing its power rating by the operating voltage (115 V): I100W = 100/115 = 896.6 mA I60W = 60/115 = 521.7 mA I40W = 347.8 mA Thus, the total current draw is 1.739 A.



11.


10 March 2006

The DMM is connected in parallel with the 3 load resistors, across which develops the voltage we wish to measure. If the DMM appears as a short, then all 5 A flows through the DMM, and none through the resistors, resulting in a (false) reading of 0 V for the circuit undergoing testing. If, instead, the DMM has an infinite internal resistance, then no current is shunted away from the load resistors of the circuit, and a true voltage reading results.



12.


10 March 2006

In either case, a bulb failure will adversely affect the sign. Still, in the parallel-connected case, at least 10 (up to 11) of the other characters will be lit, so the sign could be read and customers will know the restaurant is open for business.



13.


10 March 2006

(a) vy = 1(3vx + iz) vx = 5 V and given that iz = –3 A, we find that vy = 3(5) – 3 = 12 V (b) vy = 1(3vx + iz) = –6 = 3vx + 0.5 Solving, we find that vx = (–6 – 0.5)/3 = –2.167 V.



14.


10 March 2006

(a) ix = v1/10 + v1/10 = 5 2v1 = 50 so

v1 = 25 V.

By Ohm’s law, we see that iy = v2/10 also, using Ohm’s law in combination with KCL, we may write ix = v2/10 + v2/10 = iy + iy = 5 A Thus,

iy = 2.5 A.

(b) From part (a), ix = 2 v1/ 10. Substituting the new value for v1, we find that ix = 6/10 = 600 mA. Since we have found that iy = 0.5 ix,

iy = 300 mA.

(c) no value – this is impossible.



15.


10 March 2006

We begin by making use of the information given regarding the power generated by the 5-A and the 40-V sources. The 5-A source supplies 100 W, so it must therefore have a terminal voltage of 20 V. The 40-V source supplies 500 W, so it must therefore provide a current IX of 12.5 A.

(1) By KVL, –40 + (–110) + R(5) – 20 = 0 Thus, R = 34 Ω. (2) By KVL, -VG – (-110) + 40 = 0 So

VG = 150 V

Now that we know the voltage across the unknown conductance G, we need only to find the current flowing through it to find its value by making use of Ohm’s law. KCL provides us with the means to find this current: The current flowing into the “+” terminal of the –110-V source is 12.5 + 6 = 18.5 A. Then,

Ix = 18.5 – 5 = 13.5 A

By Ohm’s law,

I x = G · VG

So G = 13.5/ 150

or

G = 90 mS



16.


10 March 2006

(a) -1 + 2 + 10i – 3.5 + 10i = 0 Solving,

i = 125 mA

(b) +10 + 1i - 2 + 2i + 2 – 6 + i = 0 Solving, we find that 4i = -4 or i = - 1 A.



17.


10 March 2006

Circuit I. Starting at the bottom node and proceeding clockwise, we can write the KVL equation +7 – 5 – 2 – 1(i) = 0 Which results in i = 0. Circuit II. Again starting with the bottom node and proceeding in a clockwise direction, we write the KVL equation -9 +4i + 4i = 0

(no current flows through either the -3 V source or the 2 Ω resistor)

Solving, we find that i = 9/8 A = 1.125 A.



18.


10 March 2006

Begin by defining a clockwise current i. -vS + v1 + v2 = 0 and hence i =

so

vS = v1 + v2 = i(R1 + R2)

vS . R1 + R2

Thus, v1 = R1i =

R1 R2 vS and v2 = R2i = vS . R1 + R2 R1 + R2

Q.E.D.



19.


10 March 2006

Given: (1) Vd = 0 and (2) no current flows into either terminal of Vd. Calculate Vout by writing two KVL equations. Begin by defining current i1 flowing right through the 100 Ω resistor, and i2 flowing right through the 470 Ω resistor. -5 + 100i1 + Vd = 0

[1]

-5 + 100i1 + 470i2 + Vout = 0 [2] Making use of the fact that in this case Vd = 0, we find that i1 = 5/100 A. Making use of the fact that no current flows into the input terminals of the op amp, i1 = i2. Thus, Eq. [2] reduces to -5 + 570(5/100) + Vout = 0 or Vout = -23.5 V

(hence, the circuit is acting as a voltage amplifier.)



20.


10 March 2006

(a) By KVL, -2 + vx + 8 = 0 so that

vx = -6 V.

(b) By KCL at the top left node, iin = 1 + IS + vx/4 – 6 or

iin = 23 A

(c) By KCL at the top right node, IS + 4 vx = 4 - vx/4 So

IS = 29.5 A.

(d) The power provided by the dependent source is 8(4vx) = -192 W.



21.

(a) Working from left to right, v1 = 60 V v2 = 60 V i2 = 60/20 = 3 A i4 = v1/4 = 60/4 = 15 A v3 = 5i2 = 15 V By KVL, -60 + v3 + v5 = 0 v5 = 60 – 15 = 45 V v4 = v5 = 45


v1 = 60 V v2 = 60 V v3 = 15 V v4 = 45 V v5 = 45 V

10 March 2006

i1 = 27 A i2 = 3 A i3 = 24 A i4 = 15 A i5 = 9 A

i5 = v5/5 = 45/5 = 9 A i3 = i4 + i5 = 15 + 9 = 24 A i1 = i2 + i3 = 3 + 24 = 27 (b) It is now a simple matter to compute the power absorbed by each element: p1 p2 p3 p4 p5

= -v1i1 = v2i2 = v3i3 = v4i4 = v5i5

= -(60)(27) = (60)(3) = (15)(24) = (45)(15) = (45)(9)

= -1.62 kW = 180 W = 360 W = 675 W = 405 W

and it is a simple matter to check that these values indeed sum to zero as they should.



22.


10 March 2006

Refer to the labeled diagram below.

Beginning from the left, we find p20V = -(20)(4) = -80 W v1.5 = 4(1.5) = 6 V

therefore p1.5 = (v1.5)2/ 1.5 = 24 W.

v14 = 20 – v1.5 = 20 – 6 = 14 V therefore p14 = 142/ 14 = 14 W. i2 = v2/2 = v1.5/1.5 – v14/14 = 6/1.5 – 14/14 = 3 A Therefore v2 = 2(3) = 6 V and p2 = 62/2 = 18 W. v4 = v14 – v2 = 14 – 6 = 8 V therefore

p4 = 82/4 = 16 W

i2.5 = v2.5/ 2.5 = v2/2 – v4/4 = 3 – 2 = 1 A Therefore v2.5 = (2.5)(1) = 2.5 V and so

p2.5 = (2.5)2/2.5 = 2.5 W.

I2.5 = - IS, thefore IS = -1 A. KVL allows us to write

-v4 + v2.5 + vIS = 0

so VIS = v4 – v2.5 = 8 – 2.5 = 5.5 V and

pIS = -VIS IS = 5.5 W.

A quick check assures us that these power quantities sum to zero.



23.


10 March 2006

Sketching the circuit as described,

(a) v14 = 0.

v13 = v43 v23 = -v12 – v34 = -12 + 8 v24 = v23 + v34 = -4 – 8

= 8V = -4 V = -12 V

(b) v14 = 6 V. v13 = v14 + v43 = 6 + 8 v23 = v13 – v12 = 14 – 12 v24 = v23 + v34 = 2 – 8

= 14 V = 2V = -6 V

(c) v14 = -6 V. v13 = v14 + v43 = -6 + 8 v23 = v13 – v12 = 2 – 12 v24 = v23 + v34 = -10 – 8

= 2V = -10 V = -18 V



24.


10 March 2006

(a) By KVL, -12 + 5000ID + VDS + 2000ID = 0 Therefore,

VDS = 12 – 7(1.5) = 1.5 V.

(b) By KVL, - VG + VGS + 2000ID = 0 Therefore,

VGS = VG – 2(2) = -1 V.



25.


10 March 2006

Applying KVL around this series circuit, -120 + 30ix + 40ix + 20ix + vx + 20 + 10ix = 0 where vx is defined across the unknown element X, with the “+” reference on top. Simplifying, we find that 100ix + vx = 100 To solve further we require specific information about the element X and its properties. (a) if X is a 100-Ω resistor, Thus

vx = 100ix so we find that 100 ix + 100 ix = 100. ix = 500 mA and

px = vx ix = 25 W.

(b) If X is a 40-V independent voltage source such that vx = 40 V, we find that ix = (100 – 40) / 100 = 600 mA and

px = vx ix = 24 W

(c) If X is a dependent voltage source such that vx = 25ix, ix = 100/125 = 800 mA and

px = vx ix = 16 W.

(d) If X is a dependent voltage source so that vx = 0.8v1, where v1 = 40ix, we have 100 ix + 0.8(40ix) = 100 px = vx ix = 0.8(40)(0.7576)2 = 18.37 W.

or ix = 100/132 = 757.6 mA and

(e) If X is a 2-A independent current source, arrow up, 100(-2) + vx = 100 so that vx = 100 + 200 = 300 V and

px = vx ix = -600 W



26.


10 March 2006

(a) We first apply KVL: -20 + 10i1 + 90 + 40i1 + 2v2 = 0 where v2 = 10i1. Substituting,

70 + 70 i1 = 0

or

i1= -1 A.

(b) Applying KVL, where

-20 + 10i1 + 90 + 40i1 + 1.5v3 = 0

[1]

v3 = -90 – 10i1 + 20 = -70 – 10 i1 alternatively, we could write v3 = 40i1 + 1.5v3 = -80i1 Using either expression in Eq. [1], we find

i1 = 1 A.

(c) Applying KVL, Solving, i1 = - 2A.

-20 + 10i1 + 90 + 40i1 - 15 i1 = 0



27.


10 March 2006

Applying KVL, we find that -20 + 10i1 + 90 + 40i1 + 1.8v3 = 0

[1]

Also, KVL allows us to write v3 = 40i1 + 1.8v3 So that we may write Eq. [1] as

v3 = -50i1

50i1 – 1.8(50)i1 = -70 or i1 = -70/-40 = 1.75 A. Since v3 = -50i1 = -87.5 V, no further information is required to determine its value. The 90-V source is absorbing (90)(i1) = 157.5 W of power and the dependent source is absorbing (1.8v3)(i1) = -275.6 W of power. Therefore, none of the conditions specified in (a) to (d) can be met by this circuit.



28.


10 March 2006

(a) Define the charging current i as flowing clockwise in the circuit provided. By application of KVL, -13 + 0.02i + Ri + 0.035i + 10.5 = 0 We know that we need a current i = 4 A, so we may calculate the necessary resistance R = [13 – 10.5 – 0.055(4)]/ 4 = 570 mΩ (b) The total power delivered to the battery consists of the power absorbed by the 0.035-Ω resistance (0.035i2), and the power absorbed by the 10.5-V ideal battery (10.5i). Thus, we need to solve the quadratic equation 0.035i2 + 10.5i = 25 which has the solutions i = -302.4 A and i = 2.362 A. In order to determine which of these two values should be used, we must recall that the idea is to charge the battery, implying that it is absorbing power, or that i as defined is positive. Thus, we choose i = 2.362 A, and, making use of the expression developed in part (a), we find that R = [13 – 10.5 – 0.055(2.362)]/ 2.362 = 1.003 Ω (c) To obtain a voltage of 11 V across the battery, we apply KVL: 0.035i + 10.5 = 11 so that i = 14.29 A From part (a), this means we need R = [13 – 10.5 – 0.055(14.29)]/ 14.29 = 119.9 mΩ



29.


10 March 2006

Drawing the circuit described, we also define a clockwise current i.

By KVL, we find that -13 + (0.02 + 0.5 – 0.05)i + 0.035i + 10.5 = 0 or that i = (13 – 10.5)/0.505 = 4.951 A and Vbattery = 13 – (0.02 + 0.5)i = 10.43 V.



30.


10 March 2006

Applying KVL about this simple loop circuit (the dependent sources are still linear elements, by the way, as they depend only upon a sum of voltages) -40 + (5 + 25 + 20)i – (2v3 + v2) + (4v1 – v2) = 0

[1]

where we have defined i to be flowing in the clockwise direction, and v1 = 5i, v2 = 25i, and v3 = 20i. Performing the necessary substition, Eq. [1] becomes 50i - (40i + 25i) + (20i – 25i) = 40 so that i = 40/-20 = -2 A Computing the absorbed power is now a straightforward matter: p40V p5Ω p25Ω p20Ω pdepsrc1 pdepsrc2

= (40)(-i) = 5i2 = 25i2 = 20i2 = (2v3 + v2)(-i) = (40i + 25i) = (4v1 - v2)(-i) = (20i - 25i)

= 80 W = 20 W = 100 W = 80 W = -260 W = -20 W

and we can easily verify that these quantities indeed sum to zero as expected.



31.


10 March 2006

We begin by defining a clockwise current i. (a) i = 12/(40 + R) mA, with R expressed in kΩ. We want i2 · 25 = 2 2

or

⎛ 12 ⎞ ⎜ ⎟ ⋅ 25 = 2 ⎝ 40 + R ⎠

Rearranging, we find a quadratic expression involving R: R2 + 80R – 200 = 0 which has the solutions R = -82.43 kΩ and R = 2.426 kΩ. Only the latter is a physical solution, so R = 2.426 kΩ. (b) We require i · 12 = 3.6 or i = 0.3 mA From the circuit, we also see that i = 12/(15 + R + 25) mA. Substituting the desired value for i, we find that the required value of R is

R = 0.

(c)



32.


10 March 2006

By KVL, –12 + (1 + 2.3 + Rwire segment) i = 0 The wire segment is a 3000–ft section of 28–AWG solid copper wire. Using Table 2.3, we compute its resistance as (16.2 mΩ/ft)(3000 ft) = 48.6 Ω which is certainly not negligible compared to the other resistances in the circuit! Thus, i = 12/(1 + 2.3 + 48.6) = 231.2 mA



33.


10 March 2006

We can apply Ohm’s law to find an expression for vo: vo = 1000(–gm vπ) We do not have a value for vπ, but KVL will allow us to express that in terms of vo, which we do know: –10×10–3 cos 5t + (300 + 50×103) i = 0 where i is defined as flowing clockwise. Thus,

vπ = 50×103 i

= 50×103 (10×10–3 cos 5t) / (300 + 50×103)

= 9.940×10–3 cos 5t V and we by substitution we find that vo

= 1000(–25×10–3)( 9.940×10–3 cos 5t ) = –248.5 cos 5t mV



34.


10 March 2006

By KVL, we find that –3 + 100 ID + VD = 0 –3

Substituting ID = 3×10–6(eVD / 27×10 – 1), we find that –3

–3 + 300×10–6(eVD / 27×10 – 1) + VD = 0 This is a transcendental equation. Using a scientific calculator or a numerical software package such as MATLAB®, we find VD = 246.4 mV Let’s assume such software-based assistance is unavailable. In that case, we need to “guess” a value for VD, substitute it into the right hand side of our equation, and see how close the result is to the left hand side (in this case, zero). GUESS 0 1 0.5 0.25 0.245 0.248 0.246

RESULT –3 3.648×1012 3.308×104 0.4001 –0.1375 0.1732 –0.0377

oops better At this point, the error is getting much smaller, and our confidence is increasing as to the value of VD.



35.


10 March 2006

Define a voltage vx, “+” reference on the right, across the dependent current source. Note that in fact vx appears across each of the four elements. We first convert the 10 mS conductance into a 100–Ω resistor, and the 40–mS conductance into a 25–Ω resistor. (a) Applying KCL, we sum the currents flowing into the right–hand node: 5 – vx / 100 – vx / 25 + 0.8 ix = 0

[1]

This represents one equation in two unknowns. A second equation to introduce at this point is ix = vx /25 so that Eq. [1] becomes 5 – vx / 100 – vx / 25 + 0.8 (vx / 25) = 0 Solving for vx, we find vx = 277.8 V. It is a simple matter now to compute the power absorbed by each element: P5A

= –5 vx

= –1.389 kW

P100Ω

= (vx)2 / 100

=

771.7 W

P25Ω

2

=

3.087 kW

Pdep

= (vx) / 25 2

= –vx(0.8 ix) = –0.8 (vx) / 25

= –2.470 kW

A quick check assures us that the calculated values sum to zero, as they should. (b) Again summing the currents into the right–hand node, 5 – vx / 100 – vx / 25 + 0.8 iy = 0

[2]

where iy = 5 – vx/100 Thus, Eq. [2] becomes 5 – vx / 100 – vx / 25 + 0.8(5) – 0.8 (iy) / 100 =

0

Solving, we find that vx x = 155.2 V and iy = 3.448 A So that P5A

= –5 vx

= –776.0 W

P100Ω

= (vx)2 / 100

=

240.9 W

P25Ω

2

= (vx) / 25

=

963.5 W

Pdep

= –vx(0.8 iy)

= –428.1 W

A quick check shows us that the calculated values sum to 0.3, which is reasonably close to zero compared to the size of the terms (small roundoff errors accumulated). PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


36.


10 March 2006

Define a voltage v with the “+” reference at the top node. Applying KCL and summing the currents flowing out of the top node, v/5,000 + 4×10–3 + 3i1 + v/20,000 = 0 [1] This, unfortunately, is one equation in two unknowns, necessitating the search for a second suitable equation. Returning to the circuit diagram, we observe that =

i1 or

i1

=

3 i1 + v/2,000 –v/40,000

[2]

Upon substituting Eq. [2] into Eq. [1], Eq. [1] becomes, v/5,000 + 4×10–3 – 3v/40,000 + v/20,000 = 0 Solving, we find that v = –22.86 V and i1 = 571.4 μA Since ix = i1, we find that ix = 571.4 μA.



37.


10 March 2006

Define a voltage vx with its “+” reference at the center node. Applying KCL and summing the currents into the center node, 8 – vx /6 + 7 – vx /12 – vx /4 = 0 Solving, vx = 30 V. It is now a straightforward matter to compute the power absorbed by each element: P8A

= –8 vx 2

= –240 W

P6Ω

= (vx) / 6

=

P8A

= –7 vx

= –210 W

P12Ω P4Ω

150 W

2

=

75 W

2

=

225 W

= (vx) / 12 = (vx) / 4

and a quick check verifies that the computed quantities sum to zero, as expected.



38.


10 March 2006

(a) Define a voltage v across the 1–kΩ resistor with the “+” reference at the top node. Applying KCL at this top node, we find that 80×10–3 – 30×10–3 = v/1000 + v/4000 Solving,

v = (50×10–3)(4×106 / 5×103) = 40 V P4kΩ = v2/4000 = 400 mW

and

(b) Once again, we first define a voltage v across the 1–kΩ resistor with the “+” reference at the top node. Applying KCL at this top node, we find that 80×10–3 – 30×10–3 – 20×10–3 = v/1000 Solving, v = 30 V P20mA = v · 20×10–3 = 600 mW

and

(c) Once again, we first define a voltage v across the 1–kΩ resistor with the “+” reference at the top node. Applying KCL at this top node, we find that 80×10–3 – 30×10–3 – 2ix = v/1000

where so that

ix = v/1000 80×10–3 – 30×10–3 = 2v/1000 + v/1000

and Thus,

v = 50×10–3 (1000)/3 = 16.67 V Pdep = v · 2ix

= 555.8 mW

(d) We note that ix = 60/1000 = 60 mA. KCL stipulates that (viewing currents into and out of the top node) Thus, is = 10 mA and

80 – 30 + is = ix = 60

P60V = 60(–10) mW = –600 mW



39.


10 March 2006

(a) To cancel out the effects of both the 80-mA and 30-mA sources, iS must be set to iS = –50 mA. (b) Define a current is flowing out of the “+” reference terminal of the independent voltage source. Interpret “no power” to mean “zero power.” Summing the currents flowing into the top node and invoking KCL, we find that 80×10-3 - 30×10-3 - vS/1×103 + iS = 0 Simplifying slightly, this becomes 50 - vS + 103 iS = 0

[1]

We are seeking a value for vS such that vS · iS = 0. Clearly, setting vS = 0 will achieve this. From Eq. [1], we also see that setting vS = 50 V will work as well.



40.


10 March 2006

Define a voltage v9 across the 9-Ω resistor, with the “+” reference at the top node. (a) Summing the currents into the right-hand node and applying KCL, 5 + 7 = v9 / 3 + v9 / 9 Solving, we find that v9 = 27 V. Since ix = v9 / 9,

ix = 3 A.

(b) Again, we apply KCL, this time to the top left node: 2 – v8 / 8 + 2ix – 5 = 0 Since we know from part (a) that ix = 3 A, we may calculate v8 = 24 V. (c) p5A = (v9 – v8) · 5 = 15 W.



41.


10 March 2006

Define a voltage vx across the 5-A source, with the “+” reference on top. Applying KCL at the top node then yields 5 + 5v1 - vx/ (1 + 2) – vx/ 5 = 0

[1]

where v1 = 2[vx /(1 + 2)] = 2 vx / 3. Thus, Eq. [1] becomes 5 + 5(2 vx / 3) – vx / 3 – vx / 5 = 0 or 75 + 50 vx – 5 vx – 3 vx = 0, which, upon solving, yields vx = -1.786 V. The power absorbed by the 5-Ω resistor is then simply (vx)2/5 = 638.0 mW.



42.


10 March 2006

Despite the way it may appear at first glance, this is actually a simple node-pair circuit. Define a voltage v across the elements, with the “+” reference at the top node. Summing the currents leaving the top node and applying KCL, we find that 2 + 6 + 3 + v/5 + v/5 + v/5 = 0 or v = -55/3 = -18.33 V. The power supplied by each source is then computed as: p2A = -v(2) = 36.67 W p6A = -v(6) = 110 W p3A = -v(3) = 55 W The power absorbed by each resistor is simply v2/5 = 67.22 W for a total of 201.67 W, which is the total power supplied by all sources. If instead we want the “power supplied” by the resistors, we multiply by -1 to obtain -201.67 W. Thus, the sum of the supplied power of each circuit element is zero, as it should be.



43.


10 March 2006

Defining a voltage Vx across the 10-A source with the “+” reference at the top node, KCL tells us that 10 = 5 + I1Ω, where I1Ω is defined flowing downward through the 1-Ω resistor. Solving, we find that I1Ω = 5 A, so that Vx = (1)(5) = 5 V. So, we need to solve Vx = 5 = 5(0.5 + Rsegment) with Rsegment = 500 mΩ. From Table 2.3, we see that 28-AWG solid copper wire has a resistance of 65.3 mΩ/ft. Thus, the total number of miles needed of the wire is 500 mΩ (65.3 mΩ/ft)(5280 ft/mi)

= 1.450 × 10-3 miles



44.


10 March 2006

Since v = 6 V, we know the current through the 1-Ω resistor is 6 A, the current through the 2-Ω resistor is 3 A, and the current through the 5-Ω resistor is 6/5 = 1.2 A, as shown below:

By KCL, 6 + 3 + 1.2 + iS = 0

or

iS = -10.2 A.



45.


(a) Applying KCL, 1 – i – 3 + 3 = 0

10 March 2006

so i = 1 A.

(b) Looking at the left part of the circuit, we see 1 + 3 = 4 A flowing into the unknown current source, which, by virtue of KCL, must therefore be a 4-A current source. Thus, KCL at the node labeled with the “+” reference of the voltage v gives 4–2+7–i = 0

or

i = 9A



46.


10 March 2006

(a) We may redraw the circuit as

Then, we see that v = (1)(1) = 1 V. (b) We may combine all sources to the right of the 1-Ω resistor into a single 7-A current source. On the left, the two 1-A sources in series reduce to a single 1-A source. The new 1-A source and the 3-A source combine to yield a 4-A source in series with the unknown current source which, by KCL, must be a 4-A current source. At this point we have reduced the circuit to

Further simplification is possible, resulting in

From which we see clearly that v = (9)(1) = 9 V.



47.


10 March 2006

(a) Combine the 12-V and 2-V series connected sources to obtain a new 12 – 2 = 10 V source, with the “+” reference terminal at the top. The result is two 10-V sources in parallel, which is permitted by KVL. Therefore, i = 10/1000 = 10 mA. (b) No current flows through the 6-V source, so we may neglect it for this calculation. The 12-V, 10-V and 3-V sources are connected in series as a result, so we replace them with a 12 + 10 –3 = 19 V source as shown

Thus,

i = 19/5 = 3.8 A.



48.


10 March 2006

We first combine the 10-V and 5-V sources into a single 15-V source, with the “+” reference on top. The 2-A and 7-A current sources combine into a 7 – 2 = 5 A current source (arrow pointing down); although these two current sources may not appear to be in parallel at first glance, they actually are. Redrawing our circuit,

we see that v = 15 V (note that we can completely the ignore the 5-A source here, since we have a voltage source directly across the resistor). Thus, P16Ω = v2/16 =

14.06 W.

Returning to the original circuit, we see that the 2 A source is in parallel with both 16 Ω resistors, so that it has a voltage of 15 V across it as well (the same goes for the 7 A source). Thus,

P2 A abs = -15(2) = -30 W P7 A abs = -15(-7) = 105 W Each resistor draws 15/16 A, so the 5 V and 10 V sources each see a current of 30/16 + 5 = 6.875 A flowing through them. Thus, P5V abs = -5(6.875) = -34.38 W

P10V

abs

= -10(6.875) = -68.75 W

which sum to -0.01 W, close enough to zero compared to the size of the terms (roundoff error accumulated).



49.


10 March 2006

We can combine the voltage sources such that i = vS/ 14

(a) vS = 10 + 10 – 6 – 6 = 20 –12 = 8 Therefore i = 8/14 = 571.4 mA. (b) vS = 3 + 2.5 – 3 – 2.5 = 0

Therefore

i = 0.

(c) vS = -3 + 1.5 – (-0.5) – 0 = -1 V Therefore i = -1/14 = -71.43 mA.



50.


10 March 2006

We first simplify as shown, making use of the fact that we are told ix = 2 A to find the voltage across the middle and right-most 1-Ω resistors as labeled.

By KVL, then, we find that

v1 = 2 + 3 = 5 V.



51.


10 March 2006

We see that to determine the voltage v we will need vx due to the presence of the dependent current soruce. So, let’s begin with the right-hand side, where we find that vx = 1000(1 – 3) × 10-3 = -2 V. Returning to the left-hand side of the circuit, and summing currents into the top node, we find that (12 – 3.5) ×10-3 + 0.03 vx = v/10×103 or

v = -515 V.



52.


10 March 2006

(a) We first label the circuit with a focus on determining the current flowing through each voltage source:

Then the power absorbed by each voltage source is P2V P4V P-3V

= -2(-5) = -(-4)(4) = -(-9)(-3)

= 10 W = 16 W = -27 W

For the current sources,

So that the absorbed power is P-5A P-4A P3A P12A

= +(-5)(6) = -(-4)(4) = -(3)(7) = -(12)(-3)

= -30 W = 16 W = -21 W = 36 W

A quick check assures us that these absorbed powers sum to zero as they should. (b) We need to change the 4-V source such that the voltage across the –5-A source drops to zero. Define Vx across the –5-A source such that the “+” reference terminal is on the left. Then, -2 + Vx – Vneeded = 0 or Vneeded = -2 V.



53.


10 March 2006

We begin by noting several things: (1) The bottom resistor has been shorted out; (2) the far-right resistor is only connected by one terminal and therefore does not affect the equivalent resistance as seen from the indicated terminals; (3) All resistors to the right of the top left resistor have been shorted. Thus, from the indicated terminals, we only see the single 1-kΩ resistor, so that Req = 1 kΩ.



54.


10 March 2006

(a) We see 1Ω || (1 Ω + 1 Ω) || (1 Ω + 1 Ω + 1 Ω) = 1Ω || 2 Ω || 3 Ω = 545.5 mΩ (b) 1/Req = 1 + 1/2 + 1/3 + … + 1/N Thus, Req = [1 + 1/2 + 1/3 + … + 1/N]-1



55.


10 March 2006

(a) 5 kΩ = 10 kΩ || 10 kΩ (b) 57 333 Ω = 47 kΩ + 10 kΩ + 1 kΩ || 1kΩ || 1kΩ (c) 29.5 kΩ = 47 kΩ || 47 kΩ + 10 kΩ || 10 kΩ + 1 kΩ



56.


10 March 2006

(a) no simplification is possible using only source and/or resistor combination techniques. (b) We first simplify the circuit to 5Ω 0V

5Ω

1A

7Ω

and then notice that the 0-V source is shorting out one of the 5-Ω resistors, so a further simplification is possible, noting that 5 Ω || 7 Ω = 2.917 Ω:

1A

2.917 Ω



57.

Req


10 March 2006

= 1 kΩ + 2 kΩ || 2 kΩ + 3 kΩ || 3 kΩ + 4 kΩ || 4 kΩ = 1 kΩ + 1 k Ω + 1.5 kΩ + 2 kΩ = 5.5 kΩ.



58.


10 March 2006

(a) Working from right to left, we first see that we may combine several resistors as 100 Ω + 100 Ω || 100 Ω + 100 Ω = 250 Ω, yielding the following circuit:

100 Ω

Req →

100 Ω

100 Ω 100 Ω

100 Ω

100 Ω

250 Ω

100 Ω

Next, we see 100 Ω + 100 Ω || 250 Ω + 100 Ω = 271.4 Ω, and subsequently 100 Ω + 100 Ω || 271.4 Ω + 100 Ω = 273.1 Ω, and, finally, Req = 100 Ω || 273.1 Ω = 73.20 Ω. (b) First, we combine 24 Ω || (50 Ω + 40 Ω) || 60 Ω = 14.4 Ω, which leaves us with 5Ω

10 Ω

Req →

14.4 Ω 20 Ω

Thus, Req = 10 Ω + 20 Ω || (5 + 14.4 Ω) =

19.85 Ω.

(c) First combine the 10-Ω and 40-Ω resistors and redraw the circuit: 15 Ω

10 Ω

2Ω 50 Ω 8Ω

20 Ω

30 Ω

We now see we have (10 Ω + 15 Ω) || 50 Ω = 16.67 Ω. Redrawing once again, 16.67 Ω

2Ω

8Ω

50 Ω

where the equivalent resistance is seen to be 2 Ω + 50 Ω || 16.67 Ω + 8 Ω = 22.5 Ω.



59.


10 March 2006

(a) Req = [(40 Ω + 20 Ω) || 30 Ω + 80 Ω] || 100 Ω + 10 Ω

=

60 Ω.

(b) Req = 80 Ω = [(40 Ω + 20 Ω) || 30 Ω + R] || 100 Ω + 10 Ω 70 Ω = [(60 Ω || 30 Ω) + R] || 100 Ω 1/70 = 1/(20 + R) + 0.01 20+ R = 233.3 Ω therefore R = 213.3 Ω. (c) R = [(40 Ω + 20 Ω) || 30 Ω + R] || 100 Ω + 10 Ω R – 10 Ω = [20 + R] || 100 1/(R – 10) = 1/(R + 20) + 1/ 100 3000 = R2 + 10R – 200 Solving, we find R = -61.79 Ω or R = 51.79 Ω. Clearly, the first is not a physical solution, so

R = 51.79 Ω.



60.


10 March 2006

(a) 25 Ω = 100 Ω || 100 Ω || 100 Ω || 100 Ω (b) 60 Ω = [(100 Ω || 100 Ω) + 100 Ω] || 100 Ω (c) 40 Ω = (100 Ω + 100 Ω) || 100 Ω || 100 Ω



61.


10 March 2006

Req = [(5 Ω || 20 Ω) + 6 Ω] || 30 Ω + 2.5 Ω = 10 Ω The source therefore provides a total of 1000 W and a current of 100/10 = 10 A. P2.5Ω = (10)2 · 2.5 = 250 W V30Ω = 100 - 2.5(10) = 75 V P30Ω = 752/ 30 = 187.5 W I6Ω = 10 – V30Ω /30 = 10 – 75/30 = 7.5 A P6Ω = (7.5)2 · 6 = 337.5 W V5Ω = 75 – 6(7.5) = 30 V P5Ω = 302/ 5 = 180 W V20Ω = V5Ω = 30 V P20Ω = 302/20 = 45 W We check our results by verifying that the absorbed powers in fact add to 1000 W.



62.


10 March 2006

To begin with, the 10-Ω and 15-Ω resistors are in parallel ( = 6 Ω), and so are the 20-Ω and 5-Ω resistors (= 4 Ω). Also, the 4-A, 1-A and 6-A current sources are in parallel, so they can be combined into a single 4 + 6 – 1 = 9 A current source as shown: 9A

-

vx

+

14 Ω

6Ω

4Ω

6Ω

Next, we note that (14 Ω + 6 Ω) || (4 Ω + 6 Ω) = 6.667 Ω so that vx = 9(6.667) = 60 V and ix = -60/10 = -6 A.



63.


10 March 2006

(a) Working from right to left, and borrowing x || y notation from resistance calculations to indicate the operation xy/(x + y), Gin

= {[(6 || 2 || 3) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS = {[(1) + 0.5] || 1.5 || 2.5 + 0.8} || 4 || 5 mS = {1.377} || 4 || 5 = 0.8502 mS = 850.2 mS

(b) The 50-mS and 40-mS conductances are in series, equivalent to (50(40)/90 = 22.22 mS. The 30-mS and 25-mS conductances are also in series, equivalent to 13.64 mS. Redrawing for clarity, 22.22 mS 13.64 mS

Gin →

100 mS

we see that Gin = 10 + 22.22 + 13.64 = 135.9 mS.



64.


10 March 2006

The bottom four resistors between the 2-Ω resistor and the 30-V source are shorted out. The 10-Ω and 40-Ω resistors are in parallel (= 8 Ω), as are the 15-Ω and 60-Ω (=12 Ω) resistors. These combinations are in series. Define a clockwise current I through the 1-Ω resistor: I = (150 – 30)/(2 + 8 + 12 + 3 + 1 + 2) = 4.286 A P1Ω = I2 · 1 = 18.37 W To compute P10Ω, consider that since the 10-Ω and 40-Ω resistors are in parallel, the same voltage Vx (“+” reference on the left) appears across both resistors. The current I = 4.286 A flows into this combination. Thus, Vx = (8)(4.286) = 34.29 V and P10Ω = (Vx)2 / 10 = 117.6 W. P13Ω = 0 since no current flows through that resistor.



65.


10 March 2006

With the meter being a short circuit and no current flowing through it, we can write R1i1 = R2i2 R3i3 = RiR

→

R1i1 R2i2 [1] = R3i3 RiR

And since i1 = i3, and i2 = iR, Eq [1] becomes

R R1 R2 , or R = R2 3 . Q.E.D. = R3 R R1



66.


10 March 2006

The total resistance in the series string sums to 75 Ω. The voltage dropped across the 2.2 Ω resistor is 2.2 = 293.3 mV 75 and the voltage dropped across the 47 Ω resistor is V2.2Ω = 10

V47Ω = 10

47 = 6.267 mV 75



67.


10 March 2006

We first note that the 4.7 kΩ and 2.2 kΩ resistors can be combined into a single 1.5 kΩ resistor, which is then in series with the 10 kΩ resistor. Next we note that the 33 kΩ / 47 kΩ parallel combination can be replaced by a 19.39 kΩ resistance, which is in series with the remaining 33 kΩ resistor. By voltage division, then, and noting that V47kΩ is the same voltage as that across the 19.39 kΩ resistance, 19.39 V47kΩ = 2 = 607.0 mV 10 + 1.5 + 33 + 19.39



68.


10 March 2006

(a) The current downward through the 33 Ω resistor is calculated more easily if we first note that 134 Ω || 134 Ω = 67 Ω, and 67 Ω + 33 Ω = 100 Ω. Then,

I 33Ω

1 100 = 12 = 571.4 mA 1 +1 +1 10 10 100

(b) The resistor flowing downward through either 134 Ω resistor is simply 571.4/2 = 285.7 mA

(by current division).



69.


10 March 2006

We first note that 20 Ω || 60 Ω = 15 Ω, and 50 Ω || 30 Ω = 18.75 Ω. Then, 15 Ω + 22 Ω + 18.75 Ω = 55.75 Ω. Finally, we are left with two current sources, the series combination of 10 Ω + 15 Ω, and 10 Ω || 55.75 Ω = 8.479 Ω. Using current division on the simplified circuit,

I15Ω

1 10 + 15 = ( 30 − 8 ) = 22.12 A 1 1 + 10 + 15 8.479



70.


10 March 2006

One possible solution of many:

vS = 2(5.5) = 11 V R1 = R2 = 1 kΩ.



71.


10 March 2006


iS = 11 mA R1 = R2 = 1 kΩ.



72.


10 March 2006

p15Ω = (v15)2 / 15×103 A v15 = 15×103 (-0.3 v1) where v1 = [4 (5)/ (5 + 2)] · 2 = 5.714 V Therefore v15 = -25714 V and p15 = 44.08 kW.



73.


10 March 2006

Replace the top 10 kΩ, 4 kΩ and 47 kΩ resistors with 10 kΩ + 4 kΩ || 47 kΩ = 13.69 kΩ. Define vx across the 10 kΩ resistor with its “+” reference at the top node: then

vx = 5 · (10 kΩ || 13.69 kΩ) / (15 kΩ + 10 || 13.69 kΩ) = 1.391 V ix = vx/ 10 mA = 139.1 μA v15 = 5 – 1.391 = 3.609 V and p15 = (v15)2/ 15×103 = 868.3 μW.



74.


10 March 2006

We may combine the 12-A and 5-A current sources into a single 7-A current source with its arrow oriented upwards. The left three resistors may be replaced by a 3 + 6 || 13 = 7.105 Ω resistor, and the right three resistors may be replaced by a 7 + 20 || 4 = 10.33 Ω resistor. By current division, iy = 7 (7.105)/(7.105 + 10.33) = 2.853 A We must now return to the original circuit. The current into the 6 Ω, 13 Ω parallel combination is 7 – iy = 4.147 A. By current division,

ix = 4.147 . 13/ (13 + 6) = 2.837 A and px = (4.147)2 · 3 = 51.59 W



75.


10 March 2006

The controlling voltage v1, needed to obtain the power into the 47-kΩ resistor, can be found separately as that network does not depend on the left-hand network. The right-most 2 kΩ resistor can be neglected. By current division, then, in combination with Ohm’s law,

v1 = 3000[5×10-3 (2000)/ (2000 + 3000 + 7000)] = 2.5 V Voltage division gives the voltage across the 47-kΩ resistor: 0.5v1

47 47 + 100 || 20

=

0.5(2.5)(47) 47 + 16.67

= 0.9228 V

So that p47kΩ = (0.9928)2 / 47×103 = 18.12 μW



76.


10 March 2006

The temptation to write an equation such as

v1 = 10

20 20 + 20

must be fought! Voltage division only applies to resistors connected in series, meaning that the same current must flow through each resistor. In this circuit, i1 ≠ 0 , so we do not have the same current flowing through both 20 kΩ resistors.



77.


10 March 2006

R 2 || (R 3 + R 4 ) R 1 + [R 2 || (R 3 + R 4 )] R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 ) = VS R 1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 ) R 2 (R 3 + R 4 ) = VS R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )

(a) v2 = VS

R1 R 1 + [R 2 || (R 3 + R 4 )] R1 = VS R 1 + R 2 (R 3 + R 4 ) (R 2 + R 3 + R 4 ) R 1 (R 2 + R 3 + R 4 ) = VS R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )

(b) v1 = VS

⎞ ⎛ v ⎞⎛ R2 ⎟⎟ (c) i4 = ⎜⎜ 1 ⎟⎟ ⎜⎜ + + R R R R 3 4 ⎠ ⎝ 1 ⎠⎝ 2 R 1 (R 2 + R 3 + R 4 ) R 2 = VS R 1 [R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )(R 2 + R 3 + R 4 )] R2 = VS R 1 (R 2 + R 3 + R 4 ) + R 2 (R 3 + R 4 )



78.


10 March 2006

(a) With the current source open-circuited, we find that

v1 = − 40

500 = -8V 500 + 3000 || 6000

(b) With the voltage source short-circuited, we find that

i2 = ( 3 × 10−3 ) i3 =

1/3000 1/ 500 + 1/3000 + 1/6000

( 3 ×10 ) −3

500 500 + 3000 || 6000

=

400 μ A

= 600 μ A



79.


10 March 2006

(a) The current through the 5-Ω resistor is 10/5 = 2 A. Define R as 3 || (4 + 5) = 2.25 Ω. The current through the 2-Ω resistor then is given by IS

1 1 + (2 + R)

=

IS 5.25

The current through the 5-Ω resistor is IS ⎛ 3 ⎞ ⎜ ⎟ = 2A 5.25 ⎝ 3 + 9 ⎠

so that

IS = 42 A.

(b) Given that IS is now 50 A, the current through the 5-Ω resistor becomes IS ⎛ 3 ⎞ ⎜ ⎟ = 2.381 A 5.25 ⎝ 3 + 9 ⎠

Thus, vx = 5(2.381) = 11.90 V

(c)

vx IS

⎡ 5IS ⎛ 3 ⎞⎤ ⎢ 5.25 ⎜ 3 + 9 ⎟⎥ ⎝ ⎠⎦ = ⎣ = 0.2381 IS



80.


10 March 2006

First combine the 1 kΩ and 3 kΩ resistors to obtain 750 Ω. By current division, the current through resistor Rx is IRx = 10 × 10 − 3

2000 2000 + R x + 750

and we know that Rx · IRx = 9 so

9 =

20 R x 2750 + R x

9 Rx + 24750 = 20 Rx

or Rx = 2250 W. Thus,

PRx = 92/ Rx = 36 mW.



81.


10 March 2006

Define R = R3 || (R4 + R5) Then vR

⎛ R ⎞ ⎟⎟ = VS ⎜⎜ ⎝ R + R2 ⎠ ⎛ R 3 (R 4 + R 5 ) (R 3 + R 4 + R 5 ) ⎞ ⎟⎟ = VS ⎜⎜ ( ) R (R R ) R R R R + + + + 5 3 4 5 2 ⎠ ⎝ 3 4 ⎛ ⎞ R 3 (R 4 + R 5 ) ⎟⎟ = VS ⎜⎜ R (R + R + R ) + R (R + R ) 4 5 3 4 5 ⎠ ⎝ 2 3

Thus,

v5

⎛ R5 ⎞ ⎟⎟ = vR ⎜⎜ R R + 5 ⎠ ⎝ 4 ⎛ ⎞ R3 R5 ⎟⎟ = VS ⎜⎜ ⎝ R 2 (R 3 + R 4 + R 5 ) + R 3 (R 4 + R 5 ) ⎠



82.


10 March 2006

Define R1 = 10 + 15 || 30 = 20 Ω and R2 = 5 + 25 = 30 Ω. (a) Ix = I1 . 15 / (15 + 30) = 4 mA (b) I1 = Ix . 45/15 = 36 mA (c) I2 = IS R1 / (R1 + R2) and I1 = IS R2 / (R1 + R2) So I1/I2 = R2/R1 Therefore I1 = R2I2/R1 = 30(15)/20 = 22.5 mA Thus, Ix = I1 . 15/ 45 = 7.5 mA (d) I1 = IS R2/ (R1 + R2) = 60 (30)/ 50 = 36 mA Thus, Ix = I1 15/ 45 = 12 mA.



83.


10 March 2006

vout = -gm vπ (100 kΩ || 100 kΩ) = -4.762×103 gm vπ where vπ = (3 sin 10t) · 15/(15 + 0.3) = 2.941 sin 10t Thus, vout = -56.02 sin 10t V



84.


10 March 2006

vout = -1000gm vπ where vπ = 3 sin 10t therefore

15 || 3 (15 || 3) + 0.3

= 2.679 sin 10t V

vout = -(2.679)(1000)(38×10-3) sin 10t = -101.8 sin 10t V.



1. ( -0.2

a)

0.1 -0.5

-0.3 0.1 -0.3

-0.4 0 0.4

v1 = v2 v3

Chapter Four Solutions

10 March 2006

0 4 6

Solving this matrix equation using a scientific calculator, v2 = -8.387 V (b) Using a scientific calculator, the determinant is equal to 32.



2. ( -1

a)

1 2

1 2 0

1 3 4


vA = vB vC

10 March 2006

27 -16 -6

Solving this matrix equation using a scientific calculator, vA = 19.57 vB = 18.71 vC = -11.29 (b) Using a scientific calculator, -1

1 2

11 2 0

3= 4

16




10 March 2006

3. (a)

We begin by simplifying the equations prior to solution: 4 = 0.08v1 − 0.05v2 − 0.02v3 8 = −0.02v1 − 0.025v2 + 0.045v3 −2 = −0.05v1 + 0.115v2 − 0.025v3 Then, we can solve the matrix equation:

⎡ 0.08 -0.05 -0.02 ⎤ ⎡ v1 ⎤ ⎡ 4 ⎤ ⎢-0.02 -0.025 0.045 ⎥ ⎢ v ⎥ = ⎢ 8 ⎥ ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ -0.05 0.115 -0.025⎦⎥ ⎣⎢ v3 ⎥⎦ ⎢⎣ −2⎥⎦ to (b)

obtain v1 = 264.3 V, v2 = 183.9 V and v3 = 397.4 V. We may solve the matrix equation d irectly using MATLAB, but a be tter check is to invoke the symbolic processor:

>> e1 = '4 = v1/100 + (v1 - v2)/20 + (v1 - vx)/50'; >> e2 = '10 - 4 - (-2) = (vx - v1)/50 + (vx - v2)/40'; >> e3 = '-2 = v2/25 + (v2 - vx)/40 + (v2 - v1)/20'; >> a = solve(e1,e2,e3,'v1','v2','vx'); >> a.v1 ans = 82200/311 >> a.v2 ans = 57200/311 >> a.vx ans = 123600/311




10 March 2006

4. We select the bottom node as our reference terminal and define two nodal voltages:

Ref. Next, we write the two required nodal equations: v1 v1 − v2 + 2 3

Node 1:

1=

Node 2:

−3 =

v2 v2 − v1 + 1 3

Which may be simplified to: and

5v1 – 2v2 = 6 -v1 + 4v2 = -9

Solving, we find that v1 = 333.3 mV.



5. node,


10 March 2006

We begin by selecting the bottom node as the reference term inal, and defining two nodal voltages V A and V B, as shown. (Note if we cho ose the upp er right v1 becomes a nodal voltage and falls directly out of the solution.) VA

VB

Ref. We note that after completing nodal analysis, we can find v1 as v1 = VA – VB. At node A: 4 =

VA VA − VB + 10 5

At node B: −(−6) = Simplifying, Solving,

[1]

VB VB − VA [2] + 8 5

3VA – 2VB = 40 [1] –8VA + 13VB = 240 [2]

VA = 43.48 V and VB = 45.22 V, so v1 = –1.740 V.



6.


10 March 2006

By inspection, no current flows through the 2 Ω resistor, so i1 = 0. We next designate the bottom node as the reference term inal, and define V A and VB as shown: VA

VB

Ref. VA VA − VB [1] + 3 1 V V V − VA At node B: −2 = B + B + B 6 6 1

At node A: 2 =

obtain

2]

Note this y ields V A and V B, not v1, due to our choice of reference node. So, we v1 by KVL: v1 = VA – VB. Simplifying Eqs. [1] and [2],

–3V

4VA – 3VB = 6 A + 4VB = –6

[1] [2]

Solving, VA = 0.8571 V and VB = -0.8571 V, so v1 = 1.714 V.



7.


10 March 2006

The bottom node has the largest number of branch connections, so we choose that as our reference node. This also m akes vP easier to find, as it will be a nodal voltag e. Working from left to right, we name our nodes 1, P, 2, and 3. NODE 1:

10 = v1/ 20 + (v1 – vP)/ 40

[1]

NODE P:

0 = (vP – v1)/ 40 + vP/ 100 + (vP – v2)/ 50

[2]

NODE 2:

-2.5 + 2 = (v2 – vP)/ 50 + (v2 – v3)/ 10

[3]

NODE 3:

5 – 2 = v3/ 200 + (v3 – v2)/ 10

[4]

Simplifying, -50 Solving,

60v1 - 20vP = 8000 v1 + 110 vP - 40v2 =0 - vP + 6v2 - 5v3 = -25 -200v2 + 210v3 = 6000

[1] [2] [3] [4]

vP = 171.6 V



8.


10 March 2006

The logical choice for a reference node is the bottom node, as then vx will automatically become a nodal voltage. NODE 1:

4 = v1/ 100 + (v1 – v2)/ 20 + (v1 – vx)/ 50

[1]

NODE x:

10 – 4 – (-2) = (vx – v1)/ 50 + (vx – v2)/ 40

[2]

NODE 2:

-2 = v2 / 25 + (v2 – vx)/ 40 + (v2 – v1)/ 20

[3]

Simplifying,

Solving,

4 = 0.0800v1 – 0.0500v2 – 0.0200vx [1] 8 = -0.0200v1 – 0.02500v2 + 0.04500vx [2] -2 = -0.0500v1 + 0.1150v2 – 0.02500vx [3] vx = 397.4 V.



9.


10 March 2006

Designate the node between the 3-Ω and 6-Ω resistors as node X, and the right-hand node of the 6-Ω resistor as node Y. The bottom node is chosen as the reference node. (a) Writing the two nodal equations, then NODE X: –10 = (vX – 240)/ 3 + (vX – vY)/ 6 [1] NODE Y: 0 = (vY – vX)/ 6 + vY/ 30 + (vY – 60)/ 12

[2]

Simplifying, -180 + 1440 = 9 vX – 3 vY [1] 10800 = - 360 vX + 612 vY [2] Solving, Thus,

vX = 181.5 V and

vY = 124.4 V

v1 = 240 – vX = 58.50 V and (b)

v2 = vY – 60 =

64.40 V

The power absorbed by the 6-Ω resistor is (vX – vY)2 / 6 = 543.4 W



10.


10 March 2006

Only one nodal equation is required: At the node where three resistors join, 0.02v1 = (vx – 5 i2) / 45 + (vx – 100) / 30 + (vx – 0.2 v3) / 50

[1]

This, however, is one equation in four unknow ns, the other three resulting from the presence of the dependent sources. Thus, we require three additional equations: i2 = (0.2 v3 - vx) / 50

[2]

v1 = 0.2 v3 - 100

[3]

v3 = 50i2 Simplifying,

[4]

v1 – 0.2v3 = -100 – v3 + 50 i2 = 0 –vx + 0.2v3 – 50 i2 = 0 0.07556vx – 0.02v1 – 0.004v3 – 0.111i2 = 33.33

[3] [4] [2] [1]

Solving, we find that v1 = -103..8 V and i2 = -377.4 mA.



11. If


10 March 2006

v1 = 0, the dependent source is a short circuit and we may redraw the circuit as:

. 20 Ω 40 Ω

At NODE 1: Since

+ v1 = 0 -

10 Ω

4 - 6 = v1/ 40 + (v1 – 96)/ 20 + (v1 – V2)/ 10

v1 = 0, this simplifies to -2 = -96 / 20 - V2/ 10 so that V2 = -28 V.



12.


10 March 2006

We choose the bottom node as ground to make calculation of i5 easier. The left-most node is named “1”, the top node is nam ed “2”, the central node is nam ed “3” and the node between the 4-Ω and 6-Ω resistors is named “4.” NODE 1: NODE 2: NODE 3: NODE 4:

- 3 = v1/2 + (v1 – v2)/ 1 [1] 2 = (v2 – v1)/ 1 + (v2 – v3)/ 3 + (v2 – v4)/ 4 3 = v3/ 5 + (v3 – v4)/ 7 + (v3 – v2)/ 3 0 = v4/ 6 + (v4 – v3)/ 7 + (v4 – v2)/ 4

[2] [3] [4]

Rearranging and grouping terms, -12

3v1 – 2v= 2 v1 + 19v2 – 4v3 – 3v4 –35v2 + 71v3 – 15v4= -42v2 – 24v3 + 94v4=

-6 = 24 315 0

[1] [2] [3] [4]

Solving, we find that v3 = 6.760 V and so

i5 = v3/ 5 = 1.352 A.



13.


10 March 2006

We can redraw this circuit and eliminate the 2.2-kΩ resistor as no current flows through it: 5 mA

v2 470 Ω 9V

10 kΩ

vx

+ v1 -

7 mA

0.2 v1

↓

At NODE 2: 7×10-3 – 5×10-3 = (v2 + 9)/ 470 + (v2 – vx)/ 10×10-3 At NODE x: 5×10

-3

– 0.2v1 = (vx – v2)/ 10×103

[1] [2]

The additional equation required by the presence of the dependent source and the fact that its controlling variable is not one of the nodal voltages: v1 = v2 – vx

[3]

Eliminating the variable v1 and grouping terms, we obtain: and Solving, we find

10,470 v2 – 470 vx = –89,518 1999 v2 – 1999 vx = 50 vx = –8.086 V.




10 March 2006

14. We need concern ourselves with the bottom part of this circuit only. W riting a single nodal equation, -4 + 2 = v/ 50 We find that

v = -100 V.



15.


10 March 2006

We choose the bottom node as the reference terminal. Then: v1 v1 − v2 + 2 1

Node

1:

−2 =

Node

2:

4=

v2 − v1 v2 − v3 v2 − v4 [2] + + 1 2 4

Node

3:

2=

v3 − v2 v3 v3 − v4 [3] + + 2 5 10

Node

4:

0=

v4 v4 − v3 v4 − v2 [4] + + 6 10 4

Node

5:

−1 =

Node

6:

1=

v6 v6 − v7 v6 − v8 [6] + + 5 2 10

Node

7:

2=

v7 − v5 v7 − v6 v7 − v8 [7] + + 1 2 4

Node

8:

0=

v8 v8 − v6 v8 − v7 [8] + + 6 10 4

[1]

v5 v5 − v7 + 2 1

[5]

Note that Eqs. [1-4] may be solved independently of Eqs. [5-8]. Sim

plifying, 3v1 −4v1

and

−2v2 +7v2 −5v2 −15v2

−2v3 +8v3 −6v3

8v6 −2v6 −6v6

−2v7 −5v7 +7v7 −15v7

3v5 −4v5

−v4 −v4 +31v4

= −4 = 16 = 20 =0

[1] [2] to yield [3] [4]

v1 = 3.370 V v2 = 7.055 V v3 = 7.518 V v4 = 4.869 V

−v8 −v8 +31v8

= −2 = 10 =8 =0

[5] [6] to yield [7] [8]

v5 = 1.685 V v6 = 3.759 V v7 = 3.527 V v8 = 2.434 V



16.


10 March 2006

We choose the center node for our common terminal, since it connects to the largest number of branches. W e na me the left node “A”, the top node “B”, the right node “C”, and the bottom node “D”. We next form a supernode between nodes A and B. At the supernode:

5 = (VA – VD)/ 10 + VA/ 20 + (VB – VC)/ 12.5

[1]

At node C:

VC = 150

[2]

At node D:

-10 = VD/ 25 + (VD – VA)/ 10

Our supernode-related equation is VB – VA = 100

[3] [4]

Simplifiying and grouping terms, 0.15 VA + 0.08 VB - 0.08 VC – 0.1 VD VC -25 VA + 35 V= D - VA + VB =

= 5 = 150 -2500 100 [4]

[1] [2] [3]

Solving, we find that VD = -63.06 V. Since v4 = - VD, v4 = 63.06 V.



17.


10 March 2006

Choosing the bottom node as the reference terminal and naming the left node “1”, the center node “2” and the right node “3”, we next form a supe rnode about nodes 1 and 2, encompassing the dependent voltage source. At the supernode, At node 2,

5 – 8 = (v1 – v2)/ 2 + v3/ 2.5 8 = v2 / 5 + (v2 – v1)/ 2

[1] [2]

Our supernode equation is v1 - v3 = 0.8 vA [3] Since vA = v2, we can rewrite [3] as v1 – v3 = 0.8v2 Simplifying and collecting terms, 0.5 v1 - 0.5 v2 + 0.4 v3 -0.5 v1 + 0.7 v2 v1 - 0.8 v2 - v3 (a) Solving for v2 = vA, we find that

= -3 = 8 = 0

[1] [2] [3]

vA = 25.91 V

(b) The power absorbed by the 2.5-Ω resistor is = 82.66 mW. (v3)2/ 2.5 = (-0.4546)2/ 2.5



18.


10 March 2006

Selecting the bottom node as the reference terminal, we name the left node “1”, the middle node “2” and the right node “3.” NODE 1:

5 = (v1 – v2)/ 20 + (v1 – v3)/ 50

[1]

NODE

2:

v2 = 0.4 v1

[2]

NODE

3:

0.01 v1 = (v3 – v2)/ 30 + (v3 – v1)/ 50

[3]

Simplifying and collecting terms, we obtain 0.07 v1 – 0.05 v2 – 0.02 v3 0.4 v1 –= v2 -0.03 v1 – 0.03333 v2 + 0.05333 v3=

=5 0 0

[1] [2] [3]

Since our choice of r eference ter minal m akes the contr olling var iable of both dependent sources a nodal voltage, we have no need for an additional equation as we might have expected. Solving, we find that v1 = 148.2 V, v2 = 59.26 V, and v3 = 120.4 V. The power supplied by the dependent current source is therefore (0.01 v1) • v3 = 177.4 W.



19.

At node x: At node y:


vx/ 4 + (vx – vy)/ 2 + (vx – 6)/ 1 (vy – kvx)/ 3 + (vy – vx)/ 2

=0 =2

10 March 2006

[1] [2]

Our additional constraint is that vy = 0, so we may simplify Eqs. [1] and [2]: 14 -2k

vx = 48 [1] vx - 3 vx = 12 [2] Since Eq. [1] yields vx = 48/14 = 3.429 V, we find that k = (12 + 3 vx)/ (-2 vx)

= -3.250



20.


10 March 2006

Choosing the bottom node joining the 4-Ω resistor, the 2-A current sourcee and the 4-V voltage source as our reference node, we next nam e the other node of the 4- Ω resistor node “1”, and the node joining the 2- Ω r esistor and t he 2- A c urrent s ource node “2.” Finally, we create a supernode with nodes “1” and “2.” At the supernode: Our remaining equations: and

–2 = v1/ 4 + (v2 – 4)/ 2 v1 – v2 = –3 – 0.5i1 i1 = (v2 – 4)/ 2

Equation [1] simplifies to Combining Eqs. [2] and [3,

v1 + 2 v2 4 v1 – 3 v= 2

= 0 –8

[1] [2] [3]

[1] [4]

Solving these last two equations, we find that v2 = 727.3 mV. Making use of Eq. [3], we therefore find that i1 = – 1.636 A.



21.


10 March 2006

We first number the nodes as 1, 2, 3, 4, and 5 moving left to right. We next select node 5 as the reference terminal. To simplify the analysis, we form a supernode from nodes 1, 2, and 3. At the supernode, -4 – 8 + 6 = v1/ 40 + (v1 – v3)/ 10 + (v3 – v1)/ 10 + v2/ 50 + (v3 – v4)/ 20

[1]

Note that since both ends of the 10- Ω resistor are connected to the supernode, the related terms cancel each other out, and so could have been ignored. At node 4:

v4 = 200

[2]

Supernode KVL equation:

v1 – v3 = 400 + 4v20

[3]

Where the controlling voltage

v20 = v3 – v4 = v3 – 200

[4]

Thus, Eq. [1] becomes -6 = v1/ 40 + v2/ 50 + (v3 – 200)/ 20 or, more simply, 4 = v1/ 40 + v2/ 50 + v3/ 20 [1’] and Eq. [3] becomes

v1 – 5 v3 = -400

[3’]

Eqs. [1’], [3’], and [5] are not sufficient, however, as we have four unknowns. At this point we need to seek an additi onal equation, possibly in terms of v2. Referring to the circuit, v1 - v2 = 400 [5] Rewriting as a matrix equation,

⎡1 ⎢ 40 ⎢1 ⎢ ⎢⎣1

1

50 0 -1

1

⎤ 20⎥ -5⎥ ⎥ 0 ⎥ ⎦

⎡v1 ⎤ ⎡ 4⎤ ⎢v ⎥ = ⎢- 400⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣v3 ⎥⎦ ⎢⎣ 400⎥⎦

Solving, we find that v1 = 145.5 V, v2 = -254.5 V, and v3 = 109.1 V. Since v20 = v3 – 200, we find that v20 = -90.9 V.



22.


10 March 2006

We begin by naming the top left node “1”, the top right node “2”, the bottom node of the 6-V source “3” and the top node of the 2- Ω resistor “4.” Th e reference node has already been selected, and designated using a ground symbol. By inspection, v2 = 5 V. Forming a supernode with nodes 1 & 3, we find At the supernode:

-2 = v3/ 1 + (v1 – 5)/ 10

[1]

At node 4:

2 = v4/ 2 + (v4 – 5)/ 4

[2]

Our supernode KVL equation:

v1 – v3 = 6

[3]

Rearranging, simplifying and collecting terms, and

v1 + 10 v3 = -20 + 5 = -15

[1]

v1 - v3 = 6

[2]

Eq. [3] may be directly solved to obtain

v4 = 4.333 V.

Solving Eqs. [1] and [2], we find that v1 = 4.091 V

and

v3 = -1.909 V.



23.


10 March 2006

We begin by selecting the bottom node as the reference, naming the nodes as shown below, and forming a supernode with nodes 5 & 6. 2 A

v3

v2

4Ω

v5

v4 2Ω

1V

4A

v6

3V

4V

v1

1Ω

3Ω

By inspection,

v4 = 4 V.

By KVL, v3 – v4 = 1 so v3 = -1 + v4 = -1 + 4 At the supernode,

2 = v6/ 1 + (v5 – 4)/ 2

At node 1,

4 = v1/ 3

At node 2,

-4 – 2 = (v2 – 3)/ 4

Solving, we find that

or

v3 = 3 V.

[1] therefore,

v1 = 12 V.

v2 = -21 V

Our supernode KVL equation is

v5 - v6 = 3 [2]

Solving Eqs. [1] and [2], we find that v5 = 4.667 V and v6 = 1.667 V. The power supplied by the 2-A source therefore is (v6 – v2)(2) = 45.33 W.



24.


10 March 2006

We begin by selecting the bottom node as the reference, naming each node as shown below, and forming two different supernodes as indicated.

v2

v3

v5 v4

v8

v6 v7

Voltages in volts. Currents in amperes. Resistances in ohms.

v1

By inspection, v7 = 4 V At node 2:

and v1 = (3)(4) = 12 V.

-4 – 2 = (v2 – v3)/ 4

At the 3-4 supernode: 0 = (v3 – v2)/ 4 + (v4 – v5)/ 6

or or

At node 5: 0 = (v5 – v4)/ 6 + (v5 – 4)/ 7 + (v5 – v6)/ 2

-6v2 + 6v3 + 4v4 – 4v5 = 0 or

[1] [2]

-14v4 + 68v5 – 42v6 = 48 [3]

At the 6-8 supernode: 2 = (v6 – v5)/ 2 + v8/ 1 3-4 supernode KVL equation: 6-8 supernode KVL equation:

v2 -v3 = -24

or

v3 - v4 = -1 v6 – v8 = 3

-v5 + v6 + 2v8 = 4

[4]

[5] [6]

Rewriting Eqs. [1] to [6] in matrix form, 0 0 0 ⎡1 - 1 ⎢- 6 6 4 -4 0 ⎢ ⎢0 0 - 14 68 - 42 ⎢ 0 0 -1 1 ⎢0 ⎢0 1 -1 0 0 ⎢ 0 0 1 ⎢⎣0 0

0⎤ 0⎥⎥ 0⎥ ⎥ 2⎥ 0⎥ ⎥ - 1 ⎥⎦

⎡v2 ⎤ ⎡- 24⎤ ⎢v ⎥ ⎢ 0⎥ ⎢ 3⎥ ⎥ ⎢ ⎢v4 ⎥ ⎢ 48⎥ ⎢ ⎥ = ⎢ ⎥ ⎢v5 ⎥ ⎢ 4⎥ ⎢v ⎥ ⎢ -1 ⎥ ⎢ 6⎥ ⎥ ⎢ ⎢⎣ 3 ⎥⎦ ⎢⎣v8 ⎥⎦

Solving, we find that v2 = -68.9 V, v3 = -44.9 V, v4 = -43.9 V, v5 = -7.9 V, v6 = 700 mV, v8 = -2.3 V. The power generated by the 2-A source is therefore (v8 – v6)(2) = 133.2 W.



25.


10 March 2006

With the reference terminal already specified, we name the bottom terminal of the 3-mA source node “1,” the left term inal of the bottom 2.2-kΩ resistor node “2,” the top term inal of the 3-mA source node “3,” the “+” reference term inal of the 9-V source node “4,” and the “-” terminal of the 9-V source node “5.” Since we know that 1 mA flows through the top 2.2-kΩ resistor, v5 = -2.2 V. Also, we see that v4 – v5 = 9, so that v4 = 9 – 2.2 = 6.8 V. Proceeding with nodal analysis, At node 1:

-3×10-3 = v1/ 10×103 + (v1 – v2)/ 2.2×103

[1]

At node 2:

0 = (v2 – v1)/ 2.2×103 + (v2 – v3)/ 4.7×103

[2]

At node 3:

1×103 + 3×103 = (v3 – v2)/ 4.7×103 + v3/3.3×103 [3]

Solving,

v1 = -8.614 V, v2 = -3.909 V and v3 = 6.143 V.

Note that we could also have made use of the supernode approach here.



26.


10 March 2006

Mesh 1: –4 + 400i1 + 300i1 – 300i2 – 1 = 0 or 700i1 – 300i2 = 5 Mesh 2: 1 + 500i2 – 300i1 +2 – 2 = 0 or –300i1 + 500i2 = –3.2

Solving,

i1 = 5.923 mA and

i2 = -2.846 mA.




10 March 2006

27. (a) Define a clockwise m esh current i1 in th e lef t-most m esh; a clockw ise m esh curren t i2 in the central m esh, and note that iy can be used as a m esh current for the rem aining mesh. Mesh 1: -10 + 7i1 – 2i2 = 0 Mesh 2: -2i1 + 5i2 = 0 Mesh y: -2i2 + 9iy = 0 Solve the resulting matrix equation: ⎡ 7 −2 0 ⎤ ⎡ i1 ⎤ ⎡10 ⎤ ⎢ −2 5 0 ⎥ ⎢ i ⎥ = ⎢ 0 ⎥ to find that i = 1.613 A, and i = 143.4 mA. 1 y ⎢ ⎥⎢ 2⎥ ⎢ ⎥ ⎢⎣ 0 −2 9 ⎥⎦ ⎢⎣i y ⎥⎦ ⎢⎣ 0 ⎥⎦

(b) The power supplied by the 10 V source is (10)(i1) = 10(1.613) = 16.13 W.



28.


10 March 2006

Define three mesh currents as shown:

(a) The current through the 2 Ω resistor is i1. Mesh 1: 5i1 – 3i2 = 0 Mesh 2: –212 +8i2 –3i1 = 0 Mesh 3: 8i3 – 5i2 + 122 = 0 Solving,

or 5i1 – 3i2 = 0 or -3i1 +8i2 = 212 or –5i2 + 8i3 = –122

i1 = 20.52 A, i2 = 34.19 A and i3 = 6.121 A.

(b) The current through the 5 Ω resistor is i3, or 6.121 A. *** Note: since the problem statement did not specify a directi on, only the current magnitude is relevant, and its sign is arbitrary.



29. (a)


10 March 2006

We begin by defining three clockwise m esh currents i1, i2 and i3 in the lef t-most, central, and right-most meshes, respectively. Then, Note that ix = i2 – i3. Mesh 1: i1 = 5 A (by inspection) Mesh 3: i3 = –2 A (by inspection)

Mesh –125 Thus,

2: –25i1 + 75i2 – 20i3 = 0, or, making use of the above, + 75i2 + 40= 0 so that i2 = 1.133 A. ix = i2 – i3 = 1.133 – (–2) = 3.133 A. (b) The power absorbed by the 25 Ω resistor is P25Ω = 25 (i1 – i2)2 = 25 (5 – 1.133)2 = 373.8 W.



30.


10 March 2006

Define three mesh currents as shown. Then,

Mesh 1: Mesh 2: Mesh 3:

Solving,

–2 + 80i1 – 40i2 – 30i3 = 0 =0 –40i1 + 70i2 +70i3 = 0 –30i1

⎡ 80 −40 −30⎤ ⎡ i1 ⎤ ⎡ 2 ⎤ ⎢ 0 ⎥⎥ ⎢⎢i2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢ −40 70 70 ⎦⎥ ⎢⎣ i3 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎣⎢ −30 0

we find that i2 = 25.81 mA and i3 = 19.35 mA. Thus, i = i3 – i2 = –6.46 mA.



31.


10 March 2006

Moving from left to right, we name the bottom three meshes, mesh “1”, mesh “2,” and m esh “3.” In each of these th ree meshes we define a clockwise curren t. The remaining mesh current is clearly 8 A. We may then write: MESH 1:

12 i1 - 4 i2

MESH 2:

-4 i1 + 9 i2 - 3 i3 = 0

MESH 3:

= 100

-3 i2 + 18 i3 = -80

Solving this system of three (independent) equations in three unknowns, we find that i2 =

ix = 2.791 A.



32.


10 March 2006

We define four clockwise mesh currents. The top mesh current is labeled i4. The bottom left mesh current is labeled i1, the bottom right m esh current is labeled i3, and the remaining mesh current is labeled i2. Define a voltage “v4A” across the 4-A current source with the “+” reference terminal on the left. By inspection,

i3 = 5 A and ia = i4.

MESH 1: -60 + 2i1 – 2i4 + 6i4 = 0

or 2i1

MESH 2: -6i4 + v4A + 4i2 – 4(5) = 0

or

MESH 4: 2i4 – 2i1 + 5i4 + 3i4 – 3(5) – v4A = 0 or -2i1

+ 4i4

= 60

[1]

4i2 - 6i4 + v4A = 20

[2]

+ 10i4 - v4A = 15

[3]

At this point, we are short an equation. Returning to the circuit diagram, we note that i2 – i4 = 4

[4]

Collecting these equations and writing in matrix form, we have

⎡ 2 ⎢ 0 ⎢ ⎢- 2 ⎢ ⎣ 0

0 4 0 1

4 -6 10 -1

0⎤ 1 ⎥⎥ -1 ⎥ ⎥ 0⎦

⎡i1 ⎤ ⎡60⎤ ⎢i ⎥ ⎢ ⎥ ⎢ 2 ⎥ = ⎢20⎥ ⎢i4 ⎥ ⎢15 ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 4⎦ ⎣v 4 A ⎦

Solving, i1 = 16.83 A, i2 = 10.58 A, i4 = 6.583 A and v4A = 17.17 V. Thus, the power dissipated by the 2-Ω resistor is (i1 – i4)2 • (2) = 210.0 W



33.


10 March 2006

We begin our analysis by defining three clockwise mesh currents. We will call the top mesh current i3, the bottom left mesh current i1, and the bottom right mesh current i2. By inspection, i1 = 5 A [1] MESH 3: or

and

i2 = -0.01 v1

[2]

50 i3 + 30 i3 – 30 i2 + 20 i3 – 20 i1 = 0 -20 i1 – 30 i2 + 100 i3 = 0 [3]

These three equations are insufficient, how ever, to solve for the unknow ns. It would be nice to be able to express the dependent source cont rolling variable v1 in term s of the mesh currents. Returning to the diagra m, it can be seen that KVL around m esh 1 will yield - v1 + 20 i1 – 20 i3 + 0.4 v1 = 0 or v1 = (20(5)/ 0.6 - 20 i3/ 0.6 [4] or v1 = 20 i1/ 0.6 – 20 i3/ 0.6 Substituting Eq. [4] into Eq. [2] and then the modified Eq. [2] into Eq. [3], we find -20(5) – 30(-0.01)(20)(5)/0.6 + 30(-0.01)(20) i3/ 0.6 + 100 i3 = 0 Solving, we find that i3 = (100 – 50)/ 90 = 555.6 mA Thus, v1 = 148.1 V, i2 = -1.481 A, and the power generated by the dependent voltage source is 0.4 v1 (i2 – i1) = -383.9 W.



34.


10 March 2006

We begin by defining four clockwise mesh currents i1, i2, i3 and i4, in the meshes of our circuit, starting at the left-m ost m esh. W e also define a voltage vdep acros s th e dependent current source, with the “+” on the top. i1 = 2A

By inspection,

and

i4 = -5 A.

At Mesh 2:

10 i2 - 10(2) + 20 i2 + vdep = 0

[1]

At Mesh 3:

- vdep + 25 i3 + 5 i3 –5(-5) = 0

[2]

Collecting terms, we rewrite Eqs. [1] and [2] as 30 i2

+ vdep = 20

[1]

30 i3 – vdep = -25

[2]

This is only two equations but three unknowns , however, so we require an additional equation. Returning to the circuit diagram , we note that it is pos sible to express th e current of the dependent sour ce in term s of m esh currents. (W e might also choose to obtain an expression for vdep in term s of m esh currents using KVL around m esh 2 or 3.) Thus, 1.5ix = i3 - i2 where ix = i1 – i2 so -0.5 i2 - i3 = -3

[3]

In matrix form, ⎡ 30 0 1⎤ ⎡i2 ⎤ ⎡ 20⎤ ⎢ 0 30 - 1⎥ ⎢i ⎥ = ⎢- 25⎥ ⎢ ⎥ ⎢3 ⎥ ⎢ ⎥ ⎢⎣- 0.5 - 1 0⎥⎦ ⎢⎣vdep ⎥⎦ ⎢⎣ - 3 ⎥⎦

Solving, we find that i2 = -6.333 A so that

ix = i1 – i2 = 8.333 A.



35.


10 March 2006

We define a clockwise mesh current i1 in the bottom left mesh, a clockwise mesh current i2 in the top left mesh, a clockwise mesh current i3 in the top right mesh, and a clockwise mesh current i4 in the bottom right mesh.

MESH

1:

-0.1 va + 4700 i1 – 4700 i2 + 4700 i1 – 4700 i4 = 0

[1]

MESH

2:

9400 i2 – 4700 i1 – 9 = 0

[2]

MESH 3: MESH

4:

9 + 9400 i3 – 4700 i4

= 0

9400 i4 – 4700 i1 – 4700 i3 + 0.1 ix = 0

[3] [4]

The presen ce of the two de pendent sources has led to the in troduction of tw o additional unknowns (ix and va) besides our four m esh currents. In a perfect world, it would simplify the solution if we could e xpress these tw o quantities in term s of the mesh currents. Referring to the circuit diagram, we see that ix = i2 (easy enough) and that va = 4700 i3 (also straightforward). Thus, subs tituting these express ions into our four mesh equations and creating a matrix equation, we arrive at:

Solving,

⎡ 0⎤ ⎡ 9400 - 4700 - 470 - 4700⎤ ⎡i1 ⎤ ⎢i ⎥ ⎢ 9⎥ ⎥ ⎢- 4700 9400 0 0⎥ ⎢ 2 ⎥ ⎢ = ⎢ ⎥ ⎢- 9 ⎥ ⎢ 0 0 9400 - 4700 ⎥ ⎢i3 ⎥ ⎢ ⎥ ⎥ ⎢ ⎥ ⎢ 0.1 - 4700 9400 ⎦ ⎣i4 ⎦ ⎣ 0⎦ ⎣- 4700 i1 = 239.3 μA, i2 = 1.077 mA, i3 = -1.197 mA and i4 = -478.8 μA.



36.


10 March 2006

We define a clockwise mesh current i3 in the upper right mesh, a clockwise mesh current i1 in the lower left m esh, and a clockwise m esh current i2 in th e lower right mesh. MESH 1:

-6 + 6 i1 - 2 = 0

[1]

MESH 2:

2 + 15 i2 – 12 i3 – 1.5 = 0

[2]

MESH

3:

i3 = 0.1 vx

[3]

Eq. [1] may be solved directly to obtain

i1 = 1.333 A.

It would help in the solution of Eqs. [2] and [3] if we could express the dependent source controlling variable vx in term s of m esh curren ts. Referring to the circuit diagram, we see that vx = (1)( i1) = i1, so Eq. [3] reduces to i3 = 0.1 vx = 0.1 i1 = 133.3 mA. As a result, Eq. [1] reduces to

i2 = [-0.5 + 12(0.1333)]/ 15 = 73.31 mA.



37.

10 March 2006

(a) Define a mesh current i2 in the second mesh. Then KVL allows us to write: MESH 1: -9 + R i1 + 47000 i1 – 47000 i2 = 0

[1]

MESH 2: 67000 i2 – 47000 i1 – 5 = 0

[2]

Given

that i1 = 1.5 mA, we may solve Eq. [2] to find that i2 =

and


so R =

5 + 47(1.5) mA = 1.127 mA 67

9 - 47(1.5) + 47(1.127) = -5687 Ω. 1.5 × 10-3

(b) This value of R is unique; no other value will satisfy both Eqs. [1] and [2].



38.


10 March 2006

Define three clockwise mesh currents i1, i2 and i3. The bottom 1-kΩ resistor can be ignored, as no current flows through it. MESH 1:

-4 + (2700 + 1000 + 5000) i1 – 1000 i2 = 0

[1]

MESH 2:

(1000 + 1000 + 4400 + 3000) i2 – 1000 i1 – 4400 i3 + 2.2 – 3 = 0 [2]

MESH 3:

(4400 + 4000 + 3000) i3 - 4400 i2 – 1.5 = 0

[3]

Combining terms, –1000 Solving,

8700 i1 – 1000 i2 = 4 i1 + 9400 i2 – 4400 i3 = 0.8 – 4400 i2 + 11400 i3 = 1.5

[1] [2] [3]

i1 = 487.6 μA, i2 = 242.4 μA and i3 = 225.1 μA. The power absorbed by each resistor may now be calculated: 2 P5k = 5000 (i= 1) 2 = 2700 (i= P2.7k 1) 1000 (i1 – i2)2 = P1ktop = 2 = 1000 (i= P1kmiddle 2) P1kbottom = 0 = = 4400 (i2 – i3)2 = P4.4k 2 3000 (i= P3ktop = 3) 2 = 4000 (i= P4k 3) 2 = 3000 (i= P3kbottom 2)

1.189 mW 641.9 μW 60.12 μW 58.76 μW 0 1.317 μW 152.0 μW 202.7 μW 176.3 μW

Check: The sources supply a total of 4(487.6) + (3 – 2.2)(242.4) + 1.5(225.1) = 2482 μW. The absorbed powers add to 2482 μW.



39.


10 March 2006

(a) We begin by naming four mesh currents as depicted below:

Proceeding with mesh analysis, then, keeping in mind that Ix = -i4, MESH 1:

(4700 + 300) i1 - 4700 i2

= 0

[1]

MESH 2:

(4700 + 1700) i2 – 4700 i1 – 1700 i3 = 0

[2]

Since we have a current source on the perimeter of mesh 3, we do not require a KVL equation for that mesh. Instead, we may simply write i3 = -0.03 vπ [3a] MESH

4:

where

3000 i4 – 3000 i3 + 1

vπ = 4700(i1 – i2) [3b] = 0

[4]

Simplifying and combining Eqs. 3a and 3b, 5000 i1 – 4700 i2 i1 + 6400 i2 – 1700 i3 –141 i1 + 141 i2 – i3 – 3000 i3 + 3000 i4

–4700

Solving, we find that i4 = -333.3 mA, so (b) At node “π” :

= 0 = 0 = = –1

0

Ix = 333.3 μA.

0.03 vπ = vπ / 300 + vπ / 4700 + vπ /1700

Solving, we find that vπ = 0, therefore no current flows through the dependent source. Hence,

Ix = 333.3 μA as found in part (a).

(c) V s/ I x has units of resistance. It can be t hought of as the resistance “seen” by the voltage source Vs…. more on this in Chap. 5….



40.


10 March 2006

We begin by naming each mesh and the three undefined voltage sources as shown below:

MESH 1:

–Vz + 9i1 – 2i2

MESH 2:

– 7i4 = 0

–2i1 + 7i2 – 5i3

MESH 3:

Vx

MESH 4:

Vy – 7i1

= 0

– 5i2 + 8i3 – 3i4 = 0 – 3i3 + 10i4 = 0

Rearranging and setting i1 – i2 = 0, i2 – i3 = 0, i1 – i4 = 0 and i4 – i3 = 0, 9i1 - 2i2 -7i4 -2i1 + 7i2 - 5i3 -5i2 + 8i3 – 3i4 -7i1 -3i3 + 10i4 Since

= Vz = 0 = - Vx = - Vy

i1 = i2 = i3 = i4, these equations produce: Vz 0 -Vx -Vy

= = = =

0 0 0 0

This is a unique solution. Therefore, the request that nonzero values be found cannot be satisfied.



41.


10 March 2006

The “supermesh” concept is not required (or helpful) in solving this problem, as there are no current sources s hared between m eshes. Starting with the left-most m esh and moving right, we define four clockwise m esh currents i1, i2, i3 and i4. By inspection, we see that i1 = 2 mA. MESH 2:

-10 + 5000i2 + 4 + 1000i3 = 0

[1]

MESH 3:

-1000i3 + 6 + 10,000 – 10,000i4 = 0

[2]

MESH 4:

i4 = -0.5i2

[3]

Reorganising, we find 5000 i2 + 1000 i3 = 6 9000 i3 – 10,000 i4 = -6 + i4 = 0 0.5 i2

[1] [2] [3]

We could either subtitute Eq. [3] into Eq. [2] to reduce the number of equations, or simply go ahead and solve the system of Eqs. [1-3]. Either way, we find that i1 = 2 mA, i2 = 1.5 mA, i3 = -1.5 mA and i4 = -0.75 mA. The power generated by each source is: P2mA = 5000(i1 – i2)(i1) P4V = 4 (-i2) = 6 (-i3) P6V 1000 i3 (i3 – i2) PdepV = 10,000(i3 – i4)(0.5 i2) PdepI=

= 5 mW = -6 mW = 9 mW = 4.5 mW = -5.625 mW



42.


10 March 2006

This circuit does not require the supermesh technique, as it does not contain any current sources. Redrawing the circuit so its planar nature and m esh structure are clear, 1Ω

•

i2

i1 20 V

2.5 iA

• MESH 1:

2Ω

4Ω iA

3Ω i3

→

-20 + i1 – i2 + 2.5 iA = 0

[1]

MESH

2:

2 i 2 + 3 i 2 + i 2 – 3 i 3 – i1 = 0

[2]

MESH

3:

-2.5 iA + 7 i3 – 3 i2

[3]

= 0

Combining terms and making use of the fact that iA = - i3, i1 – i2 – 2.5 i3 = 20 - i1 + 6i2 – 3 i3 = 0 –3 i2 + 9.5 i3 = 0 Solving,

[1] [2] [3]

i1 = 30.97 A, i2 = 6.129 A, and i3 = 1.936 A. Since iA = - i3, iA = –1.936 A.



43.

Define four mesh currents


i1 i3

10 March 2006

i2 i4

By inspection, i1 = -4.5 A. We form a supermesh with meshes 3 and 4 as defined above. MESH 2: SUPERME Superm

SH:

2.2 + 3 i2 + 4 i2 + 5 – 4 i3 = 0

[1]

3 i 4 + 9 i4 – 9 i1 + 4 i3 – 4 i2 + 6 i3 + i3 – 3 = 0

[2]

esh KCL equation:

i4 - i3 = 2

[3]

Simplifying and combining terms, we may rewrite these three equations as: 7 i2 – 4 i3 -4 i2 + 11 i3 + 12 i4 - i3 + i4

= -7.2 = -37.5 = 2

[1] [2] [3]

Solving, we find that i2 = -2.839 A, i3 = -3.168 A, and i4 = -1.168 A. The power supplied by the 2.2-V source is then 2.2 (i1 – i2) = -3.654 W.



44.


10 March 2006

We begin by defining six mesh currents as depicted below: i5 i1 • •

i2

i6 i3

i4

We form a supermesh with meshes 1 and 2 since they share a current source. We form a second supermesh with meshes 3 and 4 since th ey also share a curren t source.

1, 2 Supermesh: (4700 + 1000 + 10,000) i1 – 2200 i5 + (2200 + 1000 + 4700) i2 – 1000 i3 = 0

[1]

3, 4 Supermesh: (4700 + 1000 + 2200) i3 – 1000 i2 – 2200 i6 + (4700 + 10,000 + 1000) i4 = 0

[2]

MESH 5:

(2200 + 4700) i5 – 2200 i2 + 3.2 – 1.5 = 0

[3]

MESH 6:

1.5 + (4700 + 4700 + 2200) c – 2200 i3 = 0

[4]

1, 2 Supermesh KCL equation:

i1 – i2 = 3×10-3

[5]

3, 4 Supermesh KCL equation:

i4 – i3 = 2×10-3

[6]

We can simplify these equations prior to solution in several ways. Choosing to retain six equations, -2200 i5 = 0 [1] 15,700 i1 + 7900 i2 - 1000 i3 -2200 i6 = 0 [2] - 1000 i2 + 7900 i3 + 15,700 i4 - 2200 i2 + 6900 i5 = -1.7 [3] [4] - 2200 i3 + 11,600 i6 = -1.5 -3 – i2 = 3×10 [5] i1 + i4 = 2×10-3 [6] - i3 Solving, we find that i4 = 540.8 mA. Thus, the voltage across the 2-mA source is (4700 + 10,000 + 1000) (540.8×10-6) = 8.491 V



45.


10 March 2006

We define a mesh current ia in the left-hand mesh, a mesh current i1 in the top right mesh, and a mesh current i2 in the bottom right mesh (all flowing clockwise). The left-most mesh can be analysed separately to determine the controlling voltage va, as KCL assures us that no current flows through either the 1-Ω or 6-Ω resistor. Thus, -1.8 + 3 ia – 1.5 + 2 ia = 0, which may be solved to find ia = 0.66 A. Hence, va = 3 ia = 1.98 V. Forming one supermesh from the remaining two meshes, we may write: -3 + 2.5 i1 + 3 i2 + 4 i2 = 0 and the supermesh KCL equation: i2 – i1 = 0.5 va = 0.5(1.98) = 0.99 Thus, we have two equations to solve: 2.5 i1 + 7 i2 = 3 -i1 + i2 = 0.99 Solving, we find that i1 = -413.7 mA and the voltage across the 2.5Ω res istor (arbitrarily assuming the left terminal is the “+” reference) is 2.5 i1 = -1.034 V.



46.


10 March 2006

There are only three meshes in this circuit, as the botton 22-mΩ resistor is not connected connected at its left term inal. Thus, we define three m esh currents, i1, i2, and i3, beginning with the left-most mesh. We next create a superm esh from meshes 1 and 2 (note that m esh 3 is independent, and can be analysed separately). Thus,

-11.8 + 10×10-3 i1 + 22×10-3 i2 + 10×10-3 i2 + 17×10-3 i1 = 0

and applying KCL to obtain an equation containing the current source, i1 – i2 = 100 Combining terms and simplifying, we obtain 27×10-3 i1 + 32×10-3 i2 = 11.8 – i2 = 100 i1 Solving, we find that i1 = 254.2 A and i2 = 154.2 A. The final mesh current is easily found: i3 = 13×103/ (14 + 11.6 + 15) = 320.2 A.



47. MESH 1: MESH 2: MESH 3:


-7 + i1 – i2 = 0 i2 – i1 + 2i2 + 3i2 – 3i3 = 0 3i3 – 3i2 + xi3 +2i3 – 7 = 0

10 March 2006

[1] [2] [3]

Grouping terms, we find that i1 – i2 -i1 + 6i2 – 3i3 -3i2 + (5 + x)i3

= 7 = =7

[1] [2] 0 [3]

This, unfortunately, is four unknowns but only three equations. However, we have not yet made use of the fact that we are trying to obtain i2 = 2.273 A. Solving these “four” equations, we find that x = (7 + 3 i2 – 5 i3)/ i3 = 4.498 Ω.



48.


10 March 2006

We begin by redrawing the circuit as instructed, and define three mesh currents: 300 mΩ

2Ω 3Ω

i2

1Ω

i1

i3

2Ω

7A

7V

B

y inspection, i3 = 7 A. MESH 1:

-7 + i1 – i2 = 0

or

i1 – i2 = 7 [1]

MESH 2:

(1 + 2 + 3) i2 – i1 –3(7) = 0 or

-i1 + 6i2 = 21 [2]

There is no need for supermesh techniques for this situation, as the only current source lies on the outside perimeter of a mesh- it is not shared between meshes. Solving, we find that

i1 = 12.6 A, i2 = 5.6 A and i3 = 7 A.



49. (


10 March 2006

a) We are asked for a voltage, and have one current source and one voltage source. Nodal analysis is probably best then- th e nodes can be nam ed so that the desired voltage is a nodal voltage, or, at worst, we have one supernode equation to solve. Name the top left node “1” a nd the top right node “x”; designate the bottom node as the reference terminal. Next, form a supernode with nodes “1” and “x.” At the supernode:

11 = v1/ 2 + vx/ 9

[1]

and the KVL Eqn:

v1 – vx = 22

[2]

Rearranging,

11(18) = 9 v1 + 2 vx 22 = v1 – vx [2]

[1]

Solving, vx = 0 (b) We are asked for a voltage, and so may suspect that nodal analysis is preferrable; with two current sources a nd only one voltage source (eas ily dea lt w ith using the supernode technique), nodal analysis does seem to have an edge over m esh analysis here. Name the top left node “x,” the top right node “y” and designate the bottom node as the reference node. Forming a supernode from nodes “x” and “y,” At the supernode: and the KVL Eqn:

6 + 9 = vx / 10 + vy/ 20 vy – vx = 12

[1] [2]

Rearranging, 15(20) = 2 vx + vy [1] and 12 = - vx + vy [2] Solving, we find that vx = 96 V. (c) W e are asked for a voltage, but would ha ve to subtract two nodal voltages (not much harder than invoking Ohm ’s law). On the other hand, the dependent current source depends on the desired unknown, whic h would lead to the need for another equation if invoking mesh analysis. Trying nodal analysis, 0.1 vx = (v1 – 50) / 2 + vx / 4

[1]

referring to the circuit w e see tha t vx = v1 – 100. Rearranging so that we m eliminate v1 in Eq. [1], we obtain v1 = vx + 100. Thus, Eq. [1] becomes

ay

0.1 vx = (vx + 100 – 50)/ 2 + vx / 4 and a little algebra yields vx = -38.46 V.



50.


10 March 2006

v1

(a) (b) Ref.

and

(a) We begin by noting that it is a voltage that is required; no current values are requested. This is a three- mesh circuit, or a four-node circuit, depending on your perspective. Either app roach requires th ree equations…. Except that applying the supernode technique reduces the number of needed equations by one. At the 1, 3 supernode: 0 = (v1 – 80)/ 10 + (v1 – v3)/ 20 + (v3 – v1)/ 20 + v3/ 40 + v3/ 30 v3 - v1 = 30 We simplify these two equations and collect terms, yielding 0.1 v1 + 0.05833 v3 = 8 - v1 + v3 = 30 Solving, we find that v3 = 69.48 V

Both ends of the resistor are connected to the supernode, so we could actually just ignore it…

(b) Mesh analysis would be straightforward, requiring 3 equations and a (trivial) application of O hm’s law to obta in the f inal answer. Nodal analysis, on the other hand, would require o nly two equations, and the desired voltage will be a nodal voltage. At the b, c, d supernode: and:

vd – vb = 30

0 = (vb – 80)/ 10 + vd/ 40 + vc/ 30 vc – vd = 9

Simplify and collect terms: Solving, vd (= v3) = 67.58 V

0.1 vb + 0.03333 vc + 0.025 vd = 80 -vb + vd = 30 vd = 9 vc -

(c) We are now faced with a dependent current source whose value depends on a mesh current. Mesh analysis in th is situation requires 1 superm esh, 1 KCL equation and Ohm’s law. Nodal analysis requires 1 supernode, 1 KVL equation, 1 other nodal equation, and one equation to express i1 in term s of nodal voltages. Thus, m esh analysis has an edg e here. Define th e left m esh as “1,” the top m esh as “2”, and the bottom mesh as “3.”

and:

Mesh 1: 2, 3 supermesh:

Rewriting,

-80 + 10 i1 + 20 i1 – 20 i2 + 30 i1 – 30 i3 = 0 20 i2 – 20 i1 – 30 + 40 i3 + 30 i3 – 30 i1 = 0 i2 - i3 = 5 i1

60

i1 – 20 i2 – 30 i3 = 80 -50 i1 + 20 i2 + 70 i3 = 30 5 i1 – i2 + i3 = 0 Solving, i3 = 4.727 A so

v3 = 40 i3 = 189 V.



51.


10 March 2006

This circuit consists of 3 meshes, and no dependent sources. Therefore 3 simultaneous equations and 1 subtraction operation would be required to solve for the two desired currents. On the other hand, if we use nodal analysis, form ing a supernode about the 30-V source would lead to 5 – 1 – 1 = 3 sim ulataneous equations as well, plus several subtraction and division opera tions to find the currents. Thus, m esh a nalysis has a slight edge here. Define three clockwis e m esh currents: ia in the lef t-most m esh, ib in the top r ight mesh, and ic in the bottom right mesh. Then our mesh equations will be: Mesh a: Mesh b: Mesh c:

-80 + (10 + 20 + 30) ia – 20 ib – 30 ic = 0 -30 + (12 + 20) ib – 12 ic – 20 ia = 0 (12 + 40 + 30) ic – 12 ib – 30 ia = 0

[1] [2] [3]

Simplifying and collecting terms, 60 ia – 20 ib – 30 ic = 80 -20 ia + 32 ib – 12 ic = 30 -30 ia – 12 ib + 82 ic = 0

[1] [2] [3]

Solving, we find that ia = 3.549 A, ib = 3.854 A, and ic = 1.863 A. Thus, i1 = ia = 3.549 A

and i2 = ia – ic = 1.686 A.



52.

10 March 2006

Approaching this problem using nodal analysis would require 3 separate nodal equations, plus one equation to deal with the dependent source, plus subtraction and division steps to actua lly find the current i10. Mesh analy sis, on the o ther hand, w ill require 2 mesh/supermesh equations, 1 KCL equation, and one subtraction step to find i10. Thus, mesh analys is has a clear edge. Define three clockwise m esh currents: i1 in the bottom left mesh, i2 in the top mesh, and i3 in the bottom right mesh. MESH 1:

i1 = 5 mA by inspection

SUPERMESH:

MESH Sim


3: plify:

i1 – i2 = 0.4 i10 i1 – i2 = 0.4(i3 – i2) i1 – 0.6 i2 – 0.4 i3 = 0

[1]

[2]

-5000 i1 – 10000 i2 + 35000 i3 = 0 [3] 0.6

i2 + 0.4 i3 = 5×10-3 [2] -10000 i2 + 35000 i3 = 25

[3]

Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that i10 = -4 mA.



53.


10 March 2006

For this circuit problem, nodal analysis will require 3 simultaneous nodal equations, then subtraction/ division steps to obtain th e desired curren ts. Mesh analysis requ ires 1 mesh equation, 1 superm esh equation, 2 simple KCL equations and one subtraction step to dete rmine the currents. If either te chnique has an edge in this situa tion, it’s probably mesh analysis. Thus, define four clockwise mesh equations: ia in the bottom left mesh, ib in the top lef t mesh, ic in the top right m esh, and id in the bottom right mesh. At the a, b, c supermesh:

-100 + 6 ia + 20 ib + 4 ic + 10 ic – 10 id = 0 [1]

Mesh d:

100 + 10 id – 10 ic + 24 id = 0

[2]

KCL: and

- ia + ib = 2 ib + ic = 3 i3 = 3 ia

[3] [4]

-

Collecting terms & simplifying,

Solving,

6 ia + 20 ib + 14 ic – 10 id = 100 -10 ic + 34 id = -100 - ia + ib = 2 = 0 -3 ia – ib + ic

[1] [2] [3] [4]

ia = 0.1206 A, ib = 2.121 A, ic = 2.482 A, and id = -2.211 A. Thus, i3 = ia = 120.6 mA and i10 = ic – id = 4.693 A.



54.


10 March 2006

With 7 nodes in this circuit, nodal analysis will require the solution of three simultaneous nodal equations (assum ing we make use of the supernode technique) and one KVL equation. Mesh analysis will require the solution of three simultaneous mesh equations (one m esh current can be found by inspecti on), plus several subtraction and m ultiplication oper ations to f inally de termine the voltage a t th e central nod e. Either will probab ly requi re a com parable am ount of algebraic manoeuvres, so we go with nodal analysis, as the desired unknown is a direct result of solving the simultaneous equations. Define the nodes as: v6 v3 v4

v1

v5

v2

-2×10-3 = (v1 – 1.3)/ 1.8×103

NODE 1:

→ v1 = -2.84 V.

2, 4 Supernode: 2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103 KVL equation: Node 5:

-v2 + v4 = 5.2

0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103

Simplifying and collecting terms, 14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1] -v2 + v4 = 5.2 [2] -8.19 v2 – 6.3 v4 + 15.79 v5 = 3.38 [3] Solving, we find the voltage at the central node is v4 = 3.460 V.



55.


10 March 2006

Mesh analysis yields current values directly, so use that approach. We therefore define four clockwise mesh currents, starting with i1 in th e left-most mesh, then i2, i3 and i4 moving towards the right.

Mesh

1:

-0.8ix + (2 + 5) i1 – 5 i2 = 0 [1]

Mesh

2:

i2 = 1 A by inspection

[2]

Mesh 3:

(3 + 4) i3 – 3(1) – 4(i4) = 0 [3]

Mesh 4:

(4 + 3) i4 – 4 i3 – 5 = 0

[4]

Simplify and collect terms, noting that ix = i1 – i2 = i1 – 1 -0.8(

i1 – 1) + 7 i1 – 5(1) = 0 yields i1 = 677.4 mA Thus, [3] and [4] become:

7 i 3 – 4 i4 = 3 -4 i3 + 7 i4 = 5

[3] [4]

Solving, we find that i3 = 1.242 A and i4 = 1.424 A. A map of individual branch currents can now be drawn: 677.4 mA → -322.6 mA ↓ ↑ 677.4 mA

-242.0 mA ↓ ↑ 182.0 mA

↑ -1.424 A



56.


10 March 2006

If we choose to perform mesh analysis, we require 2 simultaneous equations (there are four m eshes, but one m esh current is known, and we can em ploy the superm esh technique around the left two m eshes). In order to find the voltage across the 2-m A source we will need to write a KVL equation, however. Using nodal analysis is less desirable in this case, as there will be a large num ber of nodal equations needed. Thus, we de fine four clockwise m esh currents i1, i2, i3 and i4 star ting w ith the lef tmost mesh and moving towards the right of the circuit. At the 1,2 supermesh: and

2000 i1 + 6000 i2 – 3 + 5000 i2 = 0 i1 – i2 = 2×10-3

[1] [2]

by inspection, i4 = -1 m A. However, this as well as any equation for mesh four are unnecessary: w e already have two equations in two unknowns and i1 and i2 are sufficient to enable us to find the voltage across the current source. Simplifying, we obtain 1000 Solving,

i1 = 1.923 mA and

2000 i1 + 11000 i2 = 3 i1 - 1000 i2 = 2

[1] [2]

i2 = -76.92 μA.

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is v = -2000 i1 – 6000 (i1 – i2) = -15.85 V.



57.


10 March 2006

Nodal analysis will require 2 nodal equations (one being a “supernode” equation), 1 KVL equation, and subtraction/division operations to obtain the desired current. Mesh analysis s imply requ ires 2 “supermesh” equations and 2 KCL equations, with the desired current being a m esh current. Thus , we define four cl ockwise mesh currents ia, ib, ic, id starting with the left-most mesh and proceeding to the right of the circuit. At the a, b supermesh:

-5 + 2 ia + 2 ib + 3 ib – 3 ic = 0

[1]

At the c, d supermesh:

3 ic – 3 ib + 1 + 4 id = 0

[2]

and

ia - ib = 3 ic - id = 2

[3] [4]

Simplifying and collecting terms, we obtain 2 ia + 5 ib – 3 ic = 5 -3 ib + 3 ic + 4 id = -1 ia - ib = 3 ic - id = 2

[1] [2] [3] [4]

Solving, we find ia = 3.35 A, ib = 350 mA, ic = 1.15 A, and id = -850 mA. As i1 = ib, i1 = 350 mA.



58.


10 March 2006

Define a voltage vx at the top node of the current source I2, and a clockwise mesh current ib in the right-most mesh. We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A. Applying nodal analysis to the circuit, I1 + I2 = (vx – v1)/ 6 = 1 so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than we are told to select a nonzero value. Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V.



59.


10 March 2006

Inserting the new 2-V source with “+” reference at the bottom, and the new 7-mA source with the arrow pointing down, we define four clockwise mesh currents i1, i2, i3, i4 starting with the left-most mesh and proceeding towards the right of the circuit. Mesh 1:

(2000 + 1000 + 5000) i1 – 6000 i2 – 2 = 0 [1]

2, 3 Supermesh: 2 + (5000 + 5000 + 1000 + 6000) i2 – 6000 i1 + (3000 + 4000 + 5000) i3 – 5000 i4 = 0 [2] [3] and i2 - i3 = 7×10-3 Mesh 4:

i4 = -1 mA by inspection

[4]

Simplifying and combining terms, 8000 -6000

i1 –

6000 i2 1000 i2 – 1000 i3 i1 + 17000 i2 + 12000 i3

= 2 = 7 = -7

[1] [4] [2]

Solving, we find that i1 = 2.653 A, i2 = 3.204 A, i3 = -3.796 A, i4 = -1 mA



60.


10 March 2006

This circuit is e asily analyzed by mesh analysis; it’s planar, and af ter combining the 2A and 3 A sources into a single 1 A source, supermesh analysis is simple. First, define clockwise m esh currents ix, i1, i2 and i3 starting from the lef t-most mesh and moving to the right. Next, com bine the 2 A and 3 A sources tem porarily into a 1 A source, arrow pointing upwards. Th en, define four nodal voltages, V 1, V 2, V 3 and V4 moving from left to right along the top of the circuit. At the left-most mesh, ix = –5 i1 [1] For the supermesh, we can write 4i1 – 2ix + 2 + 2i3 = 0 and the corresponding KCL equation: i3 – i1 = 1

[2] [3]

Substituting Eq. [1] into Eq. [2] and simplifying, 14

i1 + 2i3 = –2 –i1 + i3 = 1

Solving, Then,

i1 = –250 mA and i3 = 750 mA. ix = -5 i1 = 1.35 A and i2 = i1 – 2 = –2.25 A

Nodal voltages are straightfoward to find, then: V4 = 2i3 = 1.5 V V3 = 2 + V4 = 3.5 V V2 – V3 = 2 i1 or V2 = 2 i1 + V3 = 3 V V1 – V2 = 2 ix or V1 = 2 ix + V2 = 5.5 V



61.


10 March 2006

Hand analysis:

Define three clockwise m esh currents: i1 in the bottom left mesh, i2 in the top m esh, and i3 in the bottom right mesh. MESH 1:

i1 = 5 mA by inspection

SUPERMESH:

MESH Sim

3: plify:

i1 – i2 = 0.4 i10 i1 – i2 = 0.4(i3 – i2) i1 – 0.6 i2 – 0.4 i3 = 0

[1]

[2]

-5000 i1 – 10000 i2 + 35000 i3 = 0 [3] 0.6

i2 + 0.4 i3 = 5×10-3 [2] -10000 i2 + 35000 i3 = 25

[3]

Solving, we find i2 = 6.6 mA and i3 = 2.6 mA. Since i10 = i3 – i2, we find that i10 = -4 mA. PSpice simulation results:

i10 →

Summary: The current entering the right-hand node of the 10-k Ω resistor R2 is equal to 4.000 mA. Since this current is –i10, i10 = -4.000 mA as found by hand.



62.


10 March 2006

Hand analysis: Define the nodes as: v6 v3 v4

v5

v2

v1

-2×10-3 = (v1 – 1.3)/ 1.8×103

NODE 1:

→ v1 = -2.84 V.

2, 4 Supernode: 2.3×10-3 = (v2 – v5)/ 1x103 + (v4 – 1.3)/ 7.3×103 + (v4 – v5)/ 1.3×103 + v4/ 1.5×103 KVL equation: Node 5:

-v2 + v4 = 5.2

0 = (v5 – v2)/ 1x103 + (v5 – v4)/ 1.3x103 + (v5 – 2.6)/ 6.3x103

Simplifying and collecting terms, 14.235 v2 + 22.39 v4 – 25.185 v5 = 35.275 [1] -v2 + v4 = 5.2 [2] -8.19 v2 – 6.3 v4 + 15.79 v5 = 3.38 [3] Solving, we find the voltage at the central node is v4 = 3.460 V. PSpice simulation results:

Summary: The voltage at the center node is found to be 3.460 V, which is in agreement with our hand calculation.



63.

Hand analysis: At the 1,2 supermesh: and


10 March 2006

2000 i1 + 6000 i2 – 3 + 5000 i2 = 0 i1 – i2 = 2×10-3

[1] [2]

by inspection, i4 = -1 m A. However, this as well as any equation for mesh four are unnecessary: w e already have two equations in two unknowns and i1 and i2 are sufficient to enable us to find the voltage across the current source. Simplifying, we obtain 1000 Solving,

i1 = 1.923 mA and

2000 i1 + 11000 i2 = 3 i1 - 1000 i2 = 2

[1] [2]

i2 = -76.92 μA.

Thus, the voltage across the 2-mA source (“+” reference at the top of the source) is v = -2000 i1 – 6000 (i1 – i2) = -15.85 V. PSpice simulation results:

Summary: Again arbitrarily selecting the “+” reference as the top node of the 2-mA current source, we find the voltage across it is –5.846 – 10 = -15.846 V, in agreement with our hand calculation.



64.


10 March 2006

Hand analysis: Define a voltage vx at the top node of the current source I2, and a clockwise mesh current ib in the right-most mesh.

We want 6 W dissipated in the 6-Ω resistor, which leads to the requirement ib = 1 A. Applying nodal analysis to the circuit, I1 + I2 = (vx – v1)/ 6 = 1 so our requirement is I1 + I2 = 1. There is no constraint on the value of v1 other than we are told to select a nonzero value. Thus, we choose I1 = I2 = 500 mA and v1 = 3.1415 V. PSpice simulation results:

Summary: We see from the labeled schematic above that our choice for I1, I2 and V1 lead to 1 A through the 6-Ω resistor, or 6 W dissipated in that resistor, as desired.



65.


10 March 2006

Hand analysis: Define node 1 as the top left node, and node 2 as the node joining the three 2-Ω resistors. Place the “+” reference term inal of the 2-V source at the righ t. The rightmost 2-Ω resistor has therefore been shorted out. Applying nodal analysis then,

Node 1:

-5 i1 = (v1 – v2)/ 2

[1]

Node 2:

0 = (v2 – v1)/ 2 + v2/ 2 + (v2 – 2)/ 2 [2]

and,

i1 = (v2 – 2)/ 2

[3]

Simplifying and collecting terms, v1 + v2 = 10 -v1 + 3 v2 = 2 Solving, we find that

[1] [2]

v1 = 3.143 V and v2 = 1.714 V.

Defining clockwise m esh currents i a, i b, i c, i d starting with the left-most m esh and proceeding right, we may easily determine that ia = -5 i1 = 714.3 mA ib = -142.9 mA ic = i1 – 2 = -2.143 A id = 3 + ic = 857.1 mA PSpice simulation results:

Summary:

The simulation results agree with the hand calculations.



66.


10 March 2006

(a) One possible circuit configuration of many that would satisfy the requirements: 100 Ω

+ vx 5V

50 Ω

-

20 Ω

3A

↑

2 vx

10 Ω

At node 1:

-3 = (v1 – 5)/ 100 + (v1 – v2)/ 50

[1]

At node 2:

2 vx = (v2 – v1)/ 50 + v2/ 30

[2]

and,

vx = 5 – v1 [3] Simplifying and collecting terms,

150

v1 – 100 v2 = -14750

[1]

2970

v1 + 80 v2 = 15000

[2]

Solving, we find that v1 = 1.036 V and v2 = 149.1 V. The current through the 100-Ω resistor is simply (5 – v1)/100 = 39.64 mA The current through the 50-Ω resistor is (v1 – v2)/ 50 = -2.961 A, and the current through the 20- Ω and 10- Ω ser ies com bination is v2/ 30 = 4.97 A. Finally, the dependent source generates a current of 2 vx = 7.928 A. (b) PSpice simulation results

Summary:

The simulated results agree with the hand calculations.



67.


10 March 2006

One possible solution of many: + 5V -

Choose R so that 3R = 5; then the voltage across the current source will be 5 V, and so will the voltage across the resistor R. R = 5/3 Ω. To construct this from 1-Ω resistors, note that 5/3 Ω = 1 Ω + 2/3 Ω = 1 Ω + 1 Ω || 1Ω || 1Ω + 1Ω || 1Ω || 1Ω * Solution to Problem 4.57 .OP V1 1 0 DC 10 I1 0 4 DC 3 R1 1 2 1 R2 2 3 1 R3 2 3 1 R4 2 3 1 R5 3 4 1 R6 3 4 1 R7 3 4 1 .END

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C ****************************************************************************** NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE ( 1) 10.0000 ( 2) 7.0000 ( 3) 6.0000 ( 4) 5.0000 VOLTAGE SOURCE CURRENTS NAME CURRENT V1

-3.000E+00



68.


We first name each node, resistor and voltage source: R7 R4 R6 R3 R1 2 1 3

10 March 2006

R9 5

4

R2

V1

R5

R8

R10

0

We next write an appropriate input deck for SPICE: * Solution to Problem 4.58 .OP V1 1 0 DC 20 R1 1 2 2 R2 2 0 3 R3 2 3 4 R4 2 4 10 R5 3 0 5 R6 3 4 6 R7 3 5 11 R8 4 0 7 R9 4 5 8 R10 5 0 9 .END

And obtain the following output:

We see from this simulation result that the voltage v5 = 2.847 V.



69.


10 March 2006

One possible solution of many: v1

R1

R2 v2 R3 v3

All resistors are 1 Ω, except R1, which represents 5 1-Ω resistors in series.

R4 R5

Verify:

v1 = 9(4/9) = 4 V v2 = 9(3/9) = 3 V v3 = 9(2/9) = 2 V

SPICE INPUT DECK:

* Solution to Problem 4.59 .OP V1 1 0 DC 9 R1 1 2 5 R2 2 3 1 R3 3 4 1 R4 4 5 1 R5 5 0 1 .END

**** 07/29/01 21:36:26 *********** Evaluation PSpice (Nov 1999) ************** * Solution to Problem 4.59 **** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C *********************************************************************** ******* NODE VOLTAGE VOLTAGE ( 1) 9.0000 ( ( 5) 1.0000

2)

NODE

VOLTAGE

NODE

VOLTAGE

NODE

4.0000 ( 3) 3.0000 ( 4) 2.0000 2

1 R1

R2 3 R3

4

R4 5 0

R5



70.


10 March 2006

(a) If only two bulbs are not lit (and thinking of each bulb as a resistor), the bulbs must be in parallel- otherwise, the burne d out bulbs, acting as short circuits, would prevent current from flowing to the “good” bulbs. (b) In a parallel connected circuit, each bulb “sees” 115 VAC. Therefore, the individual bulb current is 1 W / 115 V = 8.696 mA. The resistance of each “good ” bulb is V/I = 13.22 k Ω. A sim plified, elec trically-equivalent m odel f or this cir cuit would be a 115 VAC source connected in parallel to a resistor Req such that 1/Req = 1/13.22×103 + 1/13.22×103 + …. + 1/13.22×103 (400 – 2 = 398 terms) or Req = 33.22 Ω. We expect the source to provide 398 W. * Solution to Problem 4.60 .OP V1 1 0 AC 115 60 R1 1 0 33.22 .AC LIN 1 60 60 .PRINT AC VM(1)IM(V1) .END

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) ************** * Solution to Problem 4.60 **** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C ****************************************************************************** NODE VOLTAGE ( 1) 0.0000

NODE VOLTAGE

NODE VOLTAGE

VOLTAGE SOURCE CURRENTS NAME CURRENT V1

0.000E+00

TOTAL POWER DISSIPATION 0.00E+00 WATTS

NODE VOLTAGE

This calculated power is not the value sought. It is an artifact of the use of an ac s ource, which re quires that we perform an ac analysis. T he supplied power is th en separately computed as (1.15×102)(3.462) = 398.1 W.

**** 07/29/01 21:09:32 *********** Evaluation PSpice (Nov 1999) ************** * Solution to Problem 4.60

**** AC ANALYSIS TEMPERATURE = 27.000 DEG C ****************************************************************************** FREQ VM(1) IM(V1) 6.000E+01 1.150E+02 3.462E+00

(c) The in herent serie s res istance of th e wire connection s leads to a voltage dro p which increases the f urther one is from the voltage source. Thus, th e furthest bulbs actually have less than 115 VAC across them , so they draw slightly less current and glow more dimly.



1.

Chapter Five Solutions

10 March 2006

Define percent error as 100 [ex – (1 + x)]/ ex x 0.001 0.005 0.01 0.05 0.10 0.50 1.00 5.00

1+x 1.001 1.005 1.01 1.05 1.10 1.50 2.00 6.00

ex 1.001 1.005 1.010 1.051 1.105 1.649 2.718 148.4

% error 5×10-5 1×10-3 5×10-3 0.1 0.5 9 26 96

Of course, “reasonable” is a very subjective term. However, if we choose x < 0.1, we ensure that the error is less than 1%.



2.


10 March 2006

(a) Short-circuit the 10 V source. Note that 6 || 4 = 2.4 Ω. By voltage division, the voltage across the 6 Ω resistor is then 4

2.4 = 1.778 V 3 + 2.4

So that i1′ =

1.778 = 0.2963 A . 6

(b) Short-circuit the 4 V source. Note that 3 || 6 = 2 Ω. By voltage division, the voltage across the 6 Ω resistor is then −10

So that i1′′ =

2 = −3.333 V 6 −3.333 = −0.5556 A . 6

(c) i1 = i1′ + i2′′ = -259.3 mA



3.


10 March 2006

Open circuit the 4 A source. Then, since (7 + 2) || (5 + 5) = 4.737 Ω, we can calculate v1′ = (1)(4.737) = 4.737 V. To find the total current flowing through the 7 Ω resistor, we first determine the total voltage v1 by continuing our superposition procedure. The contribution to v1 from the 4 A source is found by first open-circuiting the 1 A source, then noting that current division yields: 5 20 4 = = 1.053 A 5 + (5 + 7 + 2) 19 Then, v1′′ = (1.053)(9) = 9.477 V. Hence, v1 = v1′ + v1′′ = 14.21 V. We may now find the total current flowing downward through the 7 Ω resistor as 14.21/7 = 2.03 A.



4.


10 March 2006

One approach to this problem is to write a set of mesh equations, leaving the voltage source and current source as variables which can be set to zero. We first rename the voltage source as Vx. We next define three clockwise mesh currents in the bottom three meshes: i1, iy and i4. Finally, we define a clockwise mesh current i3 in the top mesh, noting that it is equal to –4 A. Our general mesh equations are then: -Vx + 18i1 – 10iy = 0 –10i1 + 15iy – 3i4 = 0 –3iy + 16i4 – 5i3 = 0 ** Set Vx = 10 V, i3 = 0. Our mesh equations then become: 18i1 – 10 i′y

= 10

–10i1 + 15 i′y – 3i4 = 0 – 3 i′y + 16i4 = 0 Solving, i′y = 0.6255 A. ** Set Vx = 0 V, i3 = – 4 A. Our mesh equations then become: 18i1 – 10 i′′y

=0

–10i1 + 15 i′′y – 3i4 = 0 – 3 i′′y + 16i4 = -20 Solving, i′′y = –0.4222 A. Thus, iy = i′y + i′′y = 203.3 mA.



5.


10 March 2006

We may solve this problem without writing circuit equations if we first realise that the current i1 is composed of two terms: one that depends solely on the 4 V source, and one that depends solely on the 10 V source. This may be written as i1 = 4K1 + 10K2, where K1 and K2 are constants that depend on the circuit topology and resistor values. We may not change K1 or K2, as only the source voltages may be changed. If we increase both sources by a factor of 10, then i1 increases by the same amount. Thus, 4 V → 40 V and 10 V → 100 V.



6.


iA, vB “on”, vC = 0: iA, vC “on”, vB = 0: iA, vB, vC “on” :

ix = 20 A ix = -5 A ix = 12 A

so, we can write

ix’ + ix” + ix”’ = 12 ix’ + ix” = 20 ix’ + ix”’ = - 5

In matrix form,

(a) (b) (c) (d)

10 March 2006

⎡12 ⎤ ⎡1 1 1 ⎤ ⎡i′x ⎤ ⎢1 1 0⎥ ⎢i′′⎥ = ⎢20⎥ ⎢ ⎥ ⎥ ⎢ x⎥ ⎢ ⎢⎣- 5 ⎥⎦ ⎢⎣1 0 1⎥⎦ ⎢⎣i′x′′⎥⎦

with iA on only, the response ix = ix’ = 3 A. with vB on only, the response ix = ix” = 17 A. with vC on only, the response ix = ix”’ = -8 A. iA and vC doubled, vB reversed: 2(3) + 2(-8) + (-1)(17) = -27 A.



7.


10 March 2006

One source at a time: The contribution from the 24-V source may be found by shorting the 45-V source and open-circuiting the 2-A source. Applying voltage division, vx’ = 24

20 20 = 24 = 10 V 10 + 20 + 45 || 30 10 + 20 + 18

We find the contribution of the 2-A source by shorting both voltage sources and applying current division: 10 ⎡ ⎤ vx” = 20 ⎢2 ⎥ = 8.333 V ⎣ 10 + 20 + 18 ⎦

Finally, the contribution from the 45-V source is found by open-circuiting the 2-A source and shorting the 24-V source. Defining v30 across the 30-Ω resistor with the “+” reference on top: 0

= v30/ 20 + v30/ (10 + 20) + (v30 – 45)/ 45

solving, v30 = 11.25 V and hence vx”’ = -11.25(20)/(10 + 20) = -7.5 V Adding the individual contributions, we find that vx = vx’ + vx” + vx”’ = 10.83 V.



8.


10 March 2006

The contribution of the 8-A source is found by shorting out the two voltage sources and employing simple current division: i3' = − 8

50 = -5A 50 + 30

The contribution of the voltage sources may be found collectively or individually. The contribution of the 100-V source is found by open-circuiting the 8-A source and shorting the 60-V source. Then, i3" =

100 = 6.25 A (50 + 30) || 60 || 30

The contribution of the 60-V source is found in a similar way as i3"' = -60/30 = -2 A. The total response is i3 = i3' + i3" + i3"' = -750 mA.



9.


10 March 2006

(a) By current division, the contribution of the 1-A source i2’ is i2’ = 1 (200)/ 250 = 800 mA. The contribution of the 100-V source is i2” = 100/ 250 = 400 mA. The contribution of the 0.5-A source is found by current division once the 1-A source is open-circuited and the voltage source is shorted. Thus, i2”’ = 0.5 (50)/ 250 = 100 mA Thus, i2 = i2’ + i2” + i2”’ = 1.3 A (b) P1A = (1) [(200)(1 – 1.3)] = 60 W P200 = (1 – 1.3)2 (200) = 18 W P100V = -(1.3)(100) = -130 W P50 = (1.3 – 0.5)2 (50) = 32 W P0.5A = (0.5) [(50)(1.3 – 0.5)] = 20 W Check: 60 + 18 + 32 + 20 = +130.



10.


10 March 2006

We find the contribution of the 4-A source by shorting out the 100-V source and analysing the resulting circuit:

'

'

' 4 = V1' / 20 + (V1' – V')/ 10

[1]

0.4 i1' = V1'/ 30 + (V' – V1')/ 10

[2]

where i1' = V1'/ 20 Simplifying & collecting terms, we obtain

30 V1' – 20 V' = 800 -7.2 V1' + 8 V' = 0

[1] [2]

Solving, we find that V' = 60 V. Proceeding to the contribution of the 60-V source, we analyse the following circuit after defining a clockwise mesh current ia flowing in the left mesh and a clockwise mesh current ib flowing in the right mesh.

"

"

"

30 ia – 60 + 30 ia – 30 ib = 0 [1] ib = -0.4 i1" = +0.4 ia [2] Solving, we find that ia = 1.25 A and so V" = 30(ia – ib) = 22.5 V. Thus, V = V' + V" = 82.5 V.



11.


10 March 2006

(a) Linearity allows us to consider this by viewing each source as being scaled by 25/ 10. This means that the response (v3) will be scaled by the same factor: 25 iA'/ 10 + 25 iB'/ 10 = 25 v3'/ 10 B

∴ v3 = 25v3'/ 10 = 25(80)/ 10 = 200 V (b)

iA' = 10 A, iB' = 25 A iA" = 10 A, iB" = 25 A iA = 20 A, iB = -10 A

→ v4' = 100 V → v4" = -50 V → v4 = ?

We can view this in a somewhat abstract form: the currents iA and iB multiply the same circuit parameters regardless of their value; the result is v4. ⎡10 25⎤ ⎡a ⎤ ⎡100 ⎤ Writing in matrix form, ⎢ ⎥⎢ ⎥ = ⎢ ⎥ , we can solve to find ⎣25 10⎦ ⎣b ⎦ ⎣- 50⎦ a = -4.286 and b = 5.714, so that 20a – 10b leads to v4 = -142.9 V



12.


10 March 2006

With the current source open-circuited and the 7-V source shorted, we are left with 100k || (22k + 4.7k) = 21.07 kΩ. Thus, V3V = 3 (21.07)/ (21.07 + 47) = 0.9286 V. In a similar fashion, we find that the contribution of the 7-V source is: V7V = 7 (31.97) / (31.97 + 26.7) = 3.814 V Finally, the contribution of the current source to the voltage V across it is: V5mA = (5×10-3) ( 47k || 100k || 26.7k) = 72.75 V. Adding, we find that V = 0.9286 + 3.814 + 72.75 = 77.49 V.



13.


10 March 2006

We must find the current through the 500-kΩ resistor using superposition, and then calculate the dissipated power. The contribution from the current source may be calculated by first noting that 1M || 2.7M || 5M = 636.8 kΩ. Then, 3 ⎛ ⎞ i60μA = 60 × 10 − 6 ⎜ ⎟ = 43.51 μA ⎝ 0.5 + 3 + 0.6368 ⎠

The contribution from the voltage source is found by first noting that 2.7M || 5M = 1.753 MΩ. The total current flowing from the voltage source (with the current source open-circuited) is –1.5/ (3.5 || 1.753 + 1) μA = -0.6919 μA. The current flowing through the 500-kΩ resistor due to the voltage source acting alone is then i1.5V = 0.6919 (1.753)/ (1.753 + 3.5) mA = 230.9 nA. The total current through the 500-kΩ resistor is then i60μA + i1.5V = 43.74 μA and the dissipated power is (43.74×10-9)2 (500×103) = 956.6 μW.



14.


10 March 2006

We first determine the contribution of the voltage source: ' ' ↑ I1'

Via mesh analysis, we write: 5 = 18000 I1' – 17000 Ix' -6 Ix' = -17000 Ix' + 39000 Ix' Solving, we find I1' = 472.1 mA and Ix' = 205.8 mA, so V' = 17×103 (I1' - Ix') = 4.527 V. We proceed to find the contribution of the current source: V"

Vx"

Via supernode: -20×10-3 = Vx"/ 22×103 + V"/ 0.9444×103 and V" – Vx" = 6Ix" or V" – Vx" = 6 Vx"/ 22×103

[1] [2]

Solving, we find that V" = -18.11 V. Thus, V = V' + V" = -13.58 V. The maximum power is V2/ 17×103 = V2/ 17 mW = 250 mW, so ′ = 83.3 V. V = (17)(250) = 65.19 = V′ - ( -18.11) . Solving, we find Vmax The 5-V source may then be increased by a factor of 83.3/ 4.527, so that its maximum positive value is 92 V; past this value, and the resistor will overheat.



15.


10 March 2006

It is impossible to identify the individual contribution of each source to the power dissipated in the resistor; superposition cannot be used for such a purpose. Simplifying the circuit, we may at least determine the total power dissipated in the resistor: i→

Via superposition in one step, we may write i = Thus,

5 2.1 - 2 = 195.1 mA 2 + 2.1 2 + 2.1

P2Ω = i2 . 2 = 76.15 mW



16.


10 March 2006

We will analyse this circuit by first considering the combined effect of both dc sources (left), and then finding the effect of the single ac source acting alone (right). V2 IB vx'

V3

V1 IB

1, 3 supernode:

V1/ 100 + V1/ 17×103 + (V1 – 15)/ 33×103 + V3/ 103 = 20 IB [1]

and:

V1 – V3 = 0.7

Node 2:

[2]

-20 IB = (V2 – 15)/ 1000

[3]

We require one additional equation if we wish to have IB as an unknown: 20 IB + IB = V3/ 1000

[4]

Simplifying and collecting terms, 10.08912 V1 + V3 – 20×103 IB = 0.4545 V1 - V3 V2 + 20×103 IB

[1]

= 0.7

[2]

= 15

[3]

-V3 + 21×103 IB = 0

[4]

Solving, we find that IB = -31.04 μA. To analyse the right-hand circuit, we first find the Thévenin equivalent to the left of the wire marked iB', noting that the 33-kΩ and 17-kΩ resistors are now in parallel. We find that VTH = 16.85 cos 6t V by voltage division, and RTH = 100 || 17k || 33k = 99.12 Ω. We may now proceed: 20 iB' = vx' / 1000 + (vx' – 16.85 cos 6t)/ 99.12 20 iB' + iB'' = vx'/ 1000 B

B

[1] [2]

Solving, we find that iB' = 798.6 cos 6t mA. Thus, adding our two results, we find the complete current is iB = iB' + IB = -31.04 + 798.6 cos 6t μA. B




10 March 2006

17.

We first consider the effect of the 2-A source separately, using the left circuit: 3 ⎤ ⎡ Vx' = 5 ⎢2 ⎥ = 1.765 V ⎣ 3 + 14 ⎦

Next we consider the effect of the 6-A source on its own using the right circuit: ⎡ 9 ⎤ Vx" = 5 ⎢6 ⎥ = 15.88 V ⎣ 9 + 8⎦

Thus, Vx = Vx' + Vx" = 17.65 V. (b) PSpice verification (DC Sweep) The DC sweep results below confirm that Vx' = 1.765 V




10 March 2006

18.

'

(a) Beginning with the circuit on the left, we find the contribution of the 2-V source to Vx: Vx′ Vx′ − 2 − 4Vx′ = + 100 50 which leads to Vx' = 9.926 mV. The circuit on the right yields the contribution of the 6-A source to Vx: − 4Vx′′ =

Vx′′ Vx′′ + 100 50

which leads to Vx" = 0. Thus, Vx = Vx' + Vx" = 9.926 mV. (b) PSpice verification. As can be seen from the two separate PSpice simulations, our hand calculations are correct; the pV-scale voltage in the second simulation is a result of numerical inaccuracy.



19.


10 March 2006

Vx − 12 Vx Vx + 15 =0 + + 1 3 2 so Vx = 2.455 V

Vx′ − 6 Vx′ Vx′ + 10 =0 + + 1 3 2 so Vx′ = 0.5455 V

Vx′′ − 6 Vx′′ Vx′′ + 5 =0 + + 1 3 2 so Vx′′ = 1.909 V Adding, we find that Vx' + Vx" = 2.455 V = Vx as promised.



20.


10 March 2006

(a) We first recognise that the two current sources are in parallel, and hence may be replaced by a single –7 A source (arrow directed downward). This source is in parallel with a 10 kΩ resistor. A simple source transformation therefore yields a 10 kΩ resistor in series with a (–7)(10,000) = –70,000 V source (+ reference on top):

(b) This circuit requires several source transformations. First, we convert the 8 V source and 3 Ω resistor to an 8/3 A current source in parallel with 3 Ω. This yields a circuit with a 3 Ω and 10 Ω parallel combination, which may be replaced with a 2.308 Ω resistor. We may now convert the 8/3 A current source and 2.308 Ω resistor to a (8/3)(2.308) = 6.155 V voltage source in series with a 2.308 Ω resistor. This modified circuit contains a series combination of 2.308 Ω and 5 Ω; performing a source transformation yet again, we obtain a current source with value (6.155)/(2.308 + 5) = 0.8422 A in parallel with 7.308 Ω and in parallel with the remaining 5 Ω resistor. Since 7.308 Ω || 5 Ω = 2.969 Ω, our solution is:



21.


10 March 2006

(a) First we note the three current sources are in parallel, and may be replaced by a single current source having value 5 – 1 + 3 = 7 A, arrow pointing upwards. This source is in parallel with the 10 Ω resistor and the 6 Ω resistor. Performing a source transformation on the current source and 6 Ω resistor, we obtain a voltage source (7)(6) = 42 V in series with a 6 Ω resistor and in series with the 10 Ω resistor:

+ v –

(b) By voltage division, v = 42(10)/16 = 26.25 V. (c) Once the 10 Ω resistor is involved in a source transformation, it disappears, only to be replaced by a resistor having the same value – but whose current and voltage can be different. Since the quantity v appearing across this resistor is of interest, we cannot involve the resistor in a transformation.



22.


10 March 2006

(a) [120 cos 400t] / 60 = 2 cos 400t A. 60 || 120 = 40 Ω. [2 cos 400t] (40) = 80 cos 400t V. 40 + 10 = 50 Ω. [80 cos 400t]/ 50 = 1.6 cos 400t A. 50 || 50 = 25 Ω.

1.6 cos 400t A

(b) 2k || 3k + 6k = 7.2 kΩ.

25 Ω

7.2k || 12k = 4.5 kΩ

3.5 kΩ 4.5 kΩ

(20)(4.5) = 90 V.

8 kΩ



23.


10 March 2006

We can ignore the 1-kΩ resistor, at least when performing a source transformation on this circuit, as the 1-mA source will pump 1 mA through whatever value resistor we place there. So, we need only combine the 1 and 2 mA sources (which are in parallel once we replace the 1-kΩ resistor with a 0-Ω resistor). The current through the 5.8kΩ resistor is then simply given by voltage division: i = 3 × 10-3

4.7 = 1.343 mA 4.7 + 5.8

The power dissipated by the 5.8-kΩ resistor is then i2 . 5.8×103 = 10.46 mW. (Note that we did not “transform” either source, but rather drew on the relevant discussion to understand why the 1-kΩ resistor could be omitted.)



24.


10 March 2006

We may ignore the 10-kΩ and 9.7-kΩ resistors, as 3-V will appear across them regardless of their value. Performing a quick source transformation on the 10-kΩ resistor/ 4-mA current source combination, we replace them with a 40-V source in series with a 10-kΩ resistor: Ω

I →

Ω

I = 43/ 15.8 mA = 2.722 mA. Therefore, P5.8Ω = I2. 5.8×103 = 42.97 mW.



25.


10 March 2006

(100 kΩ)(6 mA) = 0.6 V

0.6 V

470 k || 300 k = 183.1 kΩ (-3 – 0.6)/ 300×103 = -12 μA (183.1 kΩ)(-12 μA) = -2.197 V 183.1

-2.197 V ← I

Solving, 9 + 1183.1×103 I – 2.197 = 0, so I = -5.750 μA. Thus, P1MΩ = I2 . 106 = 33.06 μW.



26.


10 March 2006

(1)(47) = 47 V. (20)(10) = 200 V. Each voltage source “+” corresponds to its corresponding current source’s arrow head.

Using KVL on the simplified circuit above, 47 + 47×103 I1 – 4 I1 + 13.3×103 I1 + 200 = 0 Solving, we find that I1 = -247/ (60.3×103 – 4) = -4.096 mA.



27.


10 March 2006

(a) (2 V1)(17) = 34 V1

34 ←I

Analysing the simplified circuit above, 34 V1 – 0.6 + 7 I + 2 I + 17 I = 0

[1]

and V1 = 2 I

[2]

Substituting, we find that I = 0.6/ (68 + 7 + 2 + 17) = 6.383 mA. Thus, V1 = 2 I = 12.77 mV (b)

V1 is the top-labelled voltage (12.77 mV).



28.


10 March 2006

(a) 12/ 9000 = 1.333 mA. 9k || 7k = 3.938 kΩ. → (1.333 mA)(3.938 kΩ) = 5.249 V. 3.938 5.249 V

5.249/ 473.938×103 = 11.08 μA

11.08 μA

473.9 kΩ

10 kΩ

473.9 k || 10 k = 9.793 kΩ. (11.08 mA)(9.793 kΩ) = 0.1085 V

0.1085 V

11.793 kΩ 17 kΩ

Ix = 0.1085/ 28.793×103 = 3.768 μA. (b)

Ix ↓



29.


10 March 2006

First, (-7 μA)(2 MΩ) = -14 V, “+” reference down. 2 MΩ + 4 MΩ = 6 MΩ. +14 V/ 6 ΜΩ = 2.333 μA, arrow pointing up; 6 M || 10 M = 3.75 MΩ.

2.333 3.75 MΩ

(2.333)(3.75) = 8.749 V. Req = 6.75 MΩ ∴ Ix = 8.749/ (6.75 + 4.7) μA = 764.1 nA.



30.


10 March 2006

To begin, note that (1 mA)(9 Ω) = 9 mV, and 5 || 4 = 2.222 Ω. 15

9 2.222 Ω

The above circuit may not be further simplified using only source transformation techniques.



31.


10 March 2006

Label the “–” terminal of the 9-V source node x and the other terminal node x'. The 9-V source will force the voltage across these two terminals to be –9 V regardless of the value of the current source and resistor to its left. These two components may therefore be neglected from the perspective of terminals a & b. Thus, we may draw:

2Ω 7.25 A



32.


10 March 2006

Beware of the temptation to employ superposition to compute the dissipated power- it won’t work! Instead, define a current I flowing into the bottom terminal of the 1-MΩ resistor. Using superposition to compute this current, I = 1.8/ 1.840 + 0 + 0 μA = 978.3 nA. Thus,

P1MΩ = (978.3×10-9)2 (106) = 957.1 nW.



33.


10 March 2006

Let’s begin by plotting the experimental results, along with a least-squares fit to part of the data:

Least-squares fit results: Voltage (V) 1.567 1.563 1.558

Current (mA) 1.6681 6.599 12.763

We see from the figure that we cannot draw a very good line through all data points representing currents from 1 mA to 20 mA. We have therefore chosen to perform a linear fit for the three lower voltages only, as shown. Our model will not be as accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA, since that is beyond the range of the experimental data. Modeling this system as an ideal voltage source in series with a resistance (representing the internal resistance of the battery) and a varying load resistance, we may write the following two equations based on the linear fit to the data: 1.567 = Vsrc – Rs (1.6681×10-3) 1.558 = Vsrc – Rs (12.763×10-3) Solving, Vsrc = 1.568 V and Rs = 811.2 mΩ. It should be noted that depending on the line fit to the experimental data, these values can change somewhat, particularly the series resistance value.



34.


10 March 2006

Let’s begin by plotting the experimental results, along with a least-squares fit to part of the data:

Least-squares fit results: Voltage (V) 1.567 1.563 1.558

Current (mA) 1.6681 6.599 12.763

We see from the figure that we cannot draw a very good line through all data points representing currents from 1 mA to 20 mA. We have therefore chosen to perform a linear fit for the three lower voltages only, as shown. Our model will not be as accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA, since that is beyond the range of the experimental data. Modeling this system as an ideal current source in parallel with a resistance Rp (representing the internal resistance of the battery) and a varying load resistance, we may write the following two equations based on the linear fit to the data: 1.6681×10-3 = Isrc – 1.567/ Rp 12.763×10-3 = Isrc – 1.558/ Rp Solving, Isrc = 1.933 A and Rs = 811.2 mΩ. It should be noted that depending on the line fit to the experimental data, these values can change somewhat, particularly the series resistance value.



35.


10 March 2006

Working from left to right, 2 μA – 1.8 μA = 200 nA, arrow up. 1.4 MΩ + 2.7 MΩ = 4.1 MΩ A transformation to a voltage source yields (200 nA)(4.1 MΩ) = 0.82 V in series with 4.1 MΩ + 2 MΩ = 6.1 MΩ, as shown below:

Then, 0.82 V/ 6.1 MΩ = 134.4 nA, arrow up. 6.1 MΩ || 3 MΩ = 2.011 MΩ 4.1 μA + 134.4 nA = 4.234 mA, arrow up. (4.234 μA) (2.011 MΩ) = 8.515 V. The final circuit is an 8.515 V voltage source in series with a 2.011 MΩ resistor, as shown:



36.


10 March 2006

To begin, we note that the 5-V and 2-V sources are in series:

3

Next, noting that 3 V/ 1 Ω = 3 A, and 4 A – 3 A = +1 A (arrow down), we obtain: The left-hand resistor and the current source are easily transformed into a 1-V source in series with a 1-Ω resistor: By voltage division, the voltage across the 5-Ω resistor in the circuit to the right is: (-1)

2 || 5 = -0.4167 V. 2 || 5 + 2

-1 V

±

Thus, the power dissipated by the 5-Ω resistor is (-0.4167)2 / 5 = 34.73 mW.



37.


10 March 2006

(a) We may omit the 10 Ω resistor from the circuit, as it does not affect the voltage or current associated with RL since it is in parallel with the voltage source. We are thus left with an 8 V source in series with a 5 Ω resistor. These may be transformed to an 8/5 A current source in parallel with 5 Ω, in parallel with RL.

(b)

We see from simulating both circuits simultaneously that the voltage across RL is the same (4 V).



38.


10 March 2006

(a) We may begin by omitting the 7 Ω and 1 Ω resistors. Performing the indicated source transformations, we find a 6/4 A source in parallel with 4 Ω, and a 5/10 A source in parallel with 10 Ω. These are both in parallel with the series combination of the two 5 Ω resistors. Since 4 Ω || 10 Ω = 2.857 Ω, and 6/4 + 5/10 = 2 A, we may further simplify the circuit to a single current source (2 A) in parallel with 2.857 Ω and the series combination of two 5 Ω resistors. Simple current division yields the current flowing through the 5 Ω resistors: I=

2(2.857) = 0.4444 A 2.857 + 10

The power dissipated in either of the 5 Ω resistors is then I2R = 987.6 mW. (b) We note that PSpice will NOT allow the 7 Ω resistor to be left floating! For both circuits simulated, we observe 987.6 mW of power dissipated for the 5 Ω resistor, confirming our analytic solution.

(c) Neither does. No current flows through the 7 Ω resistor; the 1 Ω resistor is in parallel with a voltage source and hence cannot affect any other part of the circuit.



39.


10 March 2006

We obtain a 5v3/4 A current source in parallel with 4 Ω, and a 3 A current source in parallel with 2 Ω. We now have two dependent current sources in parallel, which may be combined to yield a single –0.75v3 current source (arrow pointing upwards) in parallel with 4 Ω. Selecting the bottom node as a reference terminal, and naming the top left node Vx and the top right node Vy, we write the following equations: –0.75v3 = Vx/4 + (Vx – Vy)/3 3 = Vy/2 + (Vy – Vx)/3 v3 = Vy – Vx Solving, we find that v3 = –2 V.



40.


10 March 2006

(a) RTH = 25 || (10 + 15) = 25 || 25 = 12.5 Ω. 25 ⎛ ⎞ ⎛ 15 + 10 ⎞ VTH = Vab = 50 ⎜ ⎟ + 100 ⎜ ⎟ = 75 V. ⎝ 10 + 15 + 25 ⎠ ⎝ 15 + 10 + 25 ⎠

(b) If Rab = 50 Ω, 2

⎡ ⎛ 50 ⎞⎤ ⎛ 1 ⎞ P50Ω = ⎢75 ⎜ ⎟⎥ ⎜ ⎟ = 72 W ⎣ ⎝ 50 + 12.5 ⎠⎦ ⎝ 50 ⎠ (c) If Rab = 12.5 Ω, 2

⎡ ⎛ 12.5 ⎞⎤ ⎛ 1 ⎞ P12.5Ω = ⎢75 ⎜ ⎟⎥ ⎜ ⎟ = 112.5 W ⎣ ⎝ 12.5 + 12.5 ⎠⎦ ⎝ 12.5 ⎠



41.


10 March 2006

(a) Shorting the 14 V source, we find that RTH = 10 || 20 + 10 = 16.67 Ω. Next, we find VTH by determining VOC (recognising that the right-most 10 Ω resistor carries no current, hence we have a simple voltage divider): 10 + 10 VTH = VOC = 14 = 9.333 V 10 + 10 + 10 Thus, our Thevenin equivalent is a 9.333 V source in series with a 16.67 Ω resistor, which is in series with the 5 Ω resistor of interest. (b) I5Ω = 9.333/ (5 + 16.67) = 0.4307 A. Thus, P5Ω = (0.4307)2 . 5 = 927.5 mW (c) We see from the PSpice simulation that keeping four significant digits in calculating the Thévenin equivalent yields at least 3 digits’ agreement in the results.



42.


10 March 2006

(a) Replacing the 7 Ω resistor with a short circuit, we find ISC = 15 (8)/ 10 = 12 A. Removing the short circuit, and open-circuiting the 15 A source, we see that RTH = 2 + 8 = 10 Ω. Thus, VTH = ISCRTH = (12)(10) = 120 V. Our Thévenin equivalent is therefore a 120 V source in series with 10 Ω. (b) As found above, IN = ISC = 12 A, and RTH = 10 Ω. (c) Using the Thévenin equivalent circuit, we may find v1 using voltage division: v1 = 120 (7)/17 = 49.41 V.

Using the Norton equivalent circuit and a combination of current division and Ohm’s law, we find ⎛ 10 ⎞ v1 = 7 ⎜ 12 ⎟ = 49.41 V ⎝ 17 ⎠

As expected, the results are equal. (d) Employing the more convenient Thévenin equivalent model, v1 = 120(1)/17 = 7.059 V.



43.


10 March 2006

(a) RTH = 10 mV/ 400 μA = 25 Ω (b) RTH = 110 V/ 363.6×10–3 A = 302.5 Ω (c) Increased current leads to increased filament temperature, which results in a higher resistance (as measured). This means the Thévenin equivalent must apply to the specific current of a particular circuit – one model is not suitable for all operating conditions (the light bulb is nonlinear).



44.


10 March 2006

(a) We begin by shorting both voltage sources, and removing the 1 Ω resistor of interest. Looking into the terminals where the 1 Ω resistor had been connected, we see that the 9 Ω resistor is shorted out, so that RTH = (5 + 10) || 10 + 10 = 16 Ω. To continue, we return to the original circuit and replace the 1 Ω resistor with a short circuit. We define three clockwise mesh currents: i1 in the left-most mesh, i2 in the top-right mesh, and isc in the bottom right mesh. Writing our three mesh equations, +3=0 – 4 + 9i1 – 9i2 =0 – 9i1 + 34i2 – 10isc –3 – 10i2 + 20isc = 0 Solving using MATLAB: >> e1 = '-4 + 9*i1 - 9*i2 + 3 = 0'; >> e2 = '-9*i1 + 34*i2 - 10*isc = 0'; >> e3 = '-3 + 20*isc - 10*i2 = 0'; >> a = solve(e1,e2,e3,'i1','i2','isc'); we find isc = 0.2125 A, so IN = 212.5 mA and VTH = INRTH = (0.2125)(16) = 3.4 V. (b) Working with the Thévenin equivalent circuit, I1Ω = VTH/(RTH + 1) = 200 mA. Thus, P1Ω = (0.2)2.1 = 40 mW. Switching to the Norton equivalent, we find I1Ω by current division: I1Ω = (0.2125)(16)/(16+1) = 200 mA. Once again, P1Ω = 40 mW (as expected). (c) As we can see from simulating the original circuit simultaneously with its Thevenin and Norton equivalents, the 1 Ω resistor does in fact dissipate 40 mW, and either equivalent is equally applicable. Note all three SOURCES provide a different amount of power in total.



45.


10 March 2006

(a) Removing terminal c, we need write only one nodal equation: Vb − 2 Vb − 5 , which may be solved to + 12 15 yield Vb = 4 V. Therefore, Vab = VTH = 2 – 4 = -2 V. RTH = 12 || 15 = 6.667 Ω. We may then calculate IN as IN = VTH/ RTH

0.1 =

= -300 mA (arrow pointing upwards). (b) Removing terminal a, we again find RTH = 6.667 Ω, and only need write a single nodal equation; in fact, it is identical to that written for the circuit above, and we once again find that Vb = 4 V. In this case, VTH = Vbc = 4 – 5 = -1 V, so IN = -1/ 6.667 = –150 mA (arrow pointing upwards).



46.


10 March 2006

(a) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking into the terminals x and x' a 50-Ω resistor in parallel with 10 Ω in parallel with (20 Ω + 40 Ω), so RTH = 50 || 10 || (20 + 40) = 7.317 Ω Using superposition to determine the voltage Vxx' across the 50-Ω resistor, we find

⎡ ⎡ ⎤ 50 || (20 + 40) ⎤ 40 Vxx' = VTH = ⎢88 ⎥ + (1)(50 || 10) ⎢ ⎥ ⎣ 10 + [50 || (20 + 40)] ⎦ ⎣ 40 + 20 + (50 || 10) ⎦ 40 ⎡ 27.27 ⎤ ⎡ ⎤ = ⎢88 + (1)(8.333) ⎢ ⎥ ⎥ = 69.27 V ⎣ 37.27 ⎦ ⎣ 40 + 20 + 8.333 ⎦

(b) Shorting out the 88-V source and open-circuiting the 1-A source, we see looking into the terminals y and y' a 40-Ω resistor in parallel with [20 Ω + (10 Ω || 50 Ω)]: RTH = 40 || [20 + (10 || 50)] = 16.59 Ω Using superposition to determine the voltage Vyy' across the 1-A source, we find 27.27 ⎤ ⎛ 40 ⎞ ⎡ Vyy' = VTH = (1)(RTH) + ⎢88 ⎟ ⎥⎜ ⎣ 10 + 27.27 ⎦ ⎝ 20 + 40 ⎠

= 59.52 V



47.


10 March 2006

(a) Select terminal b as the reference terminal, and define a nodal voltage V1 at the top of the 200-Ω resistor. Then, 0 =

V1 − 20 V1 − VTH V + + 1 40 100 200

[1]

1.5 i1 = (VTH – V1)/ 100

[2]

where i1 = V1/ 200, so Eq. [2] becomes

150 V1/ 200 + V1 - VTH = 0

[2]

Simplifying and collecting terms, these equations may be re-written as: (0.25 + 0.1 + 0.05) V1 – 0.1 VTH = 5 (1 + 15/ 20) V1 – VTH = 0

[1] [2]

Solving, we find that VTH = 38.89 V. To find RTH, we short the voltage source and inject 1 A into the port: V − Vin V1 V V1 0 = 1 [1] + + 1 100 40 200 + Vin − V1 1.5 i1 + 1 = [2] 100 Vin i1 = V1/ 200 [3] 1A

-

Ref.

Combining Eqs. [2] and [3] yields

1.75 V1 – Vin = -100

[4]

Solving Eqs. [1] & [4] then results in Vin = 177.8 V, so that RTH = Vin/ 1 A = 177.8 Ω. (b) Adding a 100-Ω load to the original circuit or our Thévenin equivalent, the voltage across the load is 100 ⎛ ⎞ 2 V100Ω = VTH ⎜ ⎟ = 14.00 V , and so P100Ω = (V100Ω) / 100 = 1.96 W. ⎝ 100 + 177.8 ⎠



48.


10 March 2006

We inject a current of 1 A into the port (arrow pointing up), select the bottom terminal as our reference terminal, and define the nodal voltage Vx across the 200-Ω resistor. Then,

1 = V1/ 100 + (V1 – Vx)/ 50 -0.1 V1 = Vx/ 200 + (Vx – V1)/ 50

[1] [2]

which may be simplified to 3 V1 – 2 Vx = 100 0 16 V1 + 5 Vx =

[1] [2]

Solving, we find that V1 = 10.64 V, so RTH = V1/ (1 A) = 10.64 Ω. Since there are no independent sources present in the original network, IN = 0.



49.


10 March 2006

With no independent sources present, VTH = 0. We decide to inject a 1-A current into the port: vx

v•f A

Ref.

Node ‘x’: Supernode: and:

0.01 vab = vx/ 200 + (vx – vf)/ 50 1 = vab/ 100 + (vf – vx) vab – vf = 0.2 vab

[1] [2] [3]

Rearranging and collecting terms, -2 vab + 5 vx – 4 vf = 0 vab – 2 vx + 2 vf = 100 - vf = 0 0.8 vab

[1] [2] [3]

Solving, we find that vab = 192.3 V, so RTH = vab/ (1 A) = 192.3 Ω.



50.


10 March 2006

We first find RTH by shorting out the voltage source and open-circuiting the current source. Looking into the terminals a & b, we see RTH = 10 || [47 + (100 || 12)] = 8.523 Ω.

Returning to the original circuit, we decide to perform nodal analysis to obtain VTH: -12×103 = (V1 – 12)/ 100×103 + V1/ 12×103 + (V1 – VTH)/ 47×103

[1]

12×103 = VTH/ 10×103 + (VTH – V1)/ 47×103

[2]

Rearranging and collecting terms, 0.1146 V1 - 0.02128 VTH = -11.88 [1] [2] -0.02128 V1 + 0.02128 VTH = 12 Solving, we find that VTH = 83.48 V.



51.


10 March 2006

(a) RTH = 4 + 2 || 2 + 10 = 15 Ω. (b) same as above: 15 Ω.



52.


10 March 2006

For Fig. 5.78a, IN = 12/ ~0 → ∞ A in parallel with ~ 0 Ω. For Fig. 5.78b, VTH = (2)(~∞) → ∞ V in series with ~∞ Ω.



53.


10 March 2006

With no independent sources present, VTH = 0. Connecting a 1-V source to the port and measuring the current that flows as a result, ←I + -

I = 0.5 Vx + 0.25 Vx = 0.5 + 0.25 = 0.75 A. RTH = 1/ I = 1.333 Ω. The Norton equivalent is 0 A in parallel with 1.333 Ω.



54.


10 March 2006

Performing nodal analysis to determine VTH, 100×10-3 = Vx/ 250 + Voc/ 7.5×103 and

[1]

Vx – Voc = 5 ix

where ix = Vx/ 250. Thus, we may write the second equation as 0.98 Vx – Voc = 0

[2]

Solving, we find that Voc = VTH = 23.72 V. In order to determine RTH, we inject 1 A into the port:

Vab/ 7.5×103 + Vx/ 250

=1

[1]

and Vx – Vab = 5 ix = 5Vx/ 250

or

-Vab + (1 – 5/ 250) Vx = 0

[2]

Solving, we find that Vab = 237.2 V. Since RTH = Vab/ (1 A), RTH = 237.2 Ω. Finally, IN = VTH/ RTH = 100 mA.



55.


10 March 2006

We first note that VTH = Vx, so performing nodal analysis, -5 Vx = Vx/ 19 Thus, VTH (and hence IN) = 0.

which has the solution Vx = 0 V. (Assuming RTH ≠ 0)

To find RTH, we inject 1 A into the port, noting that RTH = Vx/ 1 A: -5 Vx + 1 = Vx/ 19 Solving, we find that Vx = 197.9 mV, so that RTH = RN = 197.9 mΩ.



56.


10 March 2006

Shorting out the voltage source, we redraw the circuit with a 1-A source in place of the 2-kΩ resistor: Ref.

1A

V

Noting that 300 Ω || 2 MΩ ≈ 300 Ω, 0 = (vgs – V)/ 300

[1]

1 – 0.02 vgs = V/ 1000 + (V – vgs)/ 300

[2]

Simplifying & collecting terms, –V = 0

[1]

0.01667 vgs + 0.00433 V = 1

[2]

vgs

Solving, we find that vgs = V = 47.62 V. Hence, RTH = V/ 1 A = 47.62 Ω.



57.


10 March 2006

We replace the source vs and the 300-Ω resistor with a 1-A source and seek its voltage: V2

V1 + 1A

V Ref.

By nodal analysis,

1 = V1/ 2×106

so V1 = 2×106 V.

Since V = V1, we have Rin = V/ 1 A = 2 MΩ.



58.


10 March 2006

Removing the voltage source and the 300-Ω resistor, we replace them with a 1-A source and seek the voltage that develops across its terminals: + V -

1A Ref.

We select the bottom node as our reference terminal, and define nodal voltages V1 and V2. Then, 1 = V1 / 2×106 + (V1 – V2)/ rπ

[1]

0.02 vπ = (V2 – V1)/ rπ + V2/ 1000 + V2/ 2000

[2]

where vπ = V1 – V2 Simplifying & collecting terms, (2×106 + rπ) V1 – 2×106 V2 = 2×106 rπ

[1]

-(2000 + 40 rπ) V1 + (2000 + 43 rπ) V2 = 0 [2]

⎛ 666.7 + 14.33 rπ Solving, we find that V1 = V = 2 × 106 ⎜⎜ 6 ⎝ 2 × 10 + 666.7 + 14.33 rπ Thus,

⎞ ⎟⎟ . ⎠

RTH = 2×106 || (666.7 + 14.33 rπ) Ω.



59.


10 March 2006

(a) We first determine vout in terms of vin and the resistor values only; in this case, VTH = vout. Performing nodal analysis, we write two equations: 0=

−vd ( −vd − vin ) ( −vd − vo ) + + Ri R1 Rf

[1]

0=

and

( vo + vd ) + ( vo − Avd ) Rf

Ro

[2]

Solving using MATLAB, we obtain: >> e1 = 'vd/Ri + (vd + vin)/R1 + (vd + vo)/Rf = 0'; >> e2 = '(vo + vd)/Rf + (vo – A*vd)/Ro = 0'; >> a = solve(e1,e2,'vo','vd'); >> pretty(a.vo) Ri vin (–Ro + Rf A) – ----------------------------------------------R1 Ro + Ri Ro + R1 Rf + Ri Rf + R1 Ri + A R1 Ri Thus, VTH =

vin Ri ( Ro − AR f )

R1 Ro + Ri Ro + R1 R f + Ri R f + R1 Ri + AR1 Ri approaches –Rf/R1.

, which in the limit of A → ∞,

To find RTH, we short out the independent source vin, and squirt 1 A into the terminal marked vout, renamed VT. Analyzing the resulting circuit, we write two nodal equations: 0=

−vd vd ( −vd − vT ) − + Ri R1 Rf

[1]

and

1=

( vT + vd ) + ( vT − Avd ) Rf

Ro

[2]

Solving using MATLAB: >> e1 = 'vd/R1 + vd/Ri + (vd + VT)/Rf = 0'; >> e2 = '1 = (VT + vd)/Rf + (VT - A*vd)/Ro'; >> a = solve(e1,e2,'vd','VT'); >> pretty(a.VT) Ro (Ri Rf + R1 Rf + R1 Ri) -------------------------------------------------------------Ri Ro + R1 Ro + Ri Rf + R1 Rf + R1 Ri + A R1 Ri Since VT/1 = VT, this is our Thévenin equivalent resistance (RTH).

+ vout –



60.


10 March 2006

Such a scheme probably would lead to maximum or at least near-maximum power transfer to our home. Since we pay the utility company based on the power we use, however, this might not be such a hot idea…



61.


10 March 2006

We need to find the Thévenin equivalent resistance of the circuit connected to RL, so we short the 20-V source and open-circuit the 2-A source; by inspection, then RTH = 12 || 8 + 5 + 6 = 15.8 Ω Analyzing the original circuit to obtain V1 and V2 with RL removed: V2

V1

+ VTH -

Ref.

V1 = 20 8/ 20 = 8 V;

V2 = -2 (6) = -12 V.

We define VTH = V1 – V2 = 8 + 12 = 20 V. Then, PRL|max

2 VTH 400 = = = 6.329 W 4 RL 4(15.8)



62.


10 March 2006

(a) RTH = 25 || (10 + 15) = 12.5 Ω Using superposition, Vab = VTH = 50

25 15 + 10 = 75 V. + 100 15 + 10 + 25 50

(b) Connecting a 50-Ω resistor, Pload

2 VTH 752 = = = 90 W R TH + R load 12.5 + 50

(c) Connecting a 12.5-Ω resistor, Pload =

2 VTH 752 = = 112.5 W 4 R TH 4 (12.5)



63.


10 March 2006

(a) By inspection, we see that i10 = 5 A, so VTH = Vab = 2(0) + 3 i10 + 10 i10 = 13 i10 = 13(5) = 65 V. To find RTH, we open-circuit the 5-A source, and connect a 1-A source between terminals a & b:

+ 1A

Vx –

A simple KVL equation yields Vx = 2(1) + 3i10 + 10 i10. Since i10 = 1 A in this circuit, Vx = 15 V. We thus find the Thevenin equivalent resistance is 15/1 = 15 Ω.

(b) Pmax =

2 VTH 4 R TH

=

652 = 70.42 W 4(15)



64. (a)


10 March 2006

Replacing the resistor RL with a 1-A source, we seek the voltage that develops across its terminals with the independent voltage source shorted:

−10i1 + 20ix + 40i1 = 0

[1] ⇒ 30i1 + 20ix = 0 [1]

and i1 − ix = 1

[2] ⇒

i1 − ix = 1

[2]

Solving, i1 = 400 mA So V = 40i1 = 16 V and RTH =

(b)

V = 16 Ω 1A

Removing the resistor RL from the original circuit, we seek the resulting open-circuit voltage: + VTH -

VTH − 10i1 VTH − 50 + [1] 20 40 V − 50 where i1 = TH 40 1 ⎛ V − 50 ⎞ ⎛ VTH − 50 ⎞ V so [1] becomes 0 = TH − ⎜ TH ⎟+⎜ ⎟ 20 2 ⎝ 40 ⎠ ⎝ 40 ⎠ V V − 50 0 = TH + TH 20 80 0 = 4VTH + VTH − 50

0=

5VTH = 50 or VTH = 10 V Thus, if

RL = RTH = 16 Ω, VRL = VTH

RL V = TH = 5 V RL + RTH 2




10 March 2006

65. (a)

I N = 2.5 A ↓ 2A RN

20 Ω

20i 2 = 80 i = 2A

By current division, RN RN + 20 Solving, RN = RTH = 80 Ω

2 = 2.5

Thus, VTH = VOC = 2.5 × 80 = 200 V 2

VTH 2002 = = 125 W 4 RTH 4 × 80

(b)

Pmax =

(c)

RL = RTH = 80 Ω




10 March 2006

66.

10 W to 250Ω corresp to 200 mA. IN

By Voltage ÷, So

RN

IR = IN

R

20 W to 80Ω corresp to 500 mA.

RN R + RN

0.2 = I N

RN 250 + RN

[1]

0.5 = I N

RN 80 + RN

[2]

Solving, I N = 1.7 A and RN = 33.33 Ω (a)

If vLiL is a maximum, RL = RN = 33.33 Ω 33.33 = 850 mA 33.33 + 33.33 vL = 33.33iL = 28.33V

iL = 1.7 ×

(b)

If vL is a maximum VL = I N ( RN RL ) So vL is a maximum when RN RL is a maximum, which occurs at RL = ∞. Then iL = 0 and vL = 1.7 × RN = 56.66 V

(c)

If iL is a maximum iL = iN

RN ; max when RL = 0 Ω RN + RL

So iL = 1.7A vL = 0 V



67.


10 March 2006

There is no conflict with our derivation concerning maximum power. While a dead short across the battery terminals will indeed result in maximum current draw from the battery, and power is indeed proportional to i2, the power delivered to the load is i2RLOAD = i2(0) = 0 watts. This is the minimum, not the maximum, power that the battery can deliver to a load.



68.


10 March 2006

Remove RE: RTH = RE Rin V − vπ V − vπ + [1] 300 70 × 103 vπ v −V v −V [2] + π + π 0= 3 10 ×10 300 70 × 103

bottom node: 1 − 3 × 10−3 vπ = at other node:

1A

V

Simplifying and collecting terms, 210×105 = 70×103 V + 300V + 63000 vπ – 70x103 vπ - 300 vπ or 70.3 × 103V − 7300vπ = 210 ×105 [1] 0 = 2100vπ + 70 × 103 vπ − 70 × 103V + 300vπ − 300V or − 69.7 × 103V + 72.4 × 103 vπ = 0

[2]

solving, V = 331.9V So RTH = RE 331.9Ω Next, we determine vTH using mesh analysis:

and:

−vs + 70.3 × 103 i1 − 70 × 103 i2 = 0

[1]

80 × 103 i2 − 70 × 103 i1 + RE i3 = 0

[2]

−3

i3 − i2 = 3 × 10 vπ −3

[3]

or i3 − i2 = 3 × 10 (10 × 10 )i2 or i3 − i2 = 30i2 or −31i2 + i3 = 0 Solving :

⎡ 70.3 × 103 ⎢ 3 ⎢ −70 ×10 ⎢ 0 ⎣

3

−70 × 103 80 × 103 −31

[3] 0 ⎤ ⎡ i1 ⎤ ⎡vs ⎤ ⎥ RE ⎥ ⎢⎢i2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ 1 ⎥⎦ ⎢⎣ i3 ⎥⎦ ⎢⎣ 0 ⎥⎦

We seek i3: −21.7 × 103 vs i3 = 7.24 ×106 + 21.79 × 103 RE So VOC = VTH

−21.7 × 103 RE = RE i3 = vs 7.24 ×106 + 21.79 ×103 RE 2

2

⎡ V ⎤ ⎡ ⎤ 8 vs 2 −21.7 ×103 RE P8Ω = 8 ⎢ TH ⎥ = ⎢ ⎥ 2 6 3 ⎣ RTH + 8 ⎦ ⎣ 7.24 ×10 + 21.79 × 10 RE ⎦ ⎡ 331.9 RE ⎤ ⎢ 331.9 + R ⎥ ⎣ E ⎦ 11.35 ×106 (331.9 + RE )2 = vs 2 6 3 2 (7.24 ×10 + 21.79 × 10 RE ) This is maximized by setting RE = ∞.



69.


10 March 2006

Thévenize the left-hand network, assigning the nodal voltage Vx at the free end of right-most 1-kΩ resistor. Vx oc A single nodal equation: 40 × 10−3 = 7 × 103 So VTH = Vx oc = 280 V RTH = 1 k + 7 k = 8 kΩ Select R1 = RTH = 8 kΩ.




10 March 2006

100 kΩ

70.

1 MΩ

850 MΩ

D = RA + RB + RC = 1 + 850 + 0.1 = 851.1× 106 RA RB 106 ×105 = = 117.5 Ω D D R R 105 × 850 × 106 = 99.87 k Ω R2 = B C = D 851.1×106 RC RA 850 ×106 ×106 = = 998.7 k Ω R3 = D 851.1×106 R1 =




10 March 2006

71. R1

R2

R3

N = R1 R2 + R2 R3 + R3 R1 = 0.1× 0.4 + 0.4 × 0.9 + 0.9 × 0.1 = 0.49Ω N = 1.225 Ω RA = R2 RB =

N = 544.4 mΩ R3

RC =

N = 4.9 Ω R1




10 March 2006

72. Δ1 : 1 + 6 + 3 = 10 Ω 6 ×1 6×3 3 ×1 = 0.6, = 1.8 , = 0.3 10 10 10 Δ 2 : 5 + 1 + 4 = 10 Ω 5 ×1 1× 4 5× 4 = 0.5 , = 0.4 , =2 10 10 10 1.8 + 2 + 0.5 = 4.3 Ω 0.3 + 0.6 + 0.4 = 1.3 Ω 1.3 4.3 = 0.9982 Ω 0.9982 + 0.6 + 2 = 3.598 Ω 3.598 6 = 2.249 Ω




10 March 2006

73. 6 × 2 + 2 × 3 + 3 × 6 = 36 Ω 2 36 36 36 = 6Ω, = 18 Ω , = 12 Ω 6 2 3 12 4 = 3 Ω, 6 12 Ω = 4 Ω 4 + 3 + 18 = 25 Ω 18 3 × = 2.16 Ω 25 18 4 × = 2.88 Ω 25 3 4 × = 0.48 Ω 25 9.48 × 2.16 + 9.48 × 2.88 + 2.88 × 2.16 = 54 Ω 2 54 54 = 18.75 Ω = 5.696 Ω 2.88 9.48 54 = 25 Ω 2.16 75 18.75 = 15 Ω

96

100 25 = 20 Ω (15 + 20) 5.696 = 4.899 Ω ∴ Rin = 5 + 4.899 = 9.899 Ω



74.


10 March 2006

We begin by converting the Δ-connected network consisting of the 4-, 6-, and 3-Ω resistors to an equivalent Y-connected network:

RA

RB RC

D = 6 + 4 + 3 = 13 Ω R R 6× 4 = 1.846 Ω R1 = A B = D 13 R R 4×3 = 923.1mΩ R2 = B C = D 13 R R 3× 6 = 1.385 Ω R3 = C A = D 13 Then network becomes: 1.846 Ω

0.9231 Ω

1.385 Ω

Then we may write Rin = 12 [13.846 + (19.385 6.9231)]

= 7.347Ω




75.

10 March 2006

1+1+ 2 = 4 Ω 1× 2 1 R1 = = Ω 4 2 2 ×1 1 = Ω R2 = 4 2 1× 1 = 0.25 Ω R3 = 4 0.5

0.25 0.5

Next, we convert the Y-connected network on the left to a Δ-connected network: 1× 0.5 + 0.5 × 2 + 2 ×1 = 3.5 Ω 2 3.5 = 7Ω 0.5 3.5 = 1.75 Ω RB = 2 3.5 = 3.5 Ω RC = 1 RA =

After this procedure, we have a 3.5-Ω resistor in parallel with the 2.5-Ω resistor. Replacing them with a 1.458-Ω resistor, we may redraw the circuit: 1.75 7

0.25 1.458

This circuit may be easily analysed to find:

12 × 1.458 = 5.454 V 1.75 + 1.458 = 0.25 + 1.458 1.75

Voc = RTH

= 1.045 Ω



76.


10 March 2006

We begin by converting the Y-network to a Δ-connected network: N = 1.1 + 1.1 + 1.1 = 3 Ω 2 3 RA = = 3 Ω 1 3 RB = = 3 Ω 1 3 RC = = 3 Ω 1

3Ω

3

3

Next, we note that 1 3 = 0.75 Ω , and hence have a simple Δ-network. This is easily converted to a Y-connected network: 0.75 + 3 + 3 = 6.75 Ω 0.75 × 3 = 0.3333 Ω 6.75 3× 3 = 1.333 Ω R2 = 6.75 3 × 0.75 = 0.3333 Ω R3 = 6.75 R1 =

Analysing this final circuit,

1A

RN = 1.333 + 0.3333 = 1.667Ω 1/ 3 1/ 3 + 1 + 1/ 3 1 1 = = 1+ 3 +1 5 = 0.2 A = 200 mA

I N = I SC = 1×



77.


10 March 2006

Since 1 V appears across the resistor associated with I1, we know that I1 = 1 V/ 10 Ω = 100 mA. From the perspective of the open terminals, the 10-Ω resistor in parallel with the voltage source has no influence if we replace the “dependent” source with a fixed 0.5-A source:

0.5 A

Then, we may write: so that ia = 200 mA.

-1 + (10 + 10 + 10) ia – 10 (0.5) = 0

We next find that VTH = Vab = 10(-0.5) + 10(ia – 0.5) + 10(-0.5) = -13 V. To determine RTH, we first recognise that with the 1-V source shorted, I1 = 0 and hence the dependent current source is dead. Thus, we may write RTH from inspection: RTH = 10 +

10 +

10 || 20 = 26.67 Ω.



78.


10 March 2006

(a) We begin by splitting the 1-kΩ resistor into two 500-Ω resistors in series. We then have two related Y-connected networks, each with a 500-Ω resistor as a leg. Converting those networks into Δ-connected networks, Σ = (17)(10) + (1)(4) + (4)(17) = 89×106 Ω2 89/0.5 = 178 kΩ;

89/ 17 = 5.236 kΩ;

89/4 = 22.25 kΩ

Following this conversion, we find that we have two 5.235 kΩ resistors in parallel, and a 178-kΩ resistor in parallel with the 4-kΩ resistor. Noting that 5.235 k || 5.235 k = 2.618 kΩ and 178 k || 4 k = 3.912 kΩ, we may draw the circuit as:

178 kΩ

27.49 kΩ

27.49 kΩ

3.912 kΩ

233.6 kΩ

We next attack the Y-connected network in the center: Σ = (22.25)(22.25) + (22.25)(2.618) + (2.618)(22.25) = 611.6×106 Ω2 611.6/ 22.25 = 27.49 kΩ; 611.6/ 2.618 = 233.6 kΩ Noting that 178 k || 27.49 k = 23.81 kΩ and 27.49 || 3.912 = 3.425 kΩ, we are left with a simple Δ-connected network. To convert this to the requested Y-network, Σ = 23.81 + 233.6 + 3.425 = 260.8 kΩ (23.81)(233.6)/ 260.8 = 21.33 kΩ (233.6)(3.425)/ 260.8 = 3.068 kΩ (3.425)(23.81)/ 260.8 = 312.6 Ω

312.6 Ω

21.33 kΩ

3.068 kΩ

(b)



79.


10 March 2006

(a) Although this network may be simplified, it is not possible to replace it with a three-resistor equivalent. (b) See (a).



80.


10 March 2006

First, replace network to left of the 0.7-V source with its Thévenin equivalent: 15 = 2.609 V 100 + 15 = 100k 15k = 13.04 k Ω

VTH = 20 × RTH

Redraw: 13.04 kΩ 2.609 V

Analysing the new circuit to find IB, we note that IC = 250 IB: −2.609 + 13.04 × 103 I B + 0.7 + 5000( I B + 250 I B ) = 0 2.609 − 0.7 = 1.505 μA 13.04 × 103 + 251× 5000 I C = 250 I B = 3.764 × 10−4 A IB =

= 376.4 μA



81.


10 March 2006

(a) Define a nodal voltage V1 at the top of the current source IS, and a nodal voltage V2 at the top of the load resistor RL. Since the load resistor can safely dissipate 1 W, and we know that V22 PRL = 1000 then V2 max = 31.62 V . This corresponds to a load resistor (and hence lamp) current of 32.62 mA, so we may treat the lamp as a 10.6-Ω resistor. Proceeding with nodal analysis, we may write: IS = V1/ 200 + (V1 – 5 Vx)/ 200

[1]

0 = V2/ 1000 + (V2 – 5 Vx)/ 10.6

[2]

Vx = V1 – 5 Vx

[3]

or Vx = V1/ 6

Substituting Eq. [3] into Eqs. [1] and [2], we find that 7 V1 = 1200 IS -5000 V1 + 6063.6 V2 = 0

[1] [2]

Substituting V2 max = 31.62 V into Eq. [2] then yields V1 = 38.35 V, so that IS| max = (7)(38.35)/ 1200 = 223.7 mA. (b)

PSpice verification.

The lamp current does not exceed 36 mA in the range of operation allowed (i.e. a load power of < 1 W.) The simulation result shows that the load will dissipate slightly more than 1 W for a source current magnitude of 224 mA, as predicted by hand analysis.



82.


10 March 2006

Short out all but the source operating at 104 rad/s, and define three clockwise mesh currents i1, i2, and i3 starting with the left-most mesh. Then 608 i1 – 300 i2 = 3.5 cos 104 t -300 i1 + 316 i2 – 8 i3 = 0 -8 i2 + 322 i3 = 0

[1] [2] [3]

Solving, we find that i1(t) = 10.84 cos 104 t mA i2(t) = 10.29 cos 104 t mA i3(t) = 255.7 cos 104 t μA Next, short out all but the 7 sin 200t V source, and and define three clockwise mesh currents ia, ib, and ic starting with the left-most mesh. Then 608 ia – 300 ib = -7 sin 200t [1] -300 ia + 316 ib – 8 ic = 7 sin 200t [2] [3] -8 ib + 322 ic = 0 Solving, we find that ia(t) = -1.084 sin 200t mA ib(t) = 21.14 sin 200t mA ic(t) = 525.1 sin 200t μA Next, short out all but the source operating at 103 rad/s, and define three clockwise mesh currents iA, iB, and iC starting with the left-most mesh. Then 608 iA – 300 iB = 0 -300 iA + 316 iB – 8 iC = 0 -8 iB + 322 iC = -8 cos 104 t B

B

[1] [2] [3]

Solving, we find that iA(t) = -584.5 cos 103 t μA iB(t) = -1.185 cos 103 t mA iC(t) = -24.87 cos 103 t mA B

We may now compute the power delivered to each of the three 8-Ω speakers: p1 = 8[i1 + ia + iA]2 = 8[10.84×10-3 cos 104 t -1.084×10-3 sin 200t -584.5×10-6 cos 103 t]2 p2 = 8[i2 + ib + iB]2 = 8[10.29×10-3 cos 104 t +21.14×10-3 sin 200t –1.185×10-3 cos 103 t]2 p3 = 8[i3 + ic + iC]2 = 8[255.7×10-6 cos 104 t +525.1×10-6 sin 200t –24.87×10-3 cos 103 t]2



83.


10 March 2006

Replacing the DMM with a possible Norton equivalent (a 1-MΩ resistor in parallel with a 1-A source):

1A

Vin

We begin by noting that 33 Ω || 1 MΩ ≈ 33 Ω. Then, and

0 = (V1 – Vin)/ 33 + V1/ 275×103

[1]

1 - 0.7 V1 = Vin/ 106 + Vin/ 33×103 + (Vin – V1)/ 33

[2]

Simplifying and collecting terms, (275×103 + 33) V1 - 275×103 Vin = 0 22.1 V1 + 1.001 Vin = 33

[1] [2]

Solving, we find that Vin = 1.429 V; in other words, the DMM sees 1.429 V across its terminals in response to the known current of 1 A it’s supplying. It therefore thinks that it is connected to a resistance of 1.429 Ω.



84.


10 March 2006

We know that the resistor R is absorbing maximum power. We might be tempted to say that the resistance of the cylinder is therefore 10 Ω, but this is wrong: The larger we make the cylinder resistance, the small the power delivery to R:

⎡ ⎤ 120 PR = 10 i = 10 ⎢ ⎥ ⎢⎣ Rcylinder + 10 ⎥⎦

2

2

Thus, if we are in fact delivering the maximum possible power to the resistor from the 120-V source, the resistance of the cylinder must be zero. This corresponds to a temperature of absolute zero using the equation given.



85.


10 March 2006

We note that the buzzer draws 15 mA at 6 V, so that it may be modeled as a 400-Ω resistor. One possible solution of many, then, is:

Note: construct the 18-V source from 12 1.5-V batteries in series, and the two 400-Ω resistors can be fabricated by soldering 400 1-Ω resistors in series, although there’s probably a much better alternative…



86.


10 March 2006

To solve this problem, we need to assume that “45 W” is a designation that applies when 120 Vac is applied directly to a particular lamp. This corresponds to a current draw of 375 mA, or a light bulb resistance of 120/ 0.375 = 320 Ω.

3

New wiring scheme

Original wiring scheme

In the original wiring scheme, Lamps 1 & 2 draw (40)2 / 320 = 5 W of power each, and Lamp 3 draws (80)2 / 320 = 20 W of power. Therefore, none of the lamps is running at its maximum rating of 45 W. We require a circuit which will deliver the same intensity after the lamps are reconnected in a Δ configuration. Thus, we need a total of 30 W from the new network of lamps. There are several ways to accomplish this, but the simplest may be to just use one 120-Vac source connected to the left port in series with a resistor whose value is chosen to obtain 30 W delivered to the three lamps.

In other words, 2

213.3 ⎤ ⎡ ⎢120 Rs + 213.3 ⎥ ⎣ ⎦ +2 320

213.3 ⎤ ⎡ ⎢60 Rs + 213.3 ⎥ ⎣ ⎦ 320

2

= 30

Solving, we find that we require Rs = 106.65 Ω, as confirmed by the PSpice simulation below, which shows that both wiring configurations lead to one lamp with 80-V across it, and two lamps with 40 V across each.




10 March 2006

87. • • • •

Maximum current rating for the LED is 35 mA. Its resistance can vary between 47 and 117 Ω. A 9-V battery must be used as a power source. Only standard resistance values may be used.

One possible current-limiting scheme is to connect a 9-V battery in series with a resistor Rlimiting and in series with the LED. From KVL, 9 ILED = R limiting + R LED The maximum value of this current will occur at the minimum LED resistance, 47 Ω. Thus, we solve 9 35×10-3 = R limiting + 47 to obtain Rlimiting ≥ 210.1 Ω to ensure an LED current of less than 35 mA. This is not a standard resistor value, however, so we select Rlimiting = 220 Ω.



1.

Chapter Six Solutions

This is an inverting amplifier, therefore, Vout = −

Rf R1

10 March 2006

Vin

So:

100 × 3 = −30V 10 1M =− × 2.5 = −2.5V 1M 4.7 =− × −1 = 1.42V 3.3

a) Vout = −

b) Vout

c) Vout



2.


10 March 2006

This is also an inverting amplifier. The loading resistance Rs only affects the output current drawn from the op-amp. Therefore, 47 × 1.5 = −7.05V 10 = 9V = −680mV

a) Vout = − b) Vout c) Vout




For this inverting amplifier, vout = −

3.

10 March 2006

10k × vin = −10vin . Therefore, 1k

a) vout = −10vin = −20 sin 5t 2π / 5

4π /5

6π / 5

8π / 5

2π

25 20 15 10 5 0 -5 -10 -15 -20 -25

b) vout = −10vin = −10 − 5 sin 5t 0

-2

-4

-5 V -6

-8

-10 V

-10

-12

-14

-15 V

-16

2π / 5

4π /5

6π / 5

8π / 5

2π




For this inverting amplifier, v out = −

4.

Rf R1

10 March 2006

vin = −0.1vin , hence,

a) vout = −0.1vin = − cos 4t 1.5

π/2

π

3π / 2

2π

1

0.5

0

-0.5

-1

-1.5

b) vout = −0.1vin = −1.5 − 0.4 cos 4t 0 -0.2 -0.4 -0.6 -0.8 -1

-1.1 V

-1.2 -1.4

-1.5 V

-1.6 -1.8

-1.9 V

-2

π/2

π

3π / 2

2π



5.


10 March 2006

One possible solution is by using an inverting amplifier design, we have

⇔

V out = −

Rf V in R in

Rf

V out 9 = V in 5

R in

=−

Using standard resistor values, we can have Rf=9.1kΩ and Rin=5.1kΩ



6.


10 March 2006

One possible solution is by using an inverting amplifier design, and a -5V input to give a positive output voltage:

The resistance values are given by: R f 20 = Rin 5 Giving possible resistor values Rf = 20 kΩ and Rin = 5.1 kΩ



7.


10 March 2006

To get a positive output that is smaller than the input, the easiest way is to use inverting amplifier with an inverted voltage supply to give a negative voltage:

The resistances are given by: R f 1.5 = Rin 5 Giving possible resistor values Rf = 1.5 kΩ and Rin = 5.1 kΩ



8.


10 March 2006

Similar to question 7, the following is proposed:

The resistances are given by: Rf 3 = Rin 9 Giving possible resistor values Rf = 3.0kΩ and Rin = 9.1 kΩ



9.


This circuit is a non-inverting amplifier, therefore, Vout = (1 +

10 March 2006

Rf R1

)Vin

So: 47 ) × 300m = 1.71 V 10 1M = (1 + ) × 1.5 = 3 V 1M 4.7 = (1 + ) × −1 = −2.42 V 3.3

a) Vout = (1 + b) Vout c) Vout



10.


10 March 2006

This is again a non-inverting amplifier. Similar to question 9, we have:

a) Vout = 200m × (1 + 4.7) = 1.14 V b) Vout = (1 + 1) × 9 = −18 V c) Vout = 7.8 × 100m = 0.78 V




10 March 2006

1 vout = (1 + )vin = 2vin for this non inverting amplifier circuit, therefore: 1 a) vout = 2vin = 8 sin 10t

11.

10

2π / 5

4π /5

6π / 5

8π / 5

2π

8 6 4 2 0 -2 -4 -6 -8 -10

b) vout = 2vin = 2 + 0.5 sin 10t 2.6

2.5 V 2.4

2.2

2V

2

1.8

1.6

1.5 V

1.4

1.2

1

2π / 5

4π /5

6π / 5

8π / 5

2π



12.

vout = (1 +

Rf Rin


10 March 2006

)vin = 1.5vin for this non inverting op-amp circuit. Hence,

a) vout = 1.5vin = 3 cos 2t 4

3

2

1

0

-1

-2

-3

π/2

2π

-4

b) vout = 1.5vin = 6 + 1.5 cos 2t 8

7.5 V

7

6V

6

5

5.5 V 4

3

2

1

0

π/2

2π



13.


10 March 2006

The first step is to perform a simple source transformation, so that a 0.15-V source in series with a 150-Ω resistor is connected to the inverting pin of the ideal op amp. Then, vout = −

2200 (0.15) = - 2.2 V 150



14.


10 March 2006

In order to deliver 150 mW to the 10-kΩ resistor, we need vout = (0.15)(10 × 103 ) = 38.73 V. Writing a nodal equation at the inverting input, we find 5 5 − vout 0 = + R 1000 Using vout = 38.73, we find that R = 148.2 Ω.



15.


10 March 2006

Since the 670-Ω switch requires 100 mA to activate, the voltage delivered to it by our op amp circuit must be (670)(0.1) = 67 V. The microphone acts as the input to the circuit, and provides 0.5 V. Thus, an amplifier circuit having a gain = 67/0.5 = 134 is required. One possible solution of many: a non-inverting op amp circuit with the microphone connected to the non-inverting input terminal, the switch connected between the op amp output pin and ground, a feedback resistor Rf = 133 Ω, and a resistor R1 = 1 Ω.



16.


10 March 2006

We begin by labeling the nodal voltages v- and v+ at the inverting and non-inverting input terminals, respectively. Since no current can flow into the non-inverting input, no current flows through the 40-kΩ resistor; hence, v+ = 0. Therefore, we know that v- = 0 as well. Writing a single nodal equation at the non-inverting input then leads to 0 =

(v- - vS ) (v - v ) + - out 100 22000

0 =

- vS - vout + 100 22000

or

Solving,

vout = -220 vS



17.


10 March 2006

We first label the nodal voltage at the output pin Vo. Then, writing a single nodal equation at the inverting input terminal of the op amp, 0 =

4 - 3 4 - Vo + 1000 17000

Solving, we find that Vo = 21 V. Since no current can flow through the 300-kΩ resistor, V1 = 21 as well.



18.


10 March 2006

A source transformation and some series combinations are well worthwhile prior to launching into the analysis. With 5 kΩ || 3 kΩ = 1.875 kΩ and (1 mA)(1.875 kΩ) = 1.875 V, we may redraw the circuit as Ω

Ω

V2

This is now a simple inverting amplifier with gain – Rf/ R1 = -75.33/ 1.975 = -38.14. Thus,

V2 = -38.14(3.975) = -151.6 V.



19.


10 March 2006

This is a simple inverting amplifier, so we may write

vout =

- 2000 (2 + 2 sin 3t ) = - 4(1 + sin 3t ) V 1000

vout(t = 3 s) = -5.648 V.



20.


10 March 2006

We first combine the 2 MΩ and 700 kΩ resistors into a 518.5 kΩ resistor. We are left with a simple non-inverting amplifier having a gain of 1 + 518.5/ 250 = 3.074. Thus,

vout = (3.074) vin = 18 so

vin = 5.856 V.



21.


10 March 2006

This is a simple non-inverting amplifier circuit, and so it has a gain of 1 + Rf/ R1. We want vout = 23.7 cos 500t V when the input is 0.1 cos 500t V, so a gain of 23.7/0.1 = 237 is required. One possible solution of many: Rf = 236 kΩ and R1 = 1 kΩ.



22.


10 March 2006

Define a nodal voltage V- at the inverting input, and a nodal voltage V+ at the noninverting input. Then, V+ At the non-inverting input: -3×10-6 = [1] 1.5 × 106 Thus, V+ = -4.5 V, and we therefore also know that V- = -4.5 V. At the inverting input:

0 =

VV -V + - out R6 R7

[2]

Solving and making use of the fact that V- = -4.5 V,

vout = −

⎞ ⎛ R7 (4.5) - 4.5 = - 4.5 ⎜⎜ R 7 + 1⎟⎟ V R6 ⎠ ⎝ R6



23.


10 March 2006

(a) B must be the non-inverting input: that yields a gain of 1 + 70/10 = 8 and an output of 8 V for a 1-V input. (b) R1 = ∞, RA = 0. We need a gain of 20/10 = 2, so choose R2 = RB = 1 Ω. (c) A is the inverting input since it has the feedback connection to the output pin.



24.


10 March 2006

It is probably best to first perform a simple source transformation: (1 mA)(2 kΩ) = 2 V. 13 kΩ V-

1 kΩ

3V

2 kΩ

V+

vout

2V

Since no current can flow into the non-inverting input pin, we know that V+ = 2 V, and therefore also that V- = 2 V. A single nodal equation at the inverting input yields: 0 =

2 - 3 2 - vout + 1000 13000

which yields vout = -11 V.



25.


10 March 2006

We begin by find the Thévenin equivalent to the left of the op amp: Vth = -3.3(3) vπ = -9.9 vπ = − 9.9

1000 vS = -9 vS 1100

Rth = 3.3 kΩ, so we can redraw the circuit as: 100 kΩ

3.3 kΩ

vout

-9 vS

which is simply a classic inverting op amp circuit with gain of -100/3.3 = -30.3. Thus, vout = (-30.3)( -9 vS) = 272.7 vS For vS = 5 sin 3t mV, vout = 1.364 sin 3t V, and vout(0.25 s) = 0.9298 V.



26.


10 March 2006

We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ. Next, a source transformation yields (3×10-6)(1300) = 3.899 mV which appears in series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit as 37.7 kΩ

vout

500 Ω 1.8 kΩ

370 Ω

23.899 mV -6 V

Since no current flows through the 1.8 kΩ resistor, V+ = 23.899 mV and hence V- = 23.899 mV as well. A single nodal equation at the inverting input terminal yields 23.899 × 10-3 23.899 × 10-3 - vout + 0 = 500 37.7 × 103 Solving,

vout = 1.826 V



27.


10 March 2006

We first combine the 4.7 MΩ and 1.3 kΩ resistors: 4.7 MΩ || 1.3 kΩ = 1.30 kΩ. Next, a source transformation yields (27×10-6)(1300) = 35.1 mV which appears in series with the 20 mV source and the 500-Ω resistor. Thus, we may redraw the circuit as 37.7 kΩ

vout

500 Ω

370 Ω

1.8 kΩ 55.1 mV

-6 V

Since no current flows through the 1.8 kΩ resistor, V+ = 55.1 mV and hence V- = 55.1 mV as well. A single nodal equation at the inverting input terminal yields 55.1 × 10-3 55.1 × 10-3 - vout + 0 = 500 37.7 × 103 Solving,

vout = 4.21 V



28.


10 March 2006

The 3 mA source, 1 kΩ resistor and 20 kΩ resistor may be replaced with a –3 V source (“+” reference up) in series with a 21 kΩ resistor. No current flows through either 1 MΩ resistor, so that the voltage at each of the four input terminals is identically zero. Considering each op amp circuit separately,

vout

LEFTOPAMP

vout

RIGH OPAMP

vx = vout

100 = 14.29 V 21 100 = - (5) = - 50 V 10

= - (-3)

LEFTOPAMP

- vout

RIGH OPAMP

= 14.29 + 50 = 64.29 V.



29.


10 March 2006

A general summing amplifier with N input sources: R1

Rf

va R2

vout

vb

v1 v2 RN vN

1. va = vb = 0 2. A single nodal equation at the inverting input leads to: 0 =

va − vout v −v v −v v − vN + a 1 + a 2 + ... + a Rf R1 R2 RN

Simplifying and making use of the fact that va = 0, we may write this as ⎡ 1 ⎢− R f ⎣ or simply

⎤ v1 N v2 N vN N v R = R + R + ... + Ri ∏ ∏ ∏ ∏ i i i ⎥ out R R R i =1 i = i = i = 1 1 1 N 2 1 ⎦ N

−

vout Rf

=

v1 v v + 2 + ... + N R1 R2 RN

Thus, vout = - R f

N

vi

∑R i =1

i



30.


A general difference amplifier:

10 March 2006

R4

R1 vout

R2

v1 v2

R3

Writing a nodal equation at the inverting input, v -v v -v 0 = a 1 + a out R1 Rf Writing a nodal equation at the non-inverting input, v v -v 0 = b + b 2 R3 R2 Simplifying and collecting terms, we may write (Rf + R1) va – R1 vout = Rf v1

[1]

(R2 + R3) vb = R3 v2

[2]

From Eqn. [2], we have vb =

R3 v2 R2 + R3

Since va = vb, we can now rewrite Eqn. [1] as − R 1 vout = R f v1 −

(R f + R1 )R 3 v R2 + R3

2

and hence vout = -

Rf R ⎛ R + R1 ⎞ ⎟ v2 v1 + 3 ⎜⎜ f R1 R1 ⎝ R 2 + R 3 ⎟⎠



31.


10 March 2006

In total darkness, the CdS cell has a resistance of 100 kΩ, and at a light intensity L of 6 candela it has a resistance of 6 kΩ. Thus, we may compute the light-dependent resistance (assuming a linear response in the range between 0 and 6 candela) as RCdS = -15L + 100 Ω. Our design requirement (using the standard inverting op amp circuit shown) is that the voltage across the load is 1.5 V at 2 candela, and less than 1.5 V for intensities greater than 2 candela. Thus, vout(2 candela) = -RCdS vS/ R1 = -70 VS/ R1 = 1.5

(R1 in kΩ).

Pick R1 = 10 kΩ. Then vS = -0.2143 V.



32.


10 March 2006

We want Rf/ Rinstrument = 2K, and Rf/ Rvocal = 1K, where K is a constant not specified. Assuming K = 1, one possible solution of many is: Rf = 2 Ω Rvocal = 1 Ω vocals microphone

Rinstruments = 2 Ω

vout

instruments microphone



33.


10 March 2006

One possible solution of many: 99 kΩ 1 kΩ vout 2V vS



34.


10 March 2006

v1 + v2 + v3 . This voltage stays 3 positive and therefore a one stage summing circuit (which inverts the voltage) is not sufficient. Using the cascade setup as shown figure 6.15 and modified for three inputs we have: To get the average voltage value, we want vout =

The nodal equation at the inverting input of the first op-amp gives v1 v2 v3 − vo + + = R1 R2 R3 R f1 If we assume R1=R2=R3=R, then

v 0 = − R f1

v1 + v 2 + v 3 R

Using the nodal equation at the inverting input of the second op-amp, we have:

− vout vo − R f1 v1 + v2 + v3 = = Rf2 R4 R4 R Or,

vout =

R f 2 R f1 v1 + v2 + v3 R4 R

For simplicity, we can take Rf2 = Rf1 = R4= Rx, then, to give a voltage average, vout = Rx

v1 + v2 + v3 v1 + v2 + v3 = 3 R

I.e. Rx/R = 3. Therefore, the circuit can be completed with R1 = R2= R3 = 30 kΩ and Rf2 = Rf1 = R4 = 10 kΩ



35.


10 March 2006

The first stage is to subtract each voltage signal from the scale by the voltage corresponding to the weight of the pallet (Vtare). This can be done by using a differential amplifier:

The resistance of R can be arbitary as long as they resistances of each resistor is the same and the current rating is not exceeded. A good choice would be R = 10 kΩ. The output voltage of the differential amps from each of the scale, V1 – V4 (now gives the weight of the items only), is then added by using a two stage summing amplifier:

The output is given by:

vout =

R f 2 R f1 v1 + v2 + v3 + v4 Rin R

Therefore, to get the sum of the voltages v1 to v4, we only need to set all resistances to be equal, so setting Rf2 = Rf1 = Rin = R =10 kΩ would give an output that is proportional to the total weight of the items



36.


10 March 2006

a) Using a difference amplifier, we can provide a voltage that is the difference between the radar gun output and police speedometer output, which is proportional to the speed difference between the targeted car and the police car. Note that since a positive voltage is required which the police car is slower, the police speedometer voltage would be feed into the inverting input:

Again, R can be arbitary as long as they are equal and doesn’t give an excessive current. 10 kΩ is a good choice here. b) To convert to kph (km per hour) from mph (miles per hour), it is noted that 1 mph = 1.609 kph. Therefore, the voltage output from each device must be multiplied by 1.609. This can be done by using a non-inverting amplifier, which has an output given by: vout = (1 +

Rf Rin

)vin = 1.609vin

This gives Rf/Rin =0.609 ≈ 61/100, i.e. Rf = 6.2 kΩ and Rin = 10 kΩ



37.


10 March 2006

vout of stage 1 is (1)(-20/ 2) = -10 V. vout of stage 2 is (-10)(-1000/ 10) = 1000 V Note: in reality, the output voltage will be limited to a value less than that used to power the op amps.



38.


10 March 2006

We have a difference amplifier as the first amplifier stage, and a simple voltage follower as the second stage. We therefore need only to find the output voltage of the first stage: vout will track this voltage. Using voltage division, then, we find that the voltage at the non-inverting input pin of the first op amp is: ⎛ R3 ⎞ ⎟⎟ V2 ⎜⎜ ⎝ R2 + R3 ⎠

and this is the voltage at the inverting input terminal also. Thus, we may write a single nodal equation at the inverting input of the first op amp: 0 =

⎤ 1 ⎡ ⎛ R3 ⎞ 1 ⎡ ⎛ R3 ⎞ ⎟⎟ - V1 ⎥ + ⎟ - Vout ⎢V2 ⎜⎜ ⎢V2 ⎜ R f ⎢⎣ ⎜⎝ R 2 + R 3 ⎟⎠ R 1 ⎢⎣ ⎝ R 2 + R 3 ⎠ ⎥⎦

⎤ Stage 1 ⎥ ⎥⎦

which may be solved to obtain: Vout = Vout

Stage 1

⎛R ⎞ R3 = ⎜⎜ f + 1⎟⎟ V2 ⎝ R1 ⎠ R2 + R3

-

Rf V1 R1



39.


10 March 2006

The output of the first op amp stage may be found by realising that the voltage at the non-inverting input (and hence the voltage at the inverting input) is 0, and writing a ingle nodal equation at the inverting input: 0 =

0 - Vout 47

stage 1

+

0-2 0-3 + which leads to Vout 1 7

steage 1

= -114.1 V

This voltage appears at the input of the second op amp stage, which has a gain of –3/ 0.3 = 10. Thus, the output of the second op amp stage is –10(-114.1) = 1141 V. This voltage appears at the input of the final op amp stage, which has a gain of –47/ 0.3 = -156.7. Thus, the output of the circuit is –156.7(1141) = -178.8 kV, which is completely and utterly ridiculous.



40.


10 March 2006

The output of the top left stage is –1(10/ 2) = -5 V. The output of the middle left stage is –2(10/ 2) = -10 V. The output of the bottom right stage is –3(10/ 2) = -15 V. These three voltages are the input to a summing amplifier such that Vout = −

R (− 5 − 10 − 15) = 10 100

Solving, we find that R = 33.33 Ω.



41.


10 March 2006

Stage 1 is configured as a voltage follower: the output voltage will be equal to the input voltage. Using voltage division, the voltage at the non-inverting input (and hence at the inverting input, as well), is 5

50 = 1.667 V 100 + 50

The second stage is wired as a voltage follower also, so

vout = 1.667 V.



42.


10 March 2006

a) Since the voltage supply is higher than the Zener voltage of the diode, the diode is operating in the breakdown region. This means V2 = 4.7 V, and assuming ideal opamp, V1 = V2= 4.7 V. This gives a nodal equation at the inverting input: 4.7 V3 − 4.7 = 1k 1.1k Solving this gives V3 = 9.87 V b) PSpice simulation gives:

It can be seen that all voltage values are very close to what was calculated. The voltage output V3 is 9.88Vinstead of 9.87 V. This can be explained by the fact that the operating voltage is slightly higher than the breakdown voltage, and also the non-ideal characteristics of the op-amp.



43.


10 March 2006

The following circuit can be used:

The circuit is governed by the equations: V3 = (1 +

Rf )V1 Rin

And

V1 = V2 = Vdiode Since the diode voltage is 5.1 V, and the desired output voltage is 5.1 V, we have Rf/Rin = 0. In other words, a voltage follower is needed with Rf = 0Ω, and Rin can be arbitary – Rin =100 kΩ would be sufficient. The resistor value of R2 is determined by:

Vs − Vdiode I ref At a voltage of 5.1 V, the current is 76 mA, as described in the problem. This gives R2≈ 51 Ω using standard resistor values. R2 =



44.


10 March 2006

For the Zener diode to operate in the breakdown region, a voltage supply greater than the breakdown voltage, in this case 10 V is needed. With only 9 V batteries, the easiest way is the stack two battery to give a 18 V power supply. Also, as the input is inverted, an inverting amplifier would be needed. Hence we have the following circuit:

Vout = −

Rf Rin

Vin

Here, the input voltage is the diode voltage = 10 V, and the desired output voltage is 2.5 V. This gives Rf/Rin = 25 / 100 = 50 / 200, or Rf = 51 kΩ and Rin = 200 kΩ using standard values. Note that large values are chosen so that most current flow through the Zener diode to provide sufficient current for breakdown condition. The resistance R is given by R = (18-10) V /25 mA = 320 Ω = 330 Ω using standard values.



45.


10 March 2006

For a 20 V Zener diode, three 9 V batteries giving a voltage of 27 V would be needed. However, because the required voltage is smaller than the Zener voltage, a noninverting amplifier can not be used. To use a inverting amplifier to give a positive voltage, we first need to invert the input to give a negative input:

In this circuit, the diode is flipped but so is the power supply, therefore keeping the diode in the breakdown region, giving Vin = -20 V. Then, using the inverting amp equation, we have Rf / Rin = 12/20 giving Rf = 120 kΩ and Rin = 200 kΩ using standard resistor values. The resistance R is then given by R = (-20 - -27) V / 12.5 mA = 560 Ω using standard resistor values.



46.


10 March 2006

a) using inverting amplifier:

Rf/Rin = 5/3.3 giving Rf = 51 kΩ and Rin = 33 kΩ. R = (9-3.3) V / 76 mA = 75 Ω using standard resistor values. b) To give a voltage output of +2.2 V instead, the same setup can be used, with supply and diode inverted:

Correspondingly, the resistor values needs to be changed: Rf / Rin = 3.3/2.2 giving Rf = 33 kΩ and Rin = 22 kΩ. R would be the same as before as the voltage difference between supply and diode stays the same i.e. R = 75 Ω.



47.


10 March 2006

The following setup can be used:

I

Is

Is = I = 10 V / R = 25 mA assuming ideal op-amp. This gives R = 400 Ω. Again, taking half of max current rating as the operating current, we get R1 = (12 – 10) V / 25 mA = 80 Ω = 82 Ω using standard values.



48.


10 March 2006

Using the following current source circuit, we have:

I

Is

Is = I = 12.5 mA = 20 V / R, assuming ideal op-amp. This gives R = 1.6 kΩ and R1 = (27 – 20) / 12.5 mA = 560 Ω.



49.


10 March 2006

In this situation, we know that there is a supply limit at ±15 V, which is lower than the zener diode voltage. Therefore, previous designs need to be modified to suit this application. One possible solution is shown here:

I

Is

Here, we have Is = I = (27-20)/ R = 75 mA, assuming infinite op-amp input resistance. This gives R = 93.3 ≈ 91 Ω using standard values. We also have R1 = (2720)/ 12.5 mA = 560 Ω. Now look at the range of possible loads. The maximum output voltage is approximately equal to the supply voltage, i.e. 15 V. Therefore, the minimum load is given by RL = (20 – 15) V / 75 mA = 66.67 Ω. Similarly, the maximum load is given by RL = (20 - -15) V/ 75 mA = 466.67 Ω. i.e. this design is suitable for 466.67 Ω > RL > 66.67 Ω.




10 March 2006

50.

(a) va = vb = 1 nV ∴ vd = 0 and vout = 0. Thus,

P8Ω = 0 W.

(b) va = 0, vb = 1 nV ∴ vd = -1 nV v2 8 = -19.28 μV. Thus, P8Ω = out = 46.46 pW. vout = (2×105)(-1×10-9) 75 + 8 8 (c) va = 2 pV, vb = 1 fV ∴ vd = 1.999 pV v2 8 vout = (2×105)(1.999×10-12) = 38.53 nV. Thus, P8Ω = out = 185.6 aW. 75 + 8 8 (c) va = 50 μV, vb = -4 μV ∴ vd = 54 μV 8 vout = (2×105)(54×10-6) = 1.041 V. 75 + 8

Thus, P8Ω =

2 vout = 135.5 mW. 8




10 March 2006

51.

Writing a nodal equation at the “-vd” node, 0 = or

-v -v - vd - vd - VS + + d out Rf R1 R in

[1]

(R1Rf + RinRf + RinR1) vd + RinR1vout = -RinRfVS

[1]

Writing a nodal equation at the “vout” node,

v - (-vd ) - vout - Avd + out [2] Rf Ro Eqn. [2] can be rewritten as: - (R f + R o ) vd = vout [2] R o - AR f so that Eqn. [1] becomes: R in (AR f - R o ) VS vout = AR in R 1 + R f R 1 + R in R f + R in R 1 + R o R 1 + R o R in 0 =

where for this circuit, A = 106, Rin = 10 TΩ, Ro = 15 Ω, Rf = 1000 kΩ, R1 = 270 kΩ. (a) –3.704 mV;

(b) 27.78 mV;

(c) –3.704 V.



52.

vout = Avd = A

(


10 March 2006

)

Ri 80 × 1015 sin 2t V 16 + R i

(a) A = 105, Ri = 100 MΩ, Ro value irrelevant. vout = 8 sin 2t nV (b) A = 106, Ri = 1 TΩ, Ro value irrelevant. vout = 80 sin 2t nV




10 March 2006

53.

(a) Find vout/ vin if Ri = ∞, Ro = 0, and A is finite. The nodal equation at the inverting input is 0 =

-v -v - vd - vin + d out 100 1

[1]

At the output, with Ro = 0 we may write vout = Avd so vd = vout/ A. Thus, Eqn. [1] becomes v v v 0 = out + vin + out + out A 100A 100 from which we find vout - 100A = [2] vin 101 + A (b) We want the value of A such that vout/ vin = -99 (the “ideal” value would be –100 if A were infinite). Substituting into Eqn. [2], we find A = 9999



54.


10 March 2006

(a) δ = 0 V ∴ vd = 0, and P8Ω = 0 W. (b) δ = 1 nV, so vd = 5 – (5 + 10-9) = -10-9 V Thus, 8 = -19.28 μV and P8Ω = (vout)2/ 8 = 46.46 pW. vout = (2×105)vd 8 + 75 (c) δ = 2.5 μV, so vd = 5 – (5 + 2.5×10-6) = -2.5×10-6V Thus, 8 = -48.19 mV and P8Ω = (vout)2/ 8 = 290.3 μW. vout = (2×105)vd 8 + 75




10 March 2006

55.

AD549

Writing a single nodal equation at the output, we find that 0 =

vout - vin v - Avd + out Ro Ri

[1]

Also, vin – vout = vd, so Eqn. [1] becomes and

0 = (vout – vin) Ro + (vout – Avin + Avout) Ri

vout =

(R o + AR i ) R o + ( A + 1) R i

vin

To within 4 significant figures (and more, actually), when vin = -16 mV, vout = -16 mV (this is, after all, a voltage follower circuit).



56.


10 March 2006

The Voltage follower with a finite op-amp model is shown below:

Nodal equation at the op-amp output gives:

Vout − Vin AVd − Vout = Ri Ro But in this circuit, Vd=Vin – Vout. Substitution gives: Vout − Vin A(Vin − Vout ) − Vout AVin − ( A + 1)Vout = = Ri Ro Ro Further rearranging gives:

⇔

RoVout + Ri ( A + 1)Vout = RoVin + Ri AVin Ro + Ri A Vout = Vin Ro + Ri ( A + 1)

This is the expression for the voltage follower in non-ideal situation. In the case of ideal op-amp, A → ∞, and so A+1 → A. This means the denominator and the numerator would cancel out to give Vout = Vin, which is exactly what we expected.



57.


10 March 2006

a) By definition, when the op-amp is at common mode, vout = ACMvin. Therefore, a model that can represent this is:

This model relies on that fact that ACM is much smaller than the differential gain A, and therefore when the inputs are different, the contribution of ACM is negligible. When the inputs are the same, however, the differential term Avd vanishes, and so vout = ACMv2, which is correct. b) The voltage source in the circuit now becomes 105vd+10v2. Assuming Ro = 0, the circuit in figure 6.25 becomes:

The circuit is governed by the following equations:

105 vd + 10vb − va va − v1 = R R vb = v2 / 2 (from voltage divider)

vd = vb − va

so

Rearranging gives:

105 (v2 / 2 − va ) + 10(v2 / 2) − va = va − v1 50000v2 − 105 va + 5v2 − va = va − v1 50005v2 + v1 va = (105 + 2) Then, the output is given by:

vout = 105 vd + 10vb = 105 × (v2 / 2 − va ) + 5v2 in this case, v1 = 5 + 2 sin t, v2 = 5. This gives va=0.50004v2+9.99980×10-6v1. Thus vout = 105(-0.00004v2 - 9.99980×10-6v1)+5v2 = 1.00008v2 - 0.99998v1 = 0.0005 – 1.99996 sin t c) If the common mode gain is 0, than the equation for va becomes

50000v2 + v1 (105 + 2) Giving va=0.49999v2+9.99980×10-6v1. The output equation becomes vout = 105 vd = 105 × (v2 / 2 − va ) =0.99998v2-0.99998v1 = 1.99996 sin t va =



58.


10 March 2006

Slew rate is the rate at which output voltage can respond to changes in the input. The higher the slew rate, the faster the op-amp responds to changes. Limitation in slew rate – i.e. when the change in input is faster than the slew rate, causes degradation in performance of the op-amp as the change is delayed and output distorted.



59.


10 March 2006

a) V2 = 4.7 V from the Zener diode, V1 = V2 = 4.7 V assuming ideal op-amp, and V3 is given by the nodal equation at the inverting input: V3 − V1 V1 = 4.7k 1k Solving gives V3 = 26.79 V b) The simulation result is shown below

There are considerable discrepancies between calculated and simulated voltages. In particular, V1 = 3.090 V is considerably lower than the expected 4.7 V. This is due to the non-ideal characteristics of uA741 which has a finite input resistance, inducing a voltage drop between the two input pins. A more severe limitation, however, is the supply voltage. Since the supply voltage is 18V, the output cannot exceed 18 V. This is consistent with the simulation result which gives V3 = 17.61 V but is quite different to the calculated value as the mathematical model does not account for supply limitations.

c) By using a DC sweep, the voltage from the diode (i.e. V2) was monitored as the battery voltage changes from 12 V to 4V.

It can be seen that the diode voltage started dropping when batteries drop below 10 V. However, the diode can still be considered as operating in the breakdown region, until it hit the knee of the curve. This occurs at Vsupply = 5.28 V



60.


10 March 2006

The ideal op amp model predicts a gain vout/ vin = -1000/ 10 = -100, regardless of the value of vin. In other words, it predicts an input-output characteristic such as: vout (V)

1 -100

vin (V)

From the PSpice simulation result shown below, we see that the ideal op amp model is reasonably accurate for |vin| × 100 < 15 V (the supply voltage, assuming both have the same magnitude), but the onset of saturation is at ±14.5 V, or |vin| ~ 145 mV. Increasing |vin| past this value does not lead to an increase in |vout|.



61.


10 March 2006

Positive voltage supply, negative voltage supply, inverting input, ground, output pin.



62.


10 March 2006

This op amp circuit is an open-loop circuit; there is no external feedback path from the output terminal to either input. Thus, the output should be the open-loop gain times the differential input voltage, minus any resistive losses. From the simulation results below, we see that all three op amps saturate at a voltage magnitude of approximately 14 V, corresponding to a differential input voltage of 50 to 100 μV, except in the interest case of the LM 324, which may be showing some unexpected input offset behavior. op amp

onset

of negat ive satur ation

μA 741 LM 324 LF 411

-92 μV 41.3 μV -31.77 μV

negative

-14.32 V -14.71 V -13.81 V

satur ation volta ge

onset of

positi ve satur ation

54.4 mV 337.2 mV 39.78 mV

positive

14.34 V 13.87 V 13.86 V

satur ation volta ge



63.


10 March 2006

This is a non-inverting op amp circuit, so we expect a gain of 1 + 1000/4.7 = 213.8. With ±15 V DC supplies, we need to sweep the input just above and just below this value divided by the gain to ensure that we see the saturation regions. Performing the indicated simulation and a DC sweep from –0.1 V to +0.1 V with 0.001 V steps, we obtain the following input-output characteristic:

Using the cursor tool, we see that the linear region is in the range of –68.2 mV < Vin < 68.5 mV. The simulation predicts a gain of 7.103 V/ 32.87 mV = 216.1, which is reasonably close to the value predicted using the ideal op amp model.



64.


10 March 2006

To give a proper simulation, the inputs are grounded to give an input of 0. This gives:

As can be seen, a current of 18.57 mA is drawn from the uA741. Assuming the output voltage from the op-amp before Ro is 0, we have Ro = (1-18.57m)/18.57m = 52.9 Ω. This is close to the value given in table 6.3. There is difference between the two as here we are still using the assumption that the voltage output is independent to the loading circuit. This is illustrated by the fact that as the supplied voltage to the 1 ohm resistor changes, the voltage at the output pin actually increases, and is always higher than the voltage provided by the battery, as long as the supplied to the op-amp is greater than the battery voltage. When the supply voltage drops to 1V, the output current increased greatly and gave an output resistance of only 8 Ω. This suggests that the inner workings of the op-amp depend on both the supply and the loading. For LF411, a current of 25.34 mA is drawn from the op-amp. This gives a output resistance of 38.4 Ω. This value is quite different to the 1 Ω figure given in the table.



65.


10 March 2006

Based on the detailed model of the LF 411 op amp , we can write the following nodal equation at the inverting input: 0 =

- vd v -v Av - v + x 4 d + 6d d R in 10 10 + R o

Substituting values for the LF 411 and simplifying, we make appropriate approximations and then solve for vd in terms of vx, finding that vd =

- 106 vx 199.9 × 106

= -

vx 199.9

With a gain of –1000/10 = -100 and supply voltage magnitudes of 15 V, we are effectively limited to values of |vx| < 150 mV. For vx = -10 mV, PSpice predicts vd = 6 μV, where the hand calculations based on the detailed model predict 50 μV, which is about one order of magnitude larger. For the same input voltage, PSpice predicts an input current of -1 μA, whereas the hand calculations predict 99.5vx mA = -995 nA (which is reasonably close).



66.


10 March 2006

(a) The gain of the inverting amplifier is –1000. At a sensor voltage of –30 mV, the predicted output voltage (assuming an ideal op amp) is +30 V. At a sensor voltage of +75 mV, the predicted output voltage (again assuming an ideal op amp) is –75 V. Since the op amp is being powered by dc sources with voltage magnitude equal to 15 V, the output voltage range will realistically be limited to the range –15 < Vout < 15 V. (b) The peak input voltage is 75 mV. Therefore, 15/ 75×10-3 = 200, and we should set the resistance ratio Rf/ R1 < 199 to ensure the op amp does not saturate.



67.


10 March 2006

(a)

We see from the simulation result that negative saturation begins at Vin = –4.72 V, and positive saturation begins at Vin = +4.67 V. (b) Using a 1 pΩ resistor between the output pin and ground, we obtain an output current of

40.61 mA, slightly larger than the expected 35 mA, but not too far off.



68.


10 March 2006

We assume that the strength of the separately-broadcast chaotic “noise” signal is received at the appropriate intensity such that it may precisely cancel out the chaotic component of the total received signal; otherwise, a variable-gain stage would need to be added so that this could be adjusted by the user. We also assume that the signal frequency is separate from the “carrier” or broadcast frequency, and has already been separated out by an appropriate circuit (in a similar fashion, a radio station transmitting at 92 MHz is sending an audio signal of between 20 and 20 kHz, which must be separated from the 92 MHz frequency.) One possible solution of many (all resistances in ohms):



69.


10 March 2006


This circuit produces an output equal to the average of V1, V2, and V3, as shown in the simulation result: Vaverage = (1.45 + 3.95 + 7.82)/ 3 = 4.407 V.



70.


10 March 2006

Assuming ideal situations (ie slew rate = infinite)

a)

20

18 V

V out (V)

15 10 5 0 -5 -5

-3

-1

1

3

5

V active (V)

b) 20

18 V

V out (V)

15 10 5 0 -5 -5

-3

-1

1

3

5

V active (V)



71.


10 March 2006

a) 15

12 V 10

V out (V)

5 0 -2

-1

0

1

2

-5 -10

-12 V

-15 V active (V)

b) The simulation is performed using the following circuit:

Where RL = load resistor which is needed for the voltage probe to perform properly. The battery is swept from -2V to +2 V and the voltage sweep is displayed on the next page. It can be seen that the sweep is very much identical to what was expected, with a discontinuity at 0V. The only difference is the voltage levels which are +11.61V and – 11.61 V instead of ±12 V. This is because the output of an op-amp or comparator can never quite reach the supplied voltage.




10 March 2006

72. a) 15 10

V out (V)

5 0 -5

-3

-1

1

3

5

-5 -10 -15 V active (V)

b) 15 10

V out (V)

5 0 -5

-3

-1

1

3

5

-5 -10 -15 V 2 (V)



73.


10 March 2006

The following comparator setup would give a logic 0 for voltages below 1.5 V and logic 1 for voltages above 1.5 V



74.


10 March 2006

The voltage output of the circuit is given by vout =

R4 1 + R2 / R1 R ( ) v + − 2 v− R3 1 + R4 / R3 R1

a) When R1 = R3 and R2 = R4, the equation reduces to R vout = 4 (v+ − v− ) R3 When v+ = v-, vout = 0, thus ACM = 0. Hence CMRR = ∞ b) If R1, R2, R3 and R4 are all different, then when v+ = v- = v, R 1 + R2 / R1 R vout = ( 4 ( ) − 2 )v R3 1 + R4 / R3 R1 Simplifying the algebra gives R R − R3 R4 vout = 1 4 v R1 R3 + R1 R4 If v+ and v- are different, it turns out that it is impossible to separate vout and vd completely. Therefore, it is not possible to obtain A or CMRR in symbolic form.



75.


10 March 2006

a) The voltage at node between R1 and R2 is

⎛ R2 V 1 = V ref ⎜⎜ ⎝ R1 + R 2

⎞ ⎟⎟ ⎠

by treating it as a voltage divider. Similarly, the voltage at node between RGauge and R3 is: ⎛ ⎞ R3 ⎟ V2 = Vref ⎜ ⎜R +R ⎟ 3 Gauge ⎝ ⎠ Therefore, the output voltage is ⎛ R2 ⎞ R3 ⎟ − Vout = V1 − V2 = Vref ⎜ ⎜R +R R +R ⎟ 2 3 Gauge ⎠ ⎝ 1 b) If R1 = R2 = R3 = Rgauge then the two terms in the bracket cancels out, giving Vout = 0. c) The amplifier has a maximum gain of 1000 and minimum gain of 2. Therefore to get a voltage of 1V at maximum loading, the voltage input into the amplifier must fall between 0.001 and 0.5, i.e. 0.5 > Vout > 0.001. To simplify the situation, let R1 = R2 = R3 = R, then at maximum loading, ⎛ R ⎞ R R ⎞ ⎟ = 12⎛⎜ 1 − − Vout = Vref ⎜ ⎟ ⎜R+R R+R ⎟ ⎝ 2 R + 5k + 50m ⎠ Gauge + ΔR ⎠ ⎝ Using this we can set up two inequalities according to the two limits. The first one is: 12 R ⎛ ⎞ ⎜6 − ⎟ ≥ 0.001 R + 5k + 50m ⎠ ⎝ Solving gives 12 R 5.999 ≥ R + 5000.05 4998.38 ≥ R

Similarly, the lower gain limit gives: 5.5 ≤

12 R R + 5000.05

⇒ 4230 ≤ R This gives 4998.38 > R > 4230. Using standard resistor values, the only possible resistor values are R = 4.3 kΩ and R = 4.7 kΩ. If we take R = 4.7 kΩ, then 4.7k ⎛1 ⎞ Vout = 12⎜ − ⎟ = 0.1855 ⎝ 2 4.7k + 5k + 50m ⎠ Giving a gain of 5.388. This means a resistor value of R = 50.5/(5.388 – 1) = 11.5 kΩ or 11kΩ using standard value is needed between pin 1 and pin 8 of the amplifier. PROPRIETARY MATERIAL . © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


1.

i=C

Chapter Seven Solutions

10 March 2006

dv dt

(a)

i=0

(DC)

(b)

i=C

dv = − 10 × 10−6 115 2 (120π ) sin120π t = −613sin120π t mA dt

(c)

i=C

dv = − (10 × 10−6 )( 4 ×10−3 ) e− t = −40e − t nA dt

(

)(

)

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


2.

i=C

v=


10 March 2006

dv dt

6−0 dv t + 6 = 6 − t , therefore i = C = −4.7 × 10−6 μA 0−6 dt

i (μA)

t (s) –4.7



3. (a)

i=C


10 March 2006

dv dt

dv = 30 ⎡⎣e − t − te− t ⎤⎦ dt

therefore i = 10−3

dv = 30 (1 − t ) e− t mA dt

(b) dv = 4 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦ dt dv therefore i = 10−3 = 4e −5t (100 cos100t − 5sin100t ) mA dt



4.


10 March 2006

1 W = CV 2 2

(a)

⎛1⎞ −6 ⎜ ⎟ 2000 × 10 1600 = 1.6 J ⎝2⎠

(b)

2 ⎛1⎞ −3 ⎜ ⎟ 25 ×10 ( 35 ) = 15.3 J ⎝2⎠

(c)

2 ⎛ 1 ⎞ −4 ⎜ ⎟ 10 ( 63) = 198 mJ ⎝2⎠

(d)

⎛1⎞ −3 ⎜ ⎟ 2.2 × 10 ( 2500 ) = 2.75 J ⎝2⎠

(e)

2 ⎛1⎞ ⎜ ⎟ ( 55 )( 2.5 ) = 171.9 J ⎝2⎠

(f)

2 ⎛1⎞ −3 ⎜ ⎟ 4.8 × 10 ( 50 ) = 6 J ⎝2⎠

(

)

(

(

(

(

)

)

)

)



εA


8.854 × 10 −12 (78.54 × 10 −6 ) = 6.954 pF 100 × 10 − 6

5. (a)

C=

(b)

2 1×10−3 1 2E 2 = = 16.96 kV Energy, E = CV ∴V = 2 6.954 × 10−12 C

(c)

E=

d

=

(

C=

εA d

10 March 2006

)

1 2 E 2(2.5 × 10 −6 ) = 500 pF CV 2 ∴ C = 2 = 2 (100 2 ) V ∴ε =

Cd (500 × 10 −12 )(100 × 10 −6 ) = 636.62 pF .m −1 = −6 A (78.54 × 10 )

\Relative permittivity :

ε 636.62 × 10−12 = = 71.9 ε 0 8.854 × 10−12



6. (a)

For VA = –1V, W =


) )

2K sε 0 × 10 −12 (Vbi − V A ) = 2(11.8) 8−.854 (0.57 + 1) qN 1.6 × 10 19 1 × 10 24

(

= 45.281 × 10 Cj =

(

10 March 2006

(

−9

m

)(

11.8 8.854 × 10 −12 1 × 10 −12 45.281 × 10

)(

−9

) = 2.307 fF (

) )

2K sε 0 2(11.8) 8.854 × 10 −12 (Vbi − V A ) = (0.57 + 5) For VA = –5V, W = qN 1.6 × 10 −19 1 × 10 24

(b)

(

)(

= 85.289 × 10 −9 m Cj = (c)

(

)(

11.8 8.854 × 10 −12 1 × 10 −12 85.289 × 10

For VA = –10V, W=

(

−9

) = 1.225 fF ) )

2K sε 0 × 10 −12 (Vbi − V A ) = 2(11.8) 8−.854 (0.57 + 10) qN 1.6 × 10 19 1 × 10 24

(

)(

= 117.491 × 10 −9 m Cj =

(

)(

11.8 8.854 × 10 −12 1 × 10 −12 117.491 × 10

−9

) = 889.239aF



7.


10 March 2006

We require a capacitor that may be manually varied between 100 and 1000 pF by rotation of a knob. Let’s choose an air di electric for simplicity of cons truction, and a series of 11 half-plates: fixed

Top view

Side view with no overlap between plates

Side view with a small overlap between plates.

Constructed as shown, the half-plates are in parallel, so that each of the 10 pairs must have a capacitance of 1000/ 10 = 100 pF when rotated such that they overlap completely. If we arbitrarily select an area of 1 cm 2 for each half-plate, then the g ap spacing between each plate is d = εA/C = (8.854×10-14 F/cm)(1 cm2)/ (100×10-12 F) = 0.8854 mm. This is tight, but not impossible to achieve. The final step is to determine the amount of overlap which corresponds to 100 pF fo r the total capacitor structure. A capacitance of 100 pF is equal to 10% of the capacitance when all of the plate areas are aligned, so we need a pieshaped wedge having an area of 0.1 cm 2. If the m iddle figure above corresponds to an angle of 0 o and the case of perfect alignm ent (maximum capacitance) corresponds to an angle of 180o, we need to set out minimum angle to be 18o.



t

8. (a)

Energy stored = ∫ v ⋅ C

(b)

Vmax = 3 V

t0


10 March 2006

t − 2×10−3 ⎛ 3 −t ⎞ dv = C∫ 3e 5 ⋅ ⎜ − e 5 ⎟ dt = −1.080 μ J 0 dt ⎝ 5 ⎠

1 CV 2 = 1.35mJ ∴ 37% E max = 499.5μJ 2 V at 37% Emax = 1.825 V

Max. energy at t=0, =

v (t ) = 1.825 = 3e

−

t 5

∴ t = 2.486s ⇒≈ 2 s

(c)

1.2 ⎞ ⎛ − dv −6 ⎜ 3 = 300 × 10 − e 5 ⎟ = −141.593μA i =C ⎜ 5 ⎟ dt ⎝ ⎠

(d)

P = vi = 2.011 − 120.658 × 10 −6 = −242.6μW

(

)



(

)2 =

9. (a)

v=

1 π . 1 × 10 −3 C 2

(b)

v=

1 ⎛π .⎜ 1 × 10 −3 C ⎝2

(c)

v=

1 ⎛π .⎜ 1 × 10 −3 C ⎝2


1 47 × 10 −6

.

10 March 2006

(3.14159) (1 × 10 −3 )2 = 33.421mV 2

(

)2 + 0 ⎞⎟⎠ =

(

)2 + π4 (1 × 10 −3 )2 ⎞⎟⎠ =

1 47 × 10 −6

.

(3.14159 ) (1 × 10 −3 )2 = 33.421mV 2

1 47 × 10 −6

(

⎛ 3π 1 × 10 −3 .⎜ ⎝ 4

)2 ⎞⎟⎠ = 50.132mV



10.

1 V= C

E=

200ms

∫0


⎞⎤ 1 ⎡⎛ 7 × 10 −3 cos πt ⎟⎥ idt = ⎢⎜ − ⎟ π C ⎣⎢⎜⎝ ⎠⎦⎥

200ms

= 0

10 March 2006

0.426 C

1 181.086 × 10 −9 181.086 × 10 −9 CV 2 = 3 × 10 −6 = ∴C = = 30181μF 2 2C 2 3 × 10 −6

(

)




10 March 2006

11. (a)

c = 0.2 μ F, vc = 5 + 3cos 2 200tV; ∴ ic = 0.2 × 10−6 (3) (−2) 200 sin 200t cos 200t

∴ ic = −0.12sin 400tmA (b)

1 1 wc = cvc2 = × 2 × 10−7 (5 + 3cos 2 200t ) 2 ∴ wc max = 10−7 × 64 = 6.4 μ J 2 2

(c)

vc =

(d)

vc = 500 − 400e −100t V

t 1 × 106 ∫ 8e −100t ×10−3 dt = 103 × 40(−0.01) (e−100t − 1) = 400(1 − e100t )V 0 0.2



12.


10 March 2006

0.1

vc (0) = 250V, c = 2mF (a) vc (0.1) = 250 + 500∫ 5dt 0

0.2

∴ vc (0.1) = 500V; vc (0.2) = 500∫ 10dt = 1000V 0.1

∴ vc (0.6) = 1750V, vc (0.9) = 2000V t

∴ 0.9 < t < 1: vc = 2000 + 500∫ 10dt = 2000 + 5000(t − 0.9) 0.9

∴ vc = 2100 = 2000 + 5000(t2 − 0.9) ∴ t2 = 0.92 ∴ 0.9 < t < 0.92s




10 March 2006

13. (a)

1 1 wc = Cv 2 = × 10−6 v 2 = 2 × 10−2 e −1000t ∴ v = ±200e −500t V 2 2 −6 i = Cv′ = 10 (±200) (−500)e−500t = m0.1e−500t −v 200 ∴R = = = 2k Ω 0.1 i

(b)

PR = i 2 R = 0.01× 2000e −1000t = 20e −1000t W ∞

∴ WR = ∫ 20e −1000t dt = −0.02e −1000t 0

∞ 0

= 0.02J



14.

(a) Left circuit:

By

Voltage division, VC =

Right

circuit:


10 March 2006

1k (5) = 0.877V 4.7k + 1k

2 V1 = 1(1 // 2 ) = V 3

1 1 Voltage Division, V2 = V ∴VC = − V 3 3

By (b)



15.

v=L


10 March 2006

di dt

(a)

v = 0 since i = constant (DC)

(b)

v = −10−8 115 2 (120π ) sin120π t = −613sin120π t μ V

(c)

v = −10−8 115 2 ( 24 ×10−3 ) e−6t = − 240e −6t pV

(

)

(

)



16.

v=L


10 March 2006

di dt

⎡ ( 6 − 0 ) × 10−9 ⎤ i=⎢ t + 6 × 10−9 = 6 × 10−9 − 10−6 t , therefore −3 ⎥ ⎢⎣ ( 0 − 6 ) × 10 ⎥⎦ di v = L = − (10−12 )(10−6 ) = −10−18 V = − 1 aV dt v (aV)

t (s) –1



17.

v=L

(a)

L


10 March 2006

di dt

(

)

di = 5 ×10−6 30 ×10−9 ⎣⎡e− t − te− t ⎦⎤ = 150 (1 − t ) e− t fV dt

(b)

(

)(

)

di = 5 ×10−6 4 × 10−3 ⎡⎣ −5e −5t sin100t + 100e −5t cos100t ⎤⎦ dt therefore v = 100e −5t ( 20 cos100t − sin100t ) pV L



18.

W=


10 March 2006

1 2 LI . Maximum energy corresponds to maximum current flow, so 2

Wmax =

1 2 5 × 10−3 (1.5 ) = 5.625 mJ 2

(

)




10 March 2006

19.

(a)

(b)

PL = vLiL ∴ PL max = (−100) (−5) = 500W at t = 40− ms

(c)

PL min = 100(−5) = −500W at t = 20+ and 40+ ms

(d)

WL =

1 2 1 Li L ∴ WL (40ms) = × 0.2(−5) 2 = 2.5J 2 2




10 March 2006

20. L = 50 × 10−3 , t < 0 : i = 0; t > 0 i = 80te −100t mA = 0.08te −100t A ∴ i′= 0.08e-100t − 8te −100t ∴ 0.08 = 8t , tm , = 0.01s, i max = 0.08 × 0.01e −1 ∴ i max = 0.2943mA; v = 0.05i′ = e −100t (0.004 − 0.4t ) 0.8 = 0.02 s 40 = 0.004V at t=0

∴ v′ = e −100t (−0.4) − 100e −100t (0.004 − 0.4t ) ∴−0.4 = 0.4 − 40t , t = v = e −2 (0.004 − 0.008) = −0.5413mV this is minimum∴ v max




10 March 2006

21. (a)

t > 0 : is = 0.4t 2 A ∴ vin = 10is + 5is′ = 4t 2 + 4tV

(b)

iin′ = 0.1vs +

1 t 40tdt + 5 = 4t + 4t 2 + 5A 5 ∫0




22.

vL = 20 cos1000tV, L = 25mH, iL (0) = 0

(a)

iL = 40 ∫ 20 cos 1000tdt = 0.8sin 1000tA ∴ p = 8sin 2000t W

(b)

1 w = × 25 ×10−3 × 0.64sin 2 1000t = 8sin 2 1000t mJ 2

10 March 2006

t

0




10 March 2006

23.

t

(a)

0 < t < 10 ms: iL = −2 + 5∫ 100dt = −2 + 500t ∴ iL (10ms) = 3A, iL (8ms) = 2A

(b)

iL (0) = 0 ∴ iL (10ms) = 500 × 0.01 = 5A ∴ iL (20ms) = 5 + 5∫

0

0.02

0.01

104 (0.02 − t )dt

4 ∴ iL (20ms) = 5 + 5 × 104 (0.02t − 0.5t )0.02 0.01 = 5 + 5 × 10 (0.0002 − 0.00015) = 7.5A

1 ∴ wL = × 0.2 × 7.52 = 5.625J 2 (c)

If the circuit has been connected for a long time, L appears like short circuit. V8Ω =

8 (100V ) = 80V 2+8

I 2Ω =

20V = 10 A 2Ω

∴i x =

80V = 1A 80Ω



24.


10 March 2006

After a very long time connected only to DC sources, the inductors act as short circuits. The circuit may thus be redrawn as

⎛ 80 ⎞ ⎜ ⎟ ⎛ 100 ⎞ And we find that ix = ⎜ 9 ⎟ ⎜ ⎟= 1A ⎜ 80 + 80 ⎟ ⎝ 2 + 8 ⎠ 9 ⎠ ⎝




25.

L = 5H, VL = 10(e −t − e −2t )V, iL (0) = 0.08A

(a)

vL (1) = 10(e −1 − e−2 ) = 2.325+ V

(b)

iL = 0.08 + 0.2 ∫ 10(e − t − e −2t ) dt = 0.08 + 2(−e − t + 0.5e −2t )t0

10 March 2006

t

0

iL = 0.08 + 2(−e − t + 0.5e −2t + 1 − 0.5) = 1.08 + e −2t − 2e − t ∴ iL (1) = 0.4796A

(c)

iL (∞) = 1.08A




10 March 2006

26. (a)

(b)

40 12 + 40 × 5 × 12 + 20 + 40 12 + 20 + 40 200 100 = + = 100V 3 3

vx = 120 ×

vx =

15 12 120 15 × × 40 + 40 × 5 12 + 15 60 15 + 60 15 12 + 60

120 1 6.667 × × 40 + 200 12 + 12 5 66.667 = 40 + 20 = 60V =



27.


(a)

1 wL = × 5 × 1.62 = 6.4J 2

(b)

1 wc = × 20 × 10−6 ×1002 = 0.1J 2

(c)

Left to right (magnitudes): 100, 0, 100, 116, 16, 16, 0 (V)

(d)

Left to right (magnitudes): 0, 0, 2, 2, 0.4, 1.6, 0 (A)

10 March 2006



28. (a)


10 March 2006

vs = 400t 2 V, t > 0; iL (0) = 0.5A; t = 0.4s 1 vc = 400 × 0.16 = 64V, wc × 10−5 × 642 = 20.48mJ 2

(b)

(c)

0.4 1 iL = 0.5 + 0.1∫ 400t 2 dt = 0.5 + 40 × × 0.43 = 1.3533A 0 3 1 ∴ wL = × 10 × 1.35332 = 9.1581J 2

0.4

iR = 4t 2 , PR = 100 × 16t 4 ∴ wR = ∫ 1600t 4 dt = 320 × 0.45 = 3.277J 0



29. (a) (b)

P7Ω = 0W ; P10Ω =


10 March 2006

V 2 (2 )2 = = 0.4W 10 R

PSpice verification We see from the PSpice simulation that the voltage across the 10-Ω resistor is –2 V, so that it is dissipating 4/10 = 400 mW.

The

7-Ω resistor has zero volts across its terminals, and hence dissipates zero power. Both results agree with the hand calculations.




10 March 2006

30.

(a) We find RTH by first short-circuiting the voltage source, removing the inductor, and looking into the open terminals.

Sim

plifying the network from the right, 3 || 6 + 4 = 6 Ω, which is in parallel with 7 Ω. 6 || 7 + 5 = 8.23 Ω. Thus, RTH = 8.23 || 8 = 4.06 Ω. To find VTH, we remove the inductor: + VTH – V1

V2

V3

REF Writing the nodal equations required: (V (V

– 9)/3 + V1/6 + (V1 – V2)/4 = 0 – V1)/4 + V2/7 + (V2 – V3)/5 = 0 V3/8 + (V3 – V2)/5 = 0

Solving, (b)

1 2

V3 = 1.592 V, therefore VTH = 9 – V3 = 7.408 V. iL = 7.408/4.06 = 1.825 A (inductor acts like a short circuit to DC).

(c)




10 March 2006

31. C equiv

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ in series with 10μ in series with 10μ + ≡ 10 μ + ⎜ 1 ⎜ 1 1 ⎟ 1 ⎟ ⎜ 10μ + 10μ ⎟ ⎜ 10μ + 10μ ⎟ ⎝ ⎠ ⎝ ⎠

≡ 4.286μF



32.


10 March 2006

Lequiv ≡ (77 p // (77 p + 77 p )) + 77 p + (77 p // (77 p + 77 p )) = 179.6& pH



33. (a) have

(b) Using


10 March 2006

Assuming all resistors have value R, all inductors have value L, and all capacitors value C,

At dc, 20μF is open circuit; 500μH is short circuit. 10k (9 ) = 3.6V voltage division, V x = 10k + 15k



34.


10 March 2006

(a) As all resistors have value R, all inductors value L, and all capacitors value C,

+ Vx -

(b)

Vx = 0 V as L is short circuit at dc.



35. C

equiv


10 March 2006

= { [(100 n + 40 n) || 12 n] + 75 n} || {7 μ + (2 μ || 12 μ)}

C equiv ≡ 85.211nF



36. L

equiv


10 March 2006

= {[ (17 p || 4 n) + 77 p] || 12 n} + {1 n || (72 p + 14 p)}

Lequiv ≡ 172.388 pH



37.


10 March 2006

C T − C x = (7 + 47 + 1 + 16 + 100) = 171μF E CT −C x =

1 (CT − C x )V 2 = 1 (171μ )(2.5)2 = 534.375μJ 2 2

E C x = E CT − E CT −C x = (534.8 − 534.375)μJ = 425nJ

1 425n (2 ) ∴ E C x = 425n = C xV 2 ⇒ C x = = 136nF 2 (2.5)2




10 March 2006

38. ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ 1 1 ⎜ ⎟ ⎜ ⎟ = 2.75H = 1.5 + + ⎜ 1 + 1 ⎟ ⎜ 1 + 1 + 1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1.5 1.5 ⎠ ⎝ 1.5 1.5 1.5 ⎠

(a)

For all L = 1.5H, Lequiv

(b)

For a general network of this type, having N stages (and all L values equiv), Lequiv =

n

LN

∑ NLN −1

N =1




10 March 2006

39. ⎞ ⎛ ⎞ ⎛ ⎟ ⎜ 1 ⎟ ⎜ 1 ⎜ ⎟ = 3H ⎜ ⎟ =1+ + ⎜1 + 1⎟ ⎜1+1+1⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 2⎠ ⎝3 3 3⎠

(a)

Lequiv

(b)

For a network of this type having 3 stages,

Lequiv = 1 +

1 1 + 2+ 2 3+3

(2 )

2

(3)

2

+

1 3

=1+

(2)2 + (3)3 2(2 ) 3(3)2

Extending for the general case of N stages, 1 1 +K+ 1 1 1 1 1 1 1 + + + +K 2 2 3 3 3 N N 1 1 1 = 1+ + +K+ = N 2(1 / 2) 3(1 / 3) N(1/N)

Lequiv = 1 +

1

+



40.

C equiv =


10 March 2006

(3 p )(0.25 p ) = 0.231 pF 3 p + 0.25 p



41.

Lequiv =


10 March 2006

(2.3& n )(0.3& n ) = 0.2916& nH 2.6& n



42.


(a)

Use 2 x 1μH in series with 4 x 1μH in parallel.

(b)

Use 2 x 1μH in parallel, in series with 4 x 1μH in parallel.

(c)

Use 5 x 1μH in parallel, in series with 4 x 1μH in parallel.

10 March 2006




10 March 2006

43. (a)

R = 10Ω :10 10 10 = ∴ R eq =

(b) (c)

10 10 55 + 10 + 10 10 = , 3 3 3

55 30 = 11.379Ω 3

L = 10H ∴ Leq = 11.379H 1 = 5.4545 1/ 30 + 1/10 + 1/ 20 10 ∴ Ceq = 5.4545 + = 8.788F 3 C = 10F :




10 March 2006

44. (a)

oc :L eq = 6 1 + 3 = 3.857H sc : L eq = (3 2 + 1) 4 = 2.2

(b)

4 = 1.4194H

1 7 1 = , ceq = = 1.3125F 1/ 4 + 1/ 2 3 3 / 7 + 1/ 2 1 5 5 = , Ceq = 4 + = 4.833F sc : 1/ 5 + 1 6 6

oc :1 +




10 March 2006

45. (a)

(b)

(c)




46.

is = 60e −200t mA, i1 (0) = 20mA

(a)

6 4 = 2.4H ∴ v = Leq is′ = 2.4 × 0.06(−200)e−200t

10 March 2006

or v = −28.8e −200t V

1 t 4.8 −200t (e −28.8e−200t dt + 0.02 = − 1) + 0.02 ∫ 6 o 200 = 24e−200t − 4mA(t > 0)

(b)

i1 =

(c)

i2 = is − i1 = 60e−200t − 24e−200t + 4 = 36e−200t + 4mA(t > 0)




47.

vs = 100e −80tV , v1 (0) = 20V

(a)

i = Ceq vs′ = 0.8 ×10−6 (−80)100e −80t = −6.4 × 10−3 e−80t A

(b)

v1 = 106 (−6.4 ×10−3 ) ∫ e −80t dt + 20 =

t

o

10 March 2006

6400 −80t (e − 1) + 20 80

∴ v1 = 80e−80t − 60V (c)

t 106 1600 −80t (−6.4 ×10−3 ) ∫ e −80t dt + 80 = (e − 1) + 80 v2 o 4 80 = 20e −80t + 60V




10 March 2006

48. (a)

vc − vs v −v + 5 ×10−6 vc′ + c L = 0 20 10 t vL − vc 1 + vL dt + 2 = 0 10 8 × 10−3 ∫o

(b)

20i20 + 1 5 × 10−6

1 5 ×10−6 t

∫ (i o

L

t

∫ (i o

20

− iL )dt + 12 = vs

− i20 )dt − 12 + 10iL + 8 × 10−3 iL′ = 0




10 March 2006

49. vc (t ): 30mA: 0.03 × 20 = 0.6V, vc = 0.6V 9V: vc = 9V, 20mA: vc = −0.02 × 20 = 0.4V 0.04 cos103 t : vc = 0 ∴ vc (t ) = 9.2V vL (t ): 30mA, 20mA, 9V: vL = 0; 0.04 cos103 t : vL = −0.06 × 0.04 (−1000) sin103 t = 2.4sin103 tV



50.


10 March 2006

We begin by selecting the bottom node as the reference and assigning four nodal voltages: V4

V2

V1

V3

Ref. 1, 4 Supernode: and: Node

V 2:

20×10-3 e-20t = 1

t V1 - V2 + 0.02 × 103 ∫ V4 − 40e − 20t dt ′ [1] 0 50

– V4 = 0.2 Vx or

0 =

(

)

0.8V1 + 0.2 V2 – V4 = 0

V2 - V1 V - 40e −20t dV2 + 2 + 10- 6 50 100 dt

[2]

[3]




10 March 2006

R i = ∞, R o = 0, A = ∞∴ vi = 0 ∴ i = Cvs′

51. (a)

also 0 + Ri + vo = 0 ∴ vo = − RCvs′ 1 idt + vi c∫ −1 1+ A vo = −Avi ∴ vi = vo ∴ i = vi A R 1 1 1 1+ A v ∴ vs = ∫ idt − vo = − vo + − o dt ∫ c A A RC A 1+ A 1+ A ∴ Avs′ = −vo′ − vo or vo′ + vo + Avs′ = 0 RC RC −vi + Ri − Avi = 0, vs =

(b)



52.


10 March 2006

Place a current source in parallel with a 1-MΩ resistor on the positive input of a buffer with output voltage, v. This feeds into an integrator stage with input resistor, R2, of 1-MΩ and feedback capacitor, Cf, of 1 μF. dv c f ions i=Cf = 1.602 × 10 −19 × dt sec

0=

0=

Va − V 1 × 10

6

+Cf

dv c f dt

=

Va − V 1 × 10

6

+ 1.602 × 10 −19

ions sec

dv c f −V −V −19 ions +Cf = 1 . 602 10 + × R2 dt sec 1 × 10 6

Integrating current with respect to t,

1 R2

∫0 vdt ' = C f (Vc t

f

)

− Vc f (0)

1.602 × 10 −19 × ions = C f Vc f R2 − R1 −1 × 1.602 × 10 −19 × ions ⇒ Vout = × 1.602 × 10 −19 × ions Vc f = Va − Vout ⇒ Vout = R2 C f Cf R1 = 1 MΩ, Cf = 1μF




10 March 2006

53.

R = 0.5MΩ, C = 2 μ F, R i = ∞, R o = 0, vo = cos10t − 1V

(a)

vo ⎞ 1⎞ 1 t⎛ ⎛ Eq. (16) is: ⎜ 1 + ⎟ vo = − ⎜ vs + ⎟ dt − vc (0) ∫ o RC ⎝ A⎠ ⎝ A⎠ vo ⎞ ⎛ 1⎞ 1⎞ 1 ⎛ 1⎞ 1 ⎛ ⎛ ∴ ⎜1 + ⎟ vo′ = − ⎜ vs + ⎟ ∴ ⎜ 1 + ⎟ (−10sin10t ) = −1⎜ vs + cos10t − ⎟ A⎠ RC ⎝ A⎠ ⎝ A⎠ A ⎝ A⎠ ⎝ 1⎞ 1 1 ⎛ ∴ vs = ⎜1 + ⎟10sin10t + − cos10t Let A = 2000 A A ⎝ A⎠ ∴ vs = 10.005sin10t + 0.0005 − 0.0005cos10t

(b)

Let A = ∞∴ vs = 10sin10tV



54.


10 March 2006

Create a op-amp based differentiator using an ideal op amp with input capacitor C1 and feedback resistor Rf followed by inverter stage with unity gain. 1mV dvs R = 60 × / min R f C1 dt rpm R RfC1=60 so choose Rf = 6 MΩ and C1 = 10 μF. Vout = +



55. (a)

0=

10 March 2006

Va − Vout 1 + vdt Rf L∫

Va = V = 0,∴ (b)


− Rf Vout 1 = ⇒ = V v dt L out L Rf L∫

t

∫0 v s dt '

In practice, capacitors are usually used as capacitor values are more readily available than inductor values.



56.


10 March 2006

One possible solution:

vout = −

1 R1C f

∫v

in

dt

we want vout = 1 V for vin = 1 mV over 1 s.

vin

In other words, 1 = −

1 R1C f

∫

1

0

10−3 dt = −

10−3 R1C f

Neglecting the sign (we can reverse terminals of output connection if needed), we therefore need R1Cf = 10–3. Arbitrarily selecting Cf = 1 μF, we find R1 = 1 kΩ.



57.


10 March 2006


dvin dt ⎛ dv ⎞ 100 mV maximum ⎜ in ⎟ . = 60s ⎝ dt ⎠

vout = − RC

vin

⎛ 100 mV ⎞ In other words, vout = 1 V = RC ⎜ ⎟ ⎝ 60s ⎠ or RC = 600

Arbitrarily selecting C = 1000 μF, we find that R = 600 kΩ.



58.


10 March 2006


dvin dt ⎛ dv ⎞ 100 mV At 1 litre/s, ⎜ in ⎟ = . s ⎝ dt ⎠

vout = − RC

vin

⎛ 100 mV ⎞ In other words, vout = 1 V = RC ⎜ ⎟ ⎝ 1s ⎠ or RC = 10

Arbitrarily selecting C = 10 μF, we find that R = 1 MΩ.



59.


10 March 2006

One possible solution:

vin

The power into a 1 Ω load is I2, therefore energy = W = I2Δt. vout =

1 R1C f

∫I

2

dt

we want vout = 1 mV for vin = 1 mV (corresponding to 1 A 2 ).

(

)

Thus, 10−3 = RC 10−3 , so RC = 1 Arbitrarily selecting C = 1 μF, we find that we need R = 1 MΩ.



60.


10 March 2006


vout = − RC

dvin dt

vin

Input: 1 mV = 1 mph, 1 mile = 1609 metres. Thus, on the input side, we see 1 mV corresponding to 1609/3600 m/s. Output: 1 mV per m/s2. Therefore, vout = 2.237 RC = 1 so RC = 0.447 Arbitrarily selecting C = 1 μF, we find that R = 447 kΩ.




10 March 2006

61. (a)

(b)

20v20 + 1 5 × 10−6

(c)

1 5 × 10−6 t

∫ (v o

c

t

∫ (v o

20

− vc )dt + 12 = is

− v20 )dt − 12 + 10vc + 8 × 10−3 vc′ = 0

iL − is i −i + 5 × 10−6 iL′ + L c = 0 20 10 t ic − iL 1 + i dt + 2 = 0 −3 ∫o c 10 8 ×10




10 March 2006

62.




10 March 2006

63.




10 March 2006

64. (a)

(b) “Let

is = 100e-80t A and i1 (0) = 20 A in the circuit of (new) Fig. 7.62.

(a) Determine v(t) for all t. (b) Find i1(t) for t ≥ 0. (c) Find v2(t) for t ≥ 0.” (c) (a)

L eq = 1 4 = 0.8μ H∴ v(t ) = Leq is′ = 0.8 × 10−6 × 100(−80)r −80t V ∴ v(t) = −6.43-80t mV t

(b)

i1 (t ) = 106 ∫ −6.4 × 10−3 e −80t dt + 20 ∴ i1 (t ) =

(c)

i2 (t ) = is − i1 (t ) ∴ i2 (t ) = 20e−80t + 60A

o

6400 −80t (e − 1) = 80e−80t − 60A 80




10 March 2006

65.

In creating the dual of the original circuit, we have lost both vs and vout. However, we may write the dual of the original transfer function: iout/ is. Performing nodal analysis, iS =

1 t V1dt ′ + G in (V1 - V2 ) L1 ∫0

iout = Aid = GfV2 + Gin (V2 – V1)

[1] [2]

Dividing, we find that

iout = iS

G in (V2 - V1 ) + G f V2

1 t V1dt ′ + G in (V1 - V2 ) L1 ∫0



66. I PSpice

L

= 4/10 = 400 mA. W =


10 March 2006

1 2 LI L = 160 mJ 2

verification:



67. I

L

= 4/(4/3) = 3 A. W =


10 March 2006

1 2 LI L = 31.5 J 2

PSpice verification:



68.


10 March 2006

We choose the bottom node as the reference node, and label the nodal voltage at the top of the dependent source VA. Then, by KCL, VA − 4 VA VA V + + = 0.8 A 100 20 25 25 Solving, we find that VA = 588 mV.

Therefore, VC, the voltage on the capacitor, is 588 mV (no DC current can flow through the 75 Ω resistor due to the presence of the capacitor.) Hence, the energy stored in the capacitor is

1 1 2 CV 2 = 10−3 ( 0.588 ) = 173 μJ 2 2

(

)



69.


10 March 2006

By inspection, noting that the capacitor is acting as an open circuit, the current through the 4 kΩ resistor is 8 mA. Thus, Vc = (8)(4) = 32 V. Hence, the energy stored in the capacitor =

PSpice

1 1 2 CV 2 = 5 × 10−6 ( 32 ) = 2.56 mJ 2 2

(

)

verification:



70. C

1


10 March 2006

= 5 nF, Rf = 100 MΩ.

vout = − R f C1

(

)( )

dvs = − 5 × 10−9 108 ( 30 cos100t ) = −15cos10t V dt

Verifying with PSpice, choosing the LF411 and ±18 V supplies:



71. PSpice


10 March 2006

verification

w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2)]2 = 220.8 μJ. This is in agreement with the PSpice simulation results shown below.



72. PSpice


10 March 2006

verification

w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2)]2 = 669.2 pJ. This is in agreement with the PSpice simulation results shown below.



73.


10 March 2006

Va − V s 1 + ∫ v L f dt R1 L − Vs 1 Va = Vb = 0, 0= + ∫ v L f dt R1 L 0=

V L f = Va − Vout = 0 − Vout =

Vout = −

(

L dVs R1 dt

)

L f dVs Lf d =− A cos 2π 10 3 t ⇒ L f = 2 R1 ; Let _R = 1 Ω and L = 1 H. R1 dt R1 dt

PSpice Verification: clearly, something rather odd is occuring in the simulation of this particular circuit, since the output is not a pure sinusoid, but a combination of several sinusoids.



74. PSpice


10 March 2006

verification

w = ½ Cv2 = 0.5 (33×10-6)[5 cos (75×10-2) - 7]2 = 184.2 μJ. This is in reasonable agreement with the PSpice simulation results shown below.



75. PSpice


10 March 2006

verification

w = ½ Li2 = 0.5 (100×10-12)[5 cos (75×10-2) - 7]2 = 558.3 pJ. This is in agreement with the PSpice simulation results shown below.



1.

i (t ) = i (0)e

−R t L

10 March 2006

9

= 2e −4.7×10 t mA

(

−4.7×109 100×10−12

(a)

i(100 ps) = 2e

(b)

i(212.8 ps) = 2e

(c)

vR = –iR

(

)

= 1.25 mA

−4.7×109 212.8×10−12

vR(75 ps) = −2 ( 4700 ) e

(d)

Chapter Eight Solutions

)

= 736 μ A

(

−4.7×109 75×10−12

)

= − 6.608 V

vL(75 ps) = vR(75 ps) = –6.608 V



2.


10 March 2006

1 2 Li = 100 mJ at t = 0. 2 Thus, i( 0) = 0.1 = 316 mA

W=

and i( t ) = i (0)e

R − t L

= 316e− t / 2 mA

(a)

At t = 1 s, i (t ) = 316e −1/ 2 mA = 192 mA

(b)

At t = 5 s, i (t ) = 316e −5/ 2 mA = 25.96 mA

(c)

At t = 10 s, i (t ) = 316e −10 / 2 mA = 2.13 mA

(d)

At t = 2 s, i (t ) = 316e −1 mA = 116.3 mA . Thus, the energy remaining is 1 W (2) = Li (2) 2 = 13.53 mJ 2



3. Thus,

We know that i (t ) = i (0)e L=

(

R − t L

−100 500 ×10−6


−3

= 2 × 10 e

−

100 t L

, and that i(500 μs) = 735.8 μA.

) = −100 ( 500 ×10 )

⎛ i (500 × 10−6 ) ⎞ ln ⎜ ⎟ −3 ⎝ 2 × 10 ⎠

10 March 2006

−6

⎛ 735.8 ×10−6 ⎞ ln ⎜ ⎟ −3 ⎝ 2 ×10 ⎠

= 50 mH



4. Thus,

We know that i (t ) = i (0)e R=−

R − t L

= 1.5e


−

R t 3×10−3

10 March 2006

, and that i(2) = 551.8 mA.

3 × 10−3 ⎛ i (2) ⎞ 3 × 10−3 ⎛ 0.5518 ⎞ ln ⎜ = − ln ⎜ ⎟ = 1.50 mΩ ⎟ 2 2 ⎝ 1.5 ⎠ ⎝ 1.5 ⎠



5. At

At

Thus,

R − t L


−

R

10 March 2006

t

We know that i (t ) = i (0)e = 1.5e , and that W(0) = 1 J; W(10–3) = 100 mJ. 2 1 t = 0, 3 ×10−3 ⎡⎣i ( 0 ) ⎤⎦ = 1 therefore i (0) = 25.82 A. 2 2 1 t = 1 ms, 3 ×10−3 ⎡⎣i 10−3 ⎤⎦ = 0.1 therefore i (10−3 ) = 8.165 A. 2

(

)

(

R=−

3×10−3

) (

)

3 × 10−3 ⎛ i (t ) ⎞ 3 × 10−3 ⎛ 8.165 ⎞ ln ⎜ ln ⎜ = − ⎟ ⎟ = 3.454 Ω t 0.001 ⎝ 25.82 ⎠ ⎝ i (0) ⎠




10 March 2006

6. (a)

Since the inductor current can’t change instantaneously, we simply need to find iL while the switch is closed. The inductor is shorting out both of the resistors, so iL(0+) = 2 A.

(b)

The instant after the switch is thrown, we know that 2 A flows through the inductor. By KCL, the simple circuit must have 2 A flowing through the 20-Ω resistor as well. Thus, v = 4(20) = 80 V.



7.


10 March 2006

(a) Prior to the switch being thrown, the 12-Ω resistor is isolated and we have a simple two-resistor current divider (the inductor is acting like a short circuit in the DC circuit, since it has been connected in this fashion long enough for any transients to have decayed). Thus, the current iL through the inductor is simply 5(8)/ (8 + 2) = 4 A. The voltage v must be 0 V.

(b) The instant just after the switch is thrown, the inductor current must remain the same, so iL = 4 A. KCL requires that the same current must now be flowing through the 12-Ω resistor, so v = 12(-4) = -48 V.




10 March 2006

8. (a)

iL (0) = 4.5mA, R/L = ∴ iL = 4.5e −10

6

t/4

103 106 = 4 × 10−3 4

mA ∴ iL (5μ s ) = 4.5e −1.25

= 1.289 mA. (b) iSW( 5 μs) = 9 – 1.289 = 7.711 mA.



9. (a)


10 March 2006

100 = 2A ∴ iL (t ) = 2e−80t / 0.2 50 −400 t = 2e A, t > 0

iL (0) =

(b)

iL (0.01) = 2e −4 = 36.63mA

(c)

2e −400t1 = 1, e 400t1 = 2, t1 = 1.7329ms




10 March 2006

10. (a) di + 5i = 0 [1] dt vR = −2i so Eq. [1] can be written as L

⎛ −v ⎞ d⎜ R ⎟ 2 ⎠ ⎛ −v − 5⎜ R L ⎝ dt ⎝ 2 dv 2.5 R + 2.5vR = 0 dt

⎞ ⎟ = 0 or ⎠

(b) Characteristic equation is 2.5s + 2.5 = 0, or s + 1 = 0 Solving, s = –1, vR (t ) = Ae −t

(c)

At t = 0–, i(0–) = 5 A = i(0+). Thus, vR (t ) = −10e − t , t > 0 2 vR (0− ) = (10 ) = 6.667 V 3 + vR (0 ) = −10 V vR (1) = −10e −1 = − 3.679 V




10 March 2006

11. (a)

I I i t t = e − t /τ , = ln o , o = 10 ∴ = ln10 = 2.303; i i Io τ τ t t Io I = 100, = 4.605; o = 1000, = 6.908 i i τ τ

(b)

i d (i / Io ) d () = e− t /τ , = −et /τ ; t t / τ = 1, =a−e −1 Io d (t / τ ) d () ⎛t ⎞ i Now, y = m( x − 1) + b = −e −1 ( x − 1) + e−1 ⎜ = x, = y ⎟ Io ⎝τ ⎠ −1 −1 At y = 0, e ( x − 1) = e ∴ x = 2 ∴ t / τ = 2



12.


10 March 2006

Reading from the graph current is at 0.37 at 2 ms ∴ τ = 2 ms I 0 = 10 A




10 March 2006

w = ½ Li2, so an initial energy of 15 mJ in a 10-mH inductor corresponds to an initial inductor current of 1.732 A. For R = 1 kΩ, τ = L/R = 10 μs, so iL(t) = 1.732 e–0.1t A. For R = 10 kΩ, τ = 1 μs so iL(t) = 1.732 e-t. For R = 100 kΩ, τ = 100 ns or 0.1 μs so iL(t) = 1.732 e-10t A. For each current expression above, it is assumed that time is expressed in microseconds.

13.

To create a sketch, we firs t realise that the m aximum current for any of the th ree cases will be 1.732 A, and af ter one time constant (10, 1, or 0.1 μs), the current will d rop to 36.79% of this value (637.2 m A); after approximately 5 tim e constants, the curren t will be close to zero.

Sketch based on hand analysis

Circuit used for PSpice verification

As can be seen by comparing the two plots, which probably should have the same x-axis scale labels for easier comparison, the PSpice simulation results obtained using a parametric sweep do in fact agree with our hand calculations.




10 March 2006

14. (a)

3.3 × 10 −6 τ= = 3.3 × 10 −12 6 1 × 10

(b)

1 2

ω = .L.I 0 2 I0 =

2 × 43 × 10− 6 = 5.1 A 3.3 × 10− 6

i (5 ps ) = 5.1e −1×10

6

× 5×10 −12 / 3.3×10 − 6

= 1.12 A

(c)

From the PSpice simulation, we see that the inductor current is 1.121 A at t = 5 ps, in agreement with the hand calculation.




10 March 2006

15. Assume the source Thévenin resistance is zero, and assume the transient is measured to 5τ. Then,

τ=

L 5L ∴ 5τ = = 100 × 10−9 secs R R (5)(125.7)10−6 ∴R > 10−7 (If 1τ assumed then R >

so R must be greater than 6.285 kΩ.

6.285 = 125.7Ω ) 5



16. For that


10 March 2006

t < 0, we have a current divider with iL(0-) = ix(0-) = 0.5 [ 10 (1/ (1 + 1.5)] mA = 2 mA. For t > 0, the resistor through which ix flows is shorted, so that ix(t > 0) = 0. The remaining 1-kΩ resistor and 1-mH inductor network exhibits a decaying current such iL(t) = 2e-t/τ mA where τ = L/R = 1 μS.

(a)

(b)

ix (mA)

2

t (μs)



17. (a) (b)


10 March 2006

2 1 2 ×10−3 C ⎡⎣ v ( 0 ) ⎤⎦ = 10−3 so v ( 0 ) = = 44.72 V 2 10−6

τ = RC = 100 s v(t ) = v(0)e −t / RC = 44.72e −0.01t V Thus, v(20) = = 44.72e −0.01(20) = 36.62 V . v 36.62 Since i = = = 366 nA R 100 × 106

(c)



18. If


10 March 2006

i(0) = 10 A, v(0) = 10 V. v(t ) = v(0)e −t / RC = 10e− t / 2 V

(a)

At t = 1 s, v(1) = 10e −1/ 2 = 6.065 V

(b)

At t = 2 s, v(2) = 10e −1 = 3.679 V

(c)

At t = 5 s, v(5) = 10e −2.5 = 821 mV

(d)

At t = 10 s, v(10) = 10e−5 = 67.4 mV



19.


10 March 2006

Referring to Fig. 8.62, we note that τ = RC = 4 s. Thus, v(t ) = v(0)e −t / RC = 5e−0.25t V.

(a)

v(1 ms) = 5e −0.25( 0.001) = 4.999 V

(b)

v(2 ms) = 5e −0.25( 0.002 ) = 4.998 V Therefore i(2 ms) = 4.998 / 1000 = 4.998 mA

(c)

v(4 ms) = 5e −0.25( 0.004 ) = 4.995 V 1 W = Cv 2 = 49.9 mJ 2



20.

Thus,


10 March 2006

(a) v(t ) = v(0)e− t / RC = 1.5e− t / RC V −t ⎡ v(t ) ⎤ = ln ⎢ ⎥ RC ⎣ 1.5 ⎦ R=

−t −2 ×10−9 = = 7.385 Ω ⎡ v(t ) ⎤ ⎡ 0.1 ⎤ −10 10 ln ⎢ ⎥ C ln ⎢ ⎣ 1.5 ⎥⎦ ⎣ 1.5 ⎦

(b)

We see from the PSpice simulation that our predicted voltage at 2 ns agrees with the information used to calculate R = 7.385 Ω.



21.


10 March 2006

The film acts as an intensity integrator. Assuming that we may model the intensity as a simple decaying exponential, φ(t) = φo e-t/ τ where the time constant τ = RTHC represents the effect of the Thévenin equivalent resistance of the equipment as it drains the energy stored in the two capacitors, then the intensity of the image on the film Φ is actually proportional to the integrated exposure: Φ = K∫

exposure time

0

φo e − t / τ dt

where K is some constant. Solving the integral, we find that

[

]

Φ = - K φoτ e − (exp osure time)/ τ - 1 The maximum value of this intensity function is –Kφoτ.

With 150 m s yielding an image intensity of approxim ately 14% of the m aximum observed and the knowledge that at 2 s no further increase is seen leads us to estim ate –3 that 1 – e–150×10 / τ = 0.14, assuming that we are observing single-exponential decay behavior and that the response speed of the fil m is not affecting the m easurement. Thus, we may extract an estimate of the circuit time constant as τ = 994.5 ms. This estim ate is consis tent with the addition al observation that a t t = 2 s, the im age appears to be saturated.

=

With two 50-m F capacitors connected in pa rallel for a total capacitance of 100 mF, we may estimate the Thévenin equivalent resistance from τ = RC as Rth = τ / C 9.945 Ω.




10 March 2006

22. (a)

vc (t ) = 192e −3000t / 24

(b)

30 = 192V 50 = 192e −125t V

vc (0) = 8(50 200) ×

0.1 = e −125t ∴ t = 18.421ms




10 March 2006

23. (a)

vc = 80e −10

(b)

wc =

6

t /100

4

4

= 80e−10 t V, t > 0; 0.5 = e−10 t ∴ t = 69.31μ s

1 1 C 802 e −20,000t = C 802 ∴ t = 34.66 μ s 2 4




10 March 2006

24. t < 0 : ic (t ) = 0, 10 = 5000is + 104 is ∴ is = ∴ vc (t ) =

20 = 6.667V 3 4

t > 0 : is = 0 ∴ vc (t ) = 6.667e − t / 2×10 ×2×10 ∴ vc (t ) = 6.667e−25t V ∴ ic (t ) =

2 mA 3

−6

−6.667 −25t e = 0.3333e−25t mA 3 20 ×10

vC(t)

iC(t)




10 March 2006

25.

( ) i (0 ) = 0.1A v 0 + = 20V +

v(1.5ms ) = 20e −1.5×10 i (1.5ms ) = 0 A

v(3ms ) = 20e −3×10 i (3ms ) = 0 A

−3

−3

/ 50× 20×10 − 6

/ 50×20×10 − 6

= 4.5V

= 1V




10 March 2006

26. (a)

1 2 iL (0− ) = × 60 = 30 mA, ix (0− ) = × 30 = 20 mA 2 3

(b)

iL (0+ ) = 30 mA, ix (0+ ) = −30 mA

(c)

iL (t ) = 30e−250t / 0.05 = 30e−5000t mA, iL (0.3ms) = 30e−1.5 = 6.694mA = −ix

Thus,

ix = –6.694 mA.




10 March 2006

27. (a)

iL (0) = 4A ∴ iL (t ) = 4e−500t A (0 ≤ t ≤ 1ms) iL (0.8ms) = 4e−0.4 = 2.681A

(b)

iL (1ms) = 4e −0.5 = 2.426A ∴ iL (t ) = 2.426e −250( t −0.001) ∴ iL (2ms) = 2.426e −0.25 = 1.8895− A




10 March 2006

28. (a)

iL = 40e−50,000t mA ∴10 = 40e−50,000t , ∴ t1 = 27.73μ s

(b)

iL (10 μ s) = 40e −0.5 = 24.26mA ∴ iL = 24.26e− (1000+ R )50t (t > 10 μ s ) −6

∴10 = 24.26e− (1000+ R )5×10 ∴ ln2.426 = 0.8863 = 0.25(1000 + R)10−3 ,1000 + R = 0.8863 × 4 ×103 ∴ R = 2545+ Ω




10 March 2006

29. (a)

i1 (0) = 20mA, i2 (0) = 15mA ∴ v(t ) = 40e −50000t + 45e −100000t V ∴ v(0) = 85V

(b)

v(15μ s ) = 40e −0.75 + 45e−1.5 = 28.94V

(c)

85 = 40e −50000t + 45e −100000t . Let e −50000t = x 10 ∴ 45 x 2 + 40 x − 8.5 = 0 −40 ± 1600 + 1530 = 0.17718, < 0 90 ∴ e −50000t = 0.17718, t = 34.61μ s ∴x =



30.

t < 0 : vR =


2R1R 2 2R1 , ↓ iL (0) = R1 +R 2 R1 + R 2

t > 0 : iL (t ) =

2R1 2R1R 2 −50 R2t e−50 R2t ∴ vR = e R1 + R 2 R1 +R 2

∴ vR (0+ ) = 10 =

2R1R 2 ∴ R1 R 2 = 5Ω. Also, vR (1ms) R1 +R 2

= 5 = 10e−50 R2 /1000 ∴ 0.05R 2 = 0.6931∴ R 2 = 13.863Ω ∴

1 1 1 + = ∴ R1 = 7.821Ω 13.863 R1 5

10 March 2006

31.

24 = 0.4A ∴ iL (t ) = 0.4e−750t A, t > 0 60

(a)

iL (0) =

(b)

5 vx = × 24 = 20V, t < 0 6 3 vx (0+ ) = 50 × 0.4 × = 7.5V 8 −750 t ∴ vx (t ) = 7.5e V, t > 0



32.


10 March 2006

3iL × 20 + 10iL = 25iL 4 v vin ∴ in = 25Ω∴ iL = 10e−25t / 0.5 = 10e −50t A, t > 0 iL

v in =



33.

iL (0) =


10 March 2006

64 40 × = 5A 4 + 40 8 48

∴ iL = 5e −24t / 8 = 5e −3t A ∴ i1 (t ) = 2.5e −3t A, t > 0; i1 (−0.1)= 2.5 A i 1 (0.03) = 2.285− A, i1 (0.1) = 1.852 A




10 March 2006

34. (a)

iL (0) = 4A ∴ iL = 4e−100t A, 0 < t < 15ms ∴ iL (15ms) = 4e −1.5 = 0.8925+ A

(b)

t >15ms: iL = 0.8925+ e−20(t −0.015) A ∴ iL (30ms) = 0.8925+ e −0.3 = 0.6612A




10 March 2006

35. (a)

i1 (0+ ) = i1 (0− ) = 10A, i2 (0+ ) = i2 (0− ) = 20A ∴ i (0+ ) = 30A

(b)

τ = L eq / R eq =

(c)

i1 (0− ) = 10A, i2 (0− ) = 20A ; i (t ) = 30e −600t A

(d)

v = −48i = −1440e−600t V

(e)

i1 = 10(−440) ∫ e −600t dt + 10 = 24e −600t t0 +10 = 24e −600t − 14A

0.08 5 = ms = 1.6667ms; 48 3

t

0

t

i2 = 2.5(−1440) ∫ e −600t dt + 20 0

= 6e −600t t0 +20 = 6e −600t + 14A

(f)

1 1 WL (0) = × 0.1×102 + × 0.4 × 202 = 5 + 80 = 85J 2 2 1 1 WL (∞) = × 0.1× 142 + × 0.4 × 142 = 9.8 + 39.2 = 49J 2 2 ∞ ∞ 900 × 48 WR = ∫ i 2 48dt = ∫ 900 × 48e−1200t dt = (−1) = 36J 0 0 −1200 ∴ 49 + 36 = 85 checks




10 March 2006

36. (a)

(b)

2 2 100 2 × = 33.33V; i1 (0− ) = × = 16.667mA 2+2 3 2+2 3 ∴ vc (9 : 59) = 33.33V, i1 (9 : 59) = 16.667mA vc (0) = 100 ×

vc (t ) = 33.33e − t / 400 , t > 10 : 00 ∴ vc (10 : 05) = 33.33e −300 / 400 = 15.745+ V, i1 (10 : 05) =

(c)

15.745 = 3.936mA 4000

τ = 400 s, so 1.2τ = 480 s. vC(1.2τ) = 33.33 e–1.2 = 10.04 V. Using Ohm’s law, we find that i1(1.2τ) = vC(1.2τ)/ 4000 = 2.51 mA.

(d) PSpice

Verification:

We see from the DC analysis of the circuit that our initial value is correct; the Probe output confirms our hand calculations, especially for part (c).




10 March 2006

37. t> 0:

25ix = 1.25ix ∴ 34 = 100(1.25ix − 0.8ix + ix ) + 25ix ∴ ix = 0.2A 20

(a)

is (0− ) = (1.25 − 0.8 + 1)0.2 = 0.290 A

(b)

ix (0− ) = 0.2A

(c)

vc (t ) = 25 × 0.2e −t = 5e− t V ∴ ix (0+ ) =

(d)

0.8ix (0+ ) = 0.04A ∴ is (0+ ) =

(e)

ix (0.4) =

5 = 0.05A 100

34 20 33.2 − 0.04 × = = 0.2767A 120 120 120

1 × 5e−0.4 = 0.03352A 100




10 March 2006

38. (a)

vc (0) = 10V ∴ vc (t ) = 10e

(b)

iA (−100μ s) = iA (0− ) =

−106 t /(10 + 50 200)

= 10e −20000t V

10 = 50mA 200 ⎛ 1 ⎞ 50 iA (100 μ s ) = 10e−2 ⎜ = 5.413mA ⎟ ⎝ 10 + 40 ⎠ 250

(c) PSpice

Verification.

From the DC simulation, we see that PSpice verifies our hand calculation of iA = 50 mA. The transient response plotted using Probe indicates that at 100 μs, the current is approximately 5.46 mA, which is within acceptable round-off error compared to the hand calculated value.



39.


10 March 2006

12 = −6mA (t < 0) 12 + 4

(a)

i1 (t ) = 8(−1)

(b)

4 12 6 = 2k Ω, vc (0) = 48V ∴ vc (t ) = 48e−10

6

t / 5× 2×103

= 48e−100t V, t > 0

∴ i (t ) = 12e−100t mA, t > 0




10 March 2006

40. (a)

vCLeft (0) = 20V, vCRIGHT (0) = 80V ∴ vCL = 20e−10

6

t /8

6

, vCR = 80e−10

t / 0.8

∴ vout = vCR − vCL = 80e−1,250,000t − 20e−125,000t V, t > 0 (b)

vout (0+ ) = 60V; vout (1μ s) = 80e−1.25 − 20e −0.125 = 5.270V vout (5μ s) = 80e−6.25 − 20e −0.625 = −10.551V



41. (a)

(c)

10 March 2006

vc − 0.25vc vc vc − 40 + + = 0 ∴ vc = 20V (t < 0) 5 10 4 1 − 0.25 + 0.1 − iin = 0.25A t > 0: Apply vc = 1V ∴ 5 1 ∴ R eq = = 4Ω 0.25

t < 0:

∴ vc (t ) = 20e −10

(b)


6

t/4

= 20e −250,000t V (t > 0)

vC(3 μs) = 9.447 V PSpice verification. Note that the switch parameters had to be changed in order to perform this simulation.

As can be seen from the simulation results, our hand calculations are accurate.



42.


10 March 2006

t < 0 : vc (0) = 60V 50 −500 /( Ro +1000) e 60 R 500 ∴ = ln 1.2 = 0.18232 ∴ o + 2 = 5.4848, Ro = 1742.4Ω R o + 1000 500 0 < t < 1ms: vc = 60e−10

6

t /( Ro +1000)

∴

∴ vc (1ms) = 60e−1000 / 2742.4 = 41.67V t > 1ms : vc = 41.67e −10

6

( t −10−3 )

/(1742.4 + R1 1000)

∴ 25 = 41.67e −1000( ) ∴ 0.5108 = = 1957.6, R1 1000 = 215.2

.1000 ,1742.4 + R1 1000 1742.4 + R1 1000

1 1 + 10−3 = ∴ R1 = 274.2Ω R1 215.2




10 March 2006

43. (a) With the switch closed, define a nodal voltage V1 at the top of the 5-kΩ resistor. Then, 0 = (V1 – 100)/ 2 + (V1 – VC)/ 3 + V1/ 5 0 = VC/ 10 + (VC – V1)/ 3 + (VC – 100)

[1] [2]

Solving, we find that VC = vC(0-) = 99.76 V. (b)

t > 0 : R eq = 10 6.5 = 3.939k Ω∴ vc = 87.59e −10

7

t / 3939

= 87.59e −2539t V (t > 0)



44.


10 March 2006

t < 0: 12 = 4i1 + 20i1 ∴ i1 = 0.5mA ∴ vc (0) = 6i1 + 20i1 = 26i1 vc (0) = 13V t > 0: Apply ← 1mA ∴1 + 0.6i1 = i1 ∴ i1 = 2.5mA; ± vin = 30i1 = 75V ∴ R eq = 75k Ω 3

−9

∴ vc (t ) = 13e −t / 75×10 ×2×10 = 13e−10 ∴ i1 (t ) =

6

t /150

= 13e−6667 t

vo = 0.4333e −6667 t mA (t > 0) 3 × 104




10 March 2006

45. (a)

v1 (0− ) = 100V. v2 (0− ) = 0, vR (0− ) = 0

(b)

v1 (0− ) = 100V. v2 (0+ ) = 0, vR (0+ ) = 100V

(c)

τ=

(d)

vR (t ) = 100e −12.5t V, t > 0

(e)

i (t ) =

(f)

v1 (t ) =

(g)

20 × 5 ×10−6 × 2 104 = 8 × 10−2 s 20 + 5

vR (t ) = 5e −12.5t mA 4 2 ×10

106 t 103 −12.5t t −3 −12.5t −12.5t − × e dt + = e + 80V 5 10 100 o +100 = −20e ∫ o 20 50 1000 t −12.5t v2 (t ) = dt + 0 = −80e −12.5t to +0 = −80e−12.5t + 80V 5e ∫ o 5 1 1 wc1 (∞) = × 20 ×10−6 × 802 = 64mJ, wc 2 (∞) × 5 × 10−6 × 802 = 16mJ 2 2 1 wc1 (0) = × 20 ×10−6 × 1002 = 100mJ, wc 2 (0) = 0 2 ∞ 25 wR = ∫ 25 × 10−6 e −25t × 2 × 104 dt = × 2 × 104 (−1)10−6 = 20mJ o −25 64 + 16 + 20 = 100 checks




10 March 2006

46. (a)

t < 0: is = 1mA ∴± vc (0) = 10V, ↓ iL (0) = −1mA ∴ vx (0) = 10V, t < 0

(b)

t > 0 : vc (t ) = 10e −t /10 ×20×10 = 10e−5000t V

4

−9

i L (t ) = −10−3 e − 103t / 0.1 = −10−3 e −10000t A ∴ ± vL (t ) = e −10000t V, t > 0 ∴ vx = vc − vL (t ) = 10e −5000t − e−10000t V, t > 0




10 March 2006

47. (a)

t < 0: vs = 20V ∴ vc = 20V, iL = 20mA ∴ ix (t ) = 20mA, t < 0

(b)

t > 0: vs = 0 ∴ iL (t ) = 0.02e −10000t A; vc (t ) = 20e− t / 2×10

−8

104

= 20e−5000t V

↓ ic (t ) = 2 × 10−8 × 20(−5000) e −5000t = −2e−5000t mA ix (t ) = iL (t ) + ic (t ) = 0.02e −10000t − 0.002e −5000t A = 20e −10000t − 2e−5000t mA



48.


10 March 2006

V 1 = = 1.1 A R 0.909 t > 0 : iL (t ) = e − 2.363t A iL ( 0 − ) =

iL (0.1s ) = 1.1e − 2.363×0.1 = 0.8685 A ∴since the current has dropped to less than 1 A prior to t = 100 ms, the fuse does not blow. PSpice verification: Note that the switch properties were changed.

We see from the simulation result that the current through the fuse (R3) is 869 mA, in agreement with our hand calculation.



49.

v(t ) = 6u (t ) − 6u (t − 2) + 3u (t − 4)


10 March 2006

V



50.


10 March 2006

i (t ) = 2u (t ) + 2u (t − 2) − 8u (t − 3) + 6u (t − 4) A



51. (a)

f(–1) = 6 + 6 – 3 = 9

(b)

f(0–) = 6 + 6 – 3 = 9

(c)

f(0+) = 6 + 6 – 3 = 9

(d)

f(1.5) = 0 + 6 – 3 = 3

(e)

f(3) = 0 + 6 – 3 = 3


10 March 2006



52. (a)

g(–1) = 0 – 6 + 3 = –3

(b)

g(0+) = 9 – 6 + 3 = 6

(c)

g(5) = 9 – 6 + 3 = 6

(d)

g(11) = 9 – 6 + 3 = 6

(e)

g(30) = 9 – 6 + 3 = 6


10 March 2006



53.


10 March 2006

(a) v A = 300u (t − 1) V, vB = −120u (t + 1) V; ic = 3u (−t )A 100 = 1A 300 −120 t = 0.5 : i1 (−0.5) = + 1 = 0.6A; 300 120 300 120 t = 0.5 : i1 = − = −0.4A; t = 1.5 : i1 = − = 0.6A 300 300 300 t = −1.5 : i1 (−1.5) = 3 ×




10 March 2006

54. ↑ v A = 600tu (t + 1) v, vB = 600 (t + 1) u (t ) V, ic = 6(t − 1) u (t − 1)A (a)

t = −1.5 : i1 = 0; t = −0.5 : i1 = 600(−0.5) / 300 = −1A 600(0.5) 600(1.5) + = 4A 300 300 600(1.5) 600(2.5) 1 t = 1.5 : i1 = + + × 6 × 0.5 = 3 + 5 + 1 = 9A 300 300 3 t = 0.5 : i1 =

(b)




10 March 2006

55. (a)

2u (−1) − 3u (1) + 4u (3) = −3 + 4 = 1

(b)

[5 − u (2)] [2 + u (1)] [1 − u (−1)]

(c)

4e −u (1)u (1) = 4e −1 = 1.4715+

= 4 × 3 ×1 = 12




10 March 2006

56. (a)

(b)

100 20 + 0 + 10 × = 6A 50 50 60 t > 0: ix = 0 + + 0 = 2A 30 t < 0: ix =

t < 0: The voltage source is shorting out the 30-Ω resistor, so ix = 0. t > 0: ix = 60/ 30 = 2 A.



57.

t = −0.5 :

50 25 = 16.667, ix =

t = 0.5 :

ix =

t = 1.5 :

ix = 3 −

t = 2.5 :

ix =

t = 3.5 :

ix = −


10 March 2006

200 1/ 50 1 −2 = 3 − = 2.5A 66.67 1/ 50 + 1/ 25 + 1/ 50 2

200 = 3A 66.67 100 1 × = 2.5A 66.67 3

200 − 100 = 2A 50 100 = −2A 50



58.


10 March 2006

v(t ) = 4 − 16u (t ) + 20u (t − 4) − 6u (t − 6)V

v(t) (V) 8 4

2 0

4

6

t (s)

-12



59. (a)


10 March 2006

7 u (t ) − 0.2 u (t ) + 8(t − 2) + 3 v (1) = 9.8 volts

(b)

Resistor of value 2Ω




10 March 2006

60. i (t ) =

R − t ⎞ Vo ⎛ L − e 1 ⎜ ⎟ u (t ) A and vR (t ) = i ( t ) R R⎝ ⎠

(a)

R − t ⎞ ⎛ vR (t ) = Vo ⎜1 − e L ⎟ u (t ) ⎝ ⎠

(b)

vR (2 ×10−3 ) = 1.2 1 − e−2

(

(

)

V = 1.2 1 − e −1000t u (t )

)

V

V = 1.038 V

(c)




10 March 2006

61. (a)

iL (t ) = (2 − 2e −200000t ) u (t )mmA

(b)

vL (t ) = LiL′ = 15 × 10−3 × 10−3 (−2) (−200000e −200000t ) u (t ) = 6e −200000t u (t )V




10 March 2006

62. (a)

iL (t ) = 2 + 2 (1 − e −2.5t ) u (t ) A ∴ i1 (−0.5) = 2A

(b)

iL (0.5) = 2 + 2 (1 − e−1.25 ) = 3.427A

(c)

iL (1.5) = 2 + 2 (1 − e−3.75 ) = 3.953A




10 March 2006

63. (a)

iL (t ) = (4 − 4e−20t / 0.02 )u (t ) ∴ iL (t ) = 4(1 − e −1000t )u (t )A

(b)

v1 (t ) = (100 − 80e−1000t )u (t )V



64.


10 March 2006

(a) 0 W (b) The total inductance is 30 || 10 = 7.5 mH. The Thévenin equivalent resistance is 12 || 11 = 5.739 kΩ. Thus, the circuit time constant is L/R = 1.307 μs. The final value of the total current flowing into the parallel inductor combination is 50/12 mA = 4.167 mA. This will be divided between the two inductors, so that i(∞) = (4.167)(30)/ (30 + 10) = 3.125 mA. We may therefore write i(t) = 3.125[1 – e-10 we find 2.810 A.

6t/ 1.307

] A. Solving at t = 3 μs,

(c) PSpice verification

We see from the Probe output that our hand calculations are correct by verifying using the cursor tool at t = 3 μs.



65. (a)

τ = L/RTH =


30 × 10−6 30 ×10−6 = = 9 ×10−6 5 ||10 3.333

10 March 2006

s

i (t ) = i f (t ) + in (t ) in = Ae

−106 t

9

and i f =

9 5

6

Thus, At so

10 − t 9 i (t ) = + Ae 9 5 t = 0, i(0 ) = i(0+) = 4.5/5. Thus, A = –4.5/5 = 0.9 6

10 − t 9 i (t ) = − 0.9e 9 A 5

(b)

At t = 1.5 μs, i = i (t ) =

1.5 − t 9 − 0.9e 9 = 1.038 A 5

(c)



66.


10 March 2006

45 ×10−3 45 ×10−3 = = 0.0135 s 10 || 5 3.333 vR (t ) = v f + vn

τ = L/Req =

(a)

vf(t) = 0 since inductor acts as a short circuit. Thus, vR (t ) = vn = Ae−74.07 t . t = 0, iL = 12/10 = 1.2 A = iL(0-) = iL(0+).

At

Writing KVL for this instant in time, 16 – 10(1.2 + vR/5) = vR 4 4 Therefore vR (0+ ) = V and hence vR (t ) = e−74.07 t V 3 3 (b)

At t = 2 ms, vR (2 ms) =

4 −74.07( 2×10−3 ) V = 1.15 V e 3

(c)



67.

τ=


10 March 2006

L 5 ×10−3 = = 50 sμ Req 100

v1 (t ) = v1 f + v1n where v1 f = 6 V since the inductor acts as a short circuit Therefore At

v1 (t ) = 6 + Ae

−

106 t 50

.

t = 0-, iL = 0 = iL(0+). Thus, v1(0+) = 0 since no current flows through the resistor. 10 t ⎞ ⎛ − v1 (t ) = 6 ⎜ 1 − e 50 ⎟ V. ⎜ ⎟ ⎝ ⎠ 6

Hence

At

27 − ⎛ ⎞ t = 27 μs, v1 (27 ×10−6 ) = 6 ⎜1 − e 50 ⎟ = 2.5 V ⎝ ⎠




10 March 2006

68. (a)

iL (t ) = 10 A, t < 0

(b)

iL (t ) = 8 + 2e−5t / 0.5 ∴ iL (t ) = 8 + 2e−10t A, t > 0




10 March 2006

69. (a)

iL (t ) = 2A, t > 0

(b)

iL (t ) = 5 − e−4t / 0.1 ∴ iL (t ) = 5 − 3e−40t A, t > 0




10 March 2006

70. (a) 0, (b)

0 0, 200V

(c) 1A, (d)

100V 50 × 10−3 1 = ms ∴ iL = 1(1 − e −4000t ) u (t )A, iL (0.2ms) = 0.5507A 200 4 v1 (t ) = (100 + 100e −4000t ) u (t )V, v1 (0.2ms) = 144.93V

τ=



71.

(a)

(b)


10 March 2006

di + Pi = Q, i = e − Pt ∫ Qe Pt dt + Ae − Pt , R = 125Ω, L = 5H dt di ∴ L LPi = LQ ∴ LP = 5P = R = 125 ∴ P = 25 dt t 10 2 = 2 ∴ i = e −25t ∫ 2e 25t dt + Ae −25t = e−25t × e 25t to + Ae−25t o L 25 2 10 2 2 ∴i = + Ae −25t , i (0) = = ∴ A = 0∴i = = 0.08A 125 25 25 25

Q(t ) =

t 10u (t ) 2 = 2u (t ) ∴ i = e −25t ∫ 2e 25t dt + Ae −25t = + Ae −25t o 5 25 2 i (0) = 0 ∴ A = − ∴ i (t ) = 0.08(1 − e −25t )A, t > 0 25

Q(t ) =

(c)

Q(t ) =

10 + 10u (t ) = 2 + 2 u (t ) ∴ i = 0.16 − 0.08e −25t A, t > 0 5

(d)

Q(t ) =

t 10u (t ) cos 50t = 2u (t ) cos 50t ∴ i = e −25t ∫ 2 cos 50t × e 25t dt + Ae −25t o 5 t

∴ i = 2e

−25t

⎡ e 25t ⎤ −25t ⎢ 502 + 252 (25cos 50t + 50sin 50t ) ⎥ + Ae ⎣ ⎦o

⎡ e 25t ⎤ 1 = 2e −25t ⎢ × 25⎥ + Ae −25t (25cos 50t + 50sin 50t ) − 3125 ⎣ 3125 ⎦ 2 4 2 −25t e + Ae −25t = cos 50t + sin 50t − 125 125 125 2 2 i (0) = 0 ∴ 0 = − + A∴A = 0 125 125 ∴ i (t ) = 0.016 cos 50t + 0.032sin 50t − 0.016e−25t A, t > 0




10 March 2006

72. 100 100 − = −15A, t < 0 20 5

(a)

iL (t ) =

(b)

iL (0+ ) = iL (0− ) = −15A

(c)

iL (∞) =

(d)

iL (t ) = 5 − 20e −40t A, t > 0

100 = 5A 20



73.


10 March 2006

18 1 × = 0.1A ∴ iL (0+ ) = 0.1A 60 + 30 2 i L (∞) = 0.1 + 0.1 = 0.2A iL (0− ) =

∴ iL (t ) = 0.2 − 0.1e −9000t A, t > 0 ∴ iL (t ) = 0.1u (−t ) + (0.2 − 0.1e −9000t ) u (t )A or, iL (t ) = 0.1 + (0.1 − 0.1e −9000t ) u (t )A



74.


10 March 2006

30 3 × = 3A, iL (0− ) = 4A 7.5 4

(a)

ix (0− ) =

(b)

ix (0+ ) = iL (0+ ) = 4A

(c)

ix (∞) = iL (∞) = 3A ∴ ix (t ) = 3 + 1e−10t / 0.5 = 3 + e−20t A ∴ ix (0.04) = 3 + e−0.8 = 3.449A




10 March 2006

75. (a) (b)

(c)

ix (0− ) = iL (0− ) = ix (0+ ) =

30 = 3A 10

30 30 15 × + 3× = 2.4A 30 + 7.5 40 10 + 15

30 30 × = 3A ∴ ix (t ) = 3 − 0.6e −6t / 0.5 7.5 40 = 3 − 0.6e −12t ∴ ix (0.04) = 3 − 0.6e −0.48 = 2.629A

ix ( ∞ ) =



76.


10 March 2006

OC : vx = 0, voc = 4 u (t )V vx − 0.2vx vx + ,12 u (t ) = 0.6vx + 2vx 40 60 v 12 u (t ) 12 u (t ) u (t ) ∴ vx = ∴ iab = x = = 2.6 60 2.6 × 60 13 4 u (t ) u (t ) (1 − e−52t / 0.2 ) u (t ) = (1 − e−260t ) u (t ) ∴ Rth = 4 × 13 = 52Ω∴ iL = 52 13 + −260 t ) u (t )V ∴ vx = 60iL = 4.615 (1 − e

SC : 0.1u (t ) =




10 March 2006

77. (a)

OC : − 100 + 30i1 + 20i1 = 0, i1 = 2A ∴ voc = 80 u (t )V 20 × 10 = 20A 20 ∴ R th = 4Ω∴ iL (t ) = 20(1 − e −40t ) u (t )A

SC : i1 = 10A, ↓ isc = 10 +

(b)

vL = 0.1× 20 × 40e −40t u (t ) = 80e −40t u (t ) ∴ i1 (t ) =

100u (t ) − 80e −40t u (t ) = 10 − 8e −40t u (t )A 10



78.


10 March 2006

τ = Req C = (5)(2) = 10 s Thus, vn (t ) = Ae−0.2t . v = vn + v f = Ae −0.1t + Be −5t .

At t = 0, v(0) = 0 since no source exists prior to t = 0. Thus, A + B = 0 [1]. As t → ∞ , v(∞) → 0 . We need another equation. i (0) = C

dv 4.7 = ( 2) dt 5

therefore

dv 4.7 4.7 or − 0.1A − 5 B = = [2] dt t =0+ 5 5

Solving our two equations, we find that A = –B = 0.192. Thus, v(t ) = 0.192 ( e−0.1t − e−5t )



79.


10 March 2006

Begin by transforming the circuit such that it contains a 9.4 cos 4t u(t) V voltage source in series with a 5 Ω resistor, in series with the 2 F capacitor. Then we find that

9.4 cos 4t = 5i + v = (5)(2)

dv +v dt

dv + 0.1v = 0.94 cos 4t , so that v(t ) = e−0.1t ∫ ( 0.94 cos 4t ) e0.1t dt + Ae−0.1t dt Performing the integration, we find that or

⎡10 cos 4t + 400sin 4t ⎤ v(t ) = 0.94 ⎢ + Ae−0.1t . ⎥ 1 + 1600 ⎣ ⎦ At t = 0, v = 0, so that A = − and v(t ) =

0.94 (10 ) 1601

0.94 ⎡ −10e−0.1t + 10 cos 4t + 400sin 4t ⎤⎦ 1601 ⎣



80. (a)


10 March 2006

6 vc (0− ) = × 3 = 2V = vc (0+ ) 9 6 vc (∞) = 2 − 6 (2 7) = −6V 7 9

3

∴ vc (t ) = −6 + 8e−10 t / 2×10 = −6 + 8e−500000t V, t > 0 vc (−2μ s) = vc (0− ) = 2V, vc (2μ s) = −6 + 8e−1 = −3.057V (b) PSpice verification.

As can be seen from the plot above, the PSpice simulation results confirm our hand calcu lations of vC(t < 0) = 2 V and vC(t = 2 μs) = -3.06 V



81.


10 March 2006

τ = RC = 2 ×10−3 ( 50 ) = 0.1 vn = Ae−10t vC (t ) = vCn + vCf = Ae −10t + 4.5

Since

since vC ( ∞ ) = 4.5 V

vC (0− ) = vC (0+ ) = 0 vC (t ) = −4.5e−10t + 4.5 = 4.5 (1 − e−10t )




10 March 2006

82. i (0− ) = 10 = 10mA, i (∞) = 2.5mA, v (0) = 0 A A c 1 10 1 iA (0+ ) = × 1.4286mA ∴ iA = 10mA, t < 0 1.75 4 8

3

iA = 2.5 + (1.4286 − 2.5) e−10 t /1.75×10 = 2.5 − 1.0714e−57140t mA, t > 0




10 March 2006

10 = 2.5mA, iA (∞) = 10mA 4 10 7.5 vc (0) = 7.5V ∴ iA (0+ ) = + = 17.5mA 1 1

− 83. iA (0 ) =

8

3

5

iA = 10 + 7.5e −10 t /10 = 10 + 7.5e−10 t mA, t > 0, iA = 2.5mA t < 0




10 March 2006

84. (a)

iin (-1.5) = 0

(b)

iin (1.5) = 0




10 March 2006

85. (a)

vs = −12u (−t ) + 24 u (t )V t < 0: vc (0− ) = −8V ∴ vc (0+ ) = −8V 2 t > 0 : vc (∞) = × 24 = 16V 3 200 RC = × 103 × 3 × 10−7 = 2 × 10−3 30 ∴ vc (t ) = 16 − 24e −500t V, t > 0 ∴ vc (t ) = −8u (−t ) + (16 − 24e −500t ) u (t )

−12 24 + 8 = −0.4mA, iin (0+ ) = = 3.2mA 30 10 24 = 0.8mA iin (∞) = 30 iin (t ) = −0.4u (t ) + (0.8 + 2.4e −500t ) u (t )mA

(b) iin (0− ) =



86.


10 March 2006

−vx vx 3 − vx − + = 0 ∴ vx = 1, voc = 3 − 1 = 2V 100 100 100 v v SC : vx = 3V ∴ isc = x + x = 0.06A 100 100 ∴ R th = voc / isc = 2 / 0.06 = 33.33Ω OC :

6

∴ vc = voc (1 − e − t / RthC ) = 2(1 − e −10 t / 33.33 ) = 2(1 − e −30,000t ) V, t > 0



87.


10 March 2006

vc (0− ) = 10V = vc (0+ ), iin (0− ) = 0 iin (0+ ) = 0 ∴ iin (t ) = 0 for all t

0 < t < 0.5s : vc = 10(1 − e −2.5t )V vc (0.4) = 6.321V, vc (0.5) = 7.135V 5 50 8 20 − 10 5 = A ∴ vc (∞) = 10 + 8 + = V, 4 8 = Ω 12 6 6 3 3 50 ⎛ 50 ⎞ + ⎜ 7.135 − ⎟ e −0.375×20( t −0.5) = 16.667 − 9.532e −7.5(t −0.5) V vc (t ) = 3 ⎝ 3 ⎠ ∴ vc (0.8) = 16.667 − 9.532e −7.5(0.3) = 15.662V t > 0.5:




10 March 2006

88. (a) For

t < 0, there are no active sources, and so vC = 0. For 0 < t < 1, only the 40-V source is active. Rth = 5k || 20 k = 4 kΩ and hence τ = Rth C = 0.4 s. The “final” value (assuming no other source is ever added) is found by voltage division to be vC(∞) = 40(20)/(20 + 5) = 32 V. Thus, we may write vC(t) = 32 + [0 – 32] e–t/ 0.4 V = 32(1 – e-2.5t) V. For t > 1, we now have two sources operating, although the circuit time constant remains unchanged. We define a new time axis temporarily: t' = t – 1. Then vC(t' = 0+) = vC(t = 1) = 29.37 V. This is the voltage across the capacitor when the second source kicks on. The new final voltage is found to be vC(∞) = 40(20)/ (20 + 5) + 100(5)/ (20 + 5) = 52 V. Thus, vC(t') = 52 + [29.37 – 52] e-2.5t' = 52 – 22.63 e-2.5(t – 1) V.

(b) For t < 0, vC = 0.

(c)

We see from the simulation results that our hand calculations and sketch are indeed correct.




10 March 2006

89. (a)

t < 0 : 8(10 + 20) = 240V = vR (t ) = 80V, t < 0

(b)

t < 0 : vc (t ) = 8 × 30 = 240V ∴ vc (0+ ) = 240V 1 t = (∞) : vc (∞) = × 8(10 + 10) = 80V 2 −6

∴ vc (t ) = 80 + 160e − t /10×10 = 80 + 160e −100000t V ∴ vR (t ) = 80 + 160e −100000t V, t > 0 (c) (d)

t < 0 : vR (t ) = 80V −6

vc (0− ) = 80V, vc (∞) = 240V ∴ vc (t ) = 240 − 160e − t / 50×10 = 240 − 160e −20000t V 20 80 ×10 + × 10 = 32 + 16 = 48V 30 + 20 50 −20000 t V, t > 0 vR (∞) = 80V ∴ vR (t ) = 80 − 32e vR (0− ) = 80V, vR (0+ ) = 8



90.


10 March 2006

t < 0 : vc = 0 0 < t < 1ms : vc = 9 (1 − e −10 ∴ 8 = 9 (1 − e −1000 /( R1 +100) ), ∴

t /( R1 +100)

)

1 = e−1000 /( R1 +100) 9

1000 = 2.197, R1 = 355.1Ω R1 + 100

t > 1ms : vc = 8e −10 ∴

6

6

t ′ /( R2 +100)

, t ′ = t − 10−3 ∴1 − 8e −1000 (R 2 + 100)

1000 = 2.079, R 2 = 480.9 − 100 = 380.9Ω R 2 + 100



91.


10 March 2006

vx , L = 200e −2000t V vx ,c = 100(1 − e −1000t )V vx = v x , L − vx , c = 0 ∴ 200e −2000t = 100 − 100e−1000t ∴100e −1000t + 200(e−1000t ) 2 − 100 = 0, −100 ± 10, 000 + 80, 000 = −0.25 ± 0.75 400 ∴ e −1000t = 0.5, t = 0.6931ms

e−1000t =




10 March 2006

92. P(t < 0 ) = I 2 R = 0.0012 × 103 = 0.001 W Vinit = I .R = 7 × 10 − 3 × 900 = 6.3 V Pinit =

V2 = 0.08 W R

V final = 7 × 10 − 3 × 900Ω // 1000Ω = 3.3 V Pfinal =

V2 = 0.02 W R Power (W) 8 6 4 2

0

7

Time (ms)



93. For


10 March 2006

t < 0, the voltage across all three capacitors is simply 9 (4.7)/ 5.7 = 7.421 V. The circuit time constant is τ = RC = 4700 (0.5455×10-6) = 2.564 ms. When the circuit was first constructed, we assume no energy was stored in any of the capacitors, and hence the voltage across each was zero. When the switch was closed, the capacitors began to charge according to ½ Cv2. The capacitors charge with the same current flowing through each, so that by KCL we may write C1

dv1 dv dv = C2 2 = C3 3 dt dt dt

With no initial energy stored, integration yields the relationship C1v1 = C2v2 = C3v3 throughout the charging (i.e. until the switch is eventually opened). Thus, just prior to the switch being thrown at what we now call t = 0, the total voltage across the capacitor string is 7.421 V, and the individual voltages may be found by solving: v1 + v2 + v3 = 7.421 =0 10-6 v1 – 2×10-6 v2 -6 -6 2×10 v2 – 3×10 v3 = 0 so

that v2 = 2.024 V. With the initial voltage across the 2-uF capacitor now known, we may write -3

v(t) = 2.024 e-t/ 2.564×10 V (a)

v(t = 5.45 ms) = 241.6 mV.

(b)

The voltage across the entire capacitor string can be written as 7.421 e-t/ 2.564×10 V. Thus, the voltage across the 4.7-kΩ resistor at t = 1.7 ms = 3.824 V and the dissipated power is therefore 3.111 mW.

(c)

Energy stored at t = 0 is ½ Cv2 = 0.5(0.5455×10-6)(7.421)2 = 15.02 μJ.

-3



94. voltage


10 March 2006

follower ∴ vo (t ) = v2 (t )

v2 (t ) = 1.25 u (t )V = vo (t ) vx (t ) = 1.25e −10 = 1.25e

−10,000 t

6

/ 0.5× 200

u (t )

u (t )V



95.


10 March 2006

This is a voltage follower ∴ vo (t ) = v2 (t ) , where v2(t) is defined at the non-inverting input. The time constant of the RC input circuit is 0.008(1000+250) = 10 s. Considering initial conditions: vC(0-) = 0 ∴ vC(0+) = 0. Applying KVL at t = 0+, 5

v250/250 + v= 2/1000.

at t = 0+ v250 = v2, we find that v2(0+) = 1 V. t → ∞, v2→ 0, so we may write,

Since As

vo(t) = v2(t) = 1.0e-t/10 u(t) V. PSpice

verification:

In plotting both the hand-derived result and the PSpice simulation result, we see that the ideal op amp approximation holds very well for this particular circuit. Although the 741 contains internal capacitors, it does not introduce any shorter time constants than that of the input circuit. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


96. For


10 March 2006

t < 0, the current source is an open circuit and so i1 = 40/ 50 = 0.8 A. The current through the 5-Ω resistor is [40 – 10(0.8)]/ 5 = 7.2 A, so the inductor current is equal to – 7.2 A PSpice Simulation

From the PSpice simulation, we see that our t < 0 calculation is indeed correct, and find that the inductor current at t = 50 ms is 7.82 A.



97. (a)

v1 = 0 (virtual gnd) ∴ i =


4 −20,000t e u (t )A 104

4 −20,000t e dt = −0.2e −20,000t o 10 4 ∴ vc (t ) = 0.2(1 − e −20,000t ) u (t ) ∴ vc = 107 ∫

10 March 2006

t

t o

∴ vR (t ) = 103 i (t ) = 0.4e −20,000t u (t )V ∴ vo (t ) = −vc (t ) − vR (t ) = (−0.2 + 0.2e −20,000t − 0.4e −20,000t ) u (t ) 3

And we may write vo(t) = -0.2[1 + e-20×10 t]u(t) V. (b) PSpice verification:

We can see from the simulation result that our ideal op amp approximation is not providing a great deal of accuracy in modeling the transient response of an op amp in this particular circuit; the output was predicted to be negative for t > 0.



98.


10 March 2006

One possible solution of many: implement a capacitor to retain charge; assuming the light is left on long enough to fully charge th e capacitor, th e stored ch arge will run the lightbulb after the wall switch is turned off. Taking a 40-W light bulb connected to 115 V, we estimate th e res istance of the light bu lb (which chan ges with its tem perature) as 330.6 Ω. We define “on” for the light bulb som ewhat arbitrarily as 50% intensity, taking intensity as proportional to the dissipated pow er. Thus, we need at least 20 W (246 mA or 81.33 V) to the light bulb for 5 seconds after the light switch is turned off.

The circuit above contains a 1-M Ω resistor in parallel with the capacitor to allow current to flow through the light bulb when the light switch is on. In order to determ ine the required capacitor size, we first recognise that it will see a Thevenin equivalent resistance of 1 M Ω || 330.6 Ω = 330.5 Ω. We want vC(t = 5s) = 81.33 = 115 e-5/τ, so we need a circuit time constant of t = 14.43 s and a capacitor value of τ/ Rth = 43.67 mF.



99.


10 March 2006

Assume at least 1 μA required otherwise alarm triggers. Add capacitor C.

vc (1) = 1 volt vc (0) =

1000 .1.5 = 1.496 volts 1002.37

∴ We have 1 = 1.496e

−

1 106 C

or C =

1 = 2.48μF 10 ln(1.496) 6



100.


10 March 2006

(a) Note that negativ e tim es are not perm itted in PSpice. The only way to m situation is to shift the time axis by a fixed amount, e.g., t ′ = t + 1 .

odel this

(b) Negative times are not permitted in PSpice. The only way to model this situation is to shift the time axis by a fixed amount, e.g., t ′ = t + 2 .



101. (a)


10 March 2006

τ = L / R = 0.1 s . This is much less than either the period or pulsewidth.

(b)



102. (a)


10 March 2006

τ = L / R = 1 s . This is much less than either the period or pulsewidth.

(b)




10 March 2006

103.



104.

τ=


10 March 2006

L 500 ×10−6 = = 34 ns Req 14.7 ×103 6

The transient response will therefore have the form Ae −29.4×10 t . (a)

(b)

(c)



1.

Chapter Nine Solutions

10 March 2006

Parallel RLC circuit: 1 1 1 = = = 175 × 103 s −1 −6 2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (2.857)(10−6 )

(a)

α=

(b)

ω0 =

1 = LC

1

( 2 ×10 )(10 ) −3

−6

=

22.4 krad/s

(c) The circuit is overdamped since α > ω0 .



2.


10 March 2006

Parallel RLC circuit: (a) For an underdamped response, we require α < ω0 , so that 1 < 2 RC

1 LC

or R >

1 L 1 2 ; R> . 2 C 2 10−12

Thus, R > 707 kΩ. (b) For critical damping, 1 L R= = 707 kΩ 2 C



3. (a)

10 March 2006

Parallel RLC circuit:

α=

ω0 =

(b)


1 1 1 = = = −6 2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (1)(10−9 )

5 × 108

1 = LC

31.6 Trad/s

1

(10 )(10 ) −12

−9

= 3.16 × 1013 rad/s =

−1

s

s1,2 = −α ± α 2 − ω02 = −0.5 × 109 ± j 1021 − (0.25)(1018 ) = −0.5 ± j 31.62 Grad/s

(c) The circuit is underdamped since α < ω0 .



4.


10 March 2006

Parallel RLC circuit: (a) For an underdamped response, we require α < ω0 , so that

1 < 2 RC

1 LC

or R >

1 L 1 10−15 . ; R> 2 C 2 2 ×10−18

Thus, R > 11.18 Ω. (b) For critical damping, 1 L R= = 11.18 Ω 2 C (c) For overdamped, R < 11.18 Ω.



5.


10 March 2006

ω o L = 10Ω, s1 = −6s −1 , s2 = −8s −1 ∴−6 = α + α 2 − ω o2 , − 8 = −α − α 2 − ω o2 adding, −14 = −2α ∴α = 7 s −1 ∴−6 = −7 + 49 − ω o2 ∴ω o2 = 48

1 , ω o = 6.928 LC

rad/s∴ 6.928 L = 10, L = 1.4434H, 1 1 C= = 14.434mF, = 7 ∴ R = 4.949Ω 48L 2RC



6.

(a)


10 March 2006

ic = 40e −100t − 30e −200t mA, C = 1mF, v(0) = −0.25V t 1 t 0.25 (40e −100t − 30e−200t ) dt − 0.25 i dt − = c ∫ ∫ o o C ∴ v(t ) = −0.4(e −100t − 1) + 0.15(e −200t − 1) − 0.25

v(t ) =

∴ v(t ) = −0.4e−100t + 0.15e−200t V (b)

s1 = −100 = −α + α 2 − ωo2 , s2 = −200 = −α − α 2 − ωo2 ∴−300 = −2α, α = 150s − 1 ∴150 +

1 500 = 3.333Ω Also, ,R = −3 2R10 150

−200 = −150 − 22500 − ωo2 ∴ωo2 = 20000 ∴ 20000 = ∴ i R (t ) =

(c)

1 100 = , L = 0.5H LC L

v = 0.12e −100t + 0.045e −200t A R

(i)t = −iR (t ) − ic (t ) = (0.12 − 0.04)e −100t + (−0.045 + 0.03)e−200t ∴ i (t ) = 80e−100t − 15e−200t mA, t > 0



7. (a)

(b)


10 March 2006

Parallel RLC with ωo = 70.71 × 1012 rad/s. L = 2 pH. 1 = (70.71× 1012 ) 2 LC 1 So 100.0 C= = 12 2 (70.71×10 ) (2 ×10−12 ) ωo2 =

aF

1 = 5 × 109 s −1 2 RC 1 So 1R = 10 = MΩ (10 ) (100 ×10−18 ) α=

(c)

α is the neper frequency: 5 Gs-1

(d)

S1 = −α + α 2 − ωo2 = −5 × 109 + j 70.71× 1012 s −1 S 2 = −α − α 2 − ωo2 = −5 × 109 − j 70.71× 1012 s −1

(e)

ζ=

α 5 ×109 = = 7.071× 10−5 12 ωo 70.71×10



8. Given:

L = 4 R 2C , α =


10 March 2006

1 2 RC

Show that v(t ) = e −αt ( A1t + A2 ) is a solution to C

d 2 v 1 dv 1 + + v=0 dt 2 R dt L

[1]

dv = e −αt ( A1 ) − αe −αt ( A1t + A2 ) dt = ( A1 − α A1t − α A2 ) e −αt

[2]

2

d v = ( A1 − α A1t − αA2 ) (−αe −αt ) − α A1e −αt 2 dt = −α ( A1 − α A2 + A1 − α A1t ) e −αt = −α (2 A1 − α A2 − α A1t )e −αt

[3]

Substituting Eqs. [2] and [3] into Eq. [1], and using the information initially provided, 2

1 1 ⎛ 1 ⎞ −αt (2 A1 ) e −αt + ⎜ ( A1 ) e −αt ⎟ ( A1t + A2 ) e + RC 2 RC ⎝ 2 RC ⎠ 1 1 ( A1t + A2 ) e −αt + 2 2 ( A1t + A2 ) e−αt − 2 RC 4R C =0 −

Thus, v(t ) = e −αt ( A1t + A2 ) is in fact a solution to the differential equation. Next, with v(0) = A2 = 16 dv = ( A1 − α A2 ) = ( A1 − 16α ) = 4 and dt t =0 we find that A1 = 4 + 16α



9.


10 March 2006

Parallel RLC with ωo = 800 rad/s, and α = 1000 s-1 when R = 100 Ω. 1 2 RC 1 ωo2 = LC α=

so C = 5μF so

L = 312.5 mH

Replace the resistor with 5 meters of 18 AWG copper wire. From Table 2.3, 18 AWG soft solid copper wire has a resistance of 6.39 Ω/1000ft. Thus, the wire has a resistance of ⎛ 100 cm ⎞ ⎛ 1in ⎞ ⎛ 1ft ⎞⎛ 6.39 Ω ⎞ (5 m) ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1m ⎠ ⎝ 2.54 cm ⎠ ⎝ 12in ⎠⎝ 1000 ft ⎠ = 0.1048 Ω or 104.8 mΩ

(a)

The resonant frequency is unchanged, so ωo = 800 rad/s

(b)

α=

(c)

ζ old =

α old ωo

ζ new =

α new ωo

1 = 954.0 × 103 s −1 2 RC

Define the percent change as =

ζ new − ζ old × 100 ζ old

α new − α old × 100 α old

= 95300%




10.

L = 5H, R = 8Ω, C = 12.5mF, v(0+ ) = 40V

(a)

i (0+ ) = 8A: α =

10 March 2006

1 1000 1 = = 5, ωo2 = = 16, 2RC 2 × 8 × 12.5 LC

ωo = 4 s1,2 = −5 ± 25 − 16 = −2, − 8 ∴ v(t ) = A1e −2t + A 2 e −8t 1000 ⎛ 40 ⎞ + ⎜ −iL (0 ) − ⎟ = 80 (−8 − 5) = −1040 12.5 ⎝ 8 ⎠ v / s = −2A1 − 8A 2 ∴−520 = − A1 − 4A 2 ∴−3A 2 = −480, A 2 = 160, A1 = −120

∴ 40 = A1 + A 2 v′(0+ ) =

∴ v(t ) = −120e −2t + 160e −8t V, t > 0

(b)

v(0+ ) 40 = = 5A R 8 ∴ i (0+ ) = A 3 + A 4 = −iR (0+ ) − ic (0+ ) = −8 − 5 = −13A; ic (0+ ) = 8A Let i( t ) = A 3e −2t + A 4 e −8t ; iR (0+ ) =

40 = 8 A / s ∴ 4 = − A 3 − 4A 4 5 ∴−3A 4 = −13 + 4, A 4 = 3, A 3 = −16 ∴ i (t ) = −16e −2t + 3e −8t A, t > 0 i (0+ ) = −2A 3 − 8A 4 =



11. (a)

RC =

Therefore


10 March 2006

1 L 1 10−3 1 10 = 1.581 Ω = = 2 C 2 10−4 2

R

= 0.1RC = 158.1 mΩ

(b)

α=

1 = 3.162 ×104 s −1 and ω0 = 2RC

1 = 3.162 × 103 rad/s LC

s1,2 = −α ± α 2 − ω02 = − 158.5 s −1 and − 6.31× 104 s −1

Thus,

So we may write i (t ) = A1e−158.5t + A2 e−6.31×10

4

t

With i(0− ) = i(0+ ) = 4 A and v(0− ) = v(0+ ) = 10 V A1+ A2 = 4

Noting

v(0+ ) = L

(

di = 10 dt t =0

[1]

)

10−3 −158.5 A1 − 6.31× 104 A2 = 10 [2] Solving Eqs. [1] and [2] yields A1 = 4.169 A and A2 = –0.169 A So that

i (t ) = 4.169e −158.5t − 0.169e −6.31×10

4

t

A



12. (a)

α=


1 = 500 s −1 and ω0 = 2RC

10 March 2006

1 = 100 rad/s LC

s1,2 = −α ± α 2 − ω02 = − 10.10 s −1 and − 989.9 s −1

Thus,

So we may write iR (t ) = A1e−10.1t + A2 e −989.9t [1] With i(0− ) = i (0+ ) = 2 mA and v(0− ) = v(0+ ) = 0 A1+ A2 = 0 We need to find

iC (0+ ) = C

Thus, Therefore,

diR dt

. Note that

[2] diR ( t )

t =0

dt

=

1 dv dv [3] and iC = C = −i − iR . dt R dt

dv v(0+ ) = −i (0+ ) − iR (0+ ) = −2 × 10−3 − = −2 × 10−3 [4] dt t =0+ R

we may write based on Eqs. [3] and [4]:

diR = (50)(−0.04) = −2 [5]. Taking the derivative of Eq. [1] and combining with dt t =0 [6]. Eq. [5] then yields: s1 A1 + s 2 A2 = −2

Solving Eqs. [2] and [6] yields A1 = −2.04 mA and A2 = 2.04 mA So that (b)

(

iR (t ) = −2.04 e −10.1t − e−989.9t (c)

)

mA We see that the simulation agrees.



13.

i (0) = 40A, v(0) = 40V, L =

(a)

α=


10 March 2006

1 H, R = 0.1Ω, C = 0.2F 80

1 80 = 25, ωo2 = = 400, 2 × 0.1× 0.2 0.2

ωo = 20, s1,2 = −25 ± 625 − 400 = 10, − 40 ∴ v(t ) = A1 e −10t + A 2 e−40t ∴ 40 = A1 + A 2 ; 1⎛ v(0) ⎞ ⎜ i (0) − ⎟ = −2200 C⎝ R ⎠ ∴−A1 − 4A 2 = −220 ∴ − 3A 2 = −180 ∴ A 2 = 60, A1 = −20 v′(0+ ) = −10A1 − 40A 2 v′(0+ ) =

∴ v(t ) = −20e −10t + 60e −40t V, t > 0

(b)

dv = 200e −10t − 600e −40t − 0.2(-20)(-10)e -10t − (0.2)(60)(-40)e −40t dt −10 t = 160e − 120e −40t A

i(t) = – v/ R – C



α=

14. (a)


1 = 6.667 × 108 s −1 and ω0 = 2RC

10 March 2006

1 = 105 rad/s LC

s1,2 = −α ± α 2 − ω02 = − 7.5 s −1 and − 1.333 × 109 s −1 . So we may write

Thus,

9

iC (t ) = A1e −7.5t + A2 e−1.333×10 t [1] With i (0− ) = i (0+ ) = 0 A and v(0− ) = v(0+ ) = 2 V , 2 iC (0+ ) = −iR (0+ ) = − = −0.133 × 106 so that −6 15 × 10 A1+ A2 = –0.133×106 [2] We need to find

C

diR dt

. We know that L t =0

di dt

di 2 = = 106 . Also, dt t =0 2 ×10−6

= 2 so t =0

(

)

9 i di di 1 dv 1 dv = iC and R = so R = C = A1e−7.5t + A2 e−1.333×10 t . dt dt R dt dt CR CR

di di diR diC + + = 0 so C dt dt dt dt

Using

= −7.5 A1 − 1.33 × 109 A2 = −106 − t =0

1 ( A1 + A2 ) [3] CR

Solving Eqs. [2] and [3] yields A1 = −0.75 mA and A2 = –0.133 MA (very different!) So that

(

9

iC (t ) = − 0.75 ×10−3 e −7.5t + 0.133 × 106 e−1.333×10

t

)

A

(b)



15. (a) Thus,

α=

1 = 0.125 s −1 and ω0 = 2RC


10 March 2006

1 = 0.112 rad/s LC

s1,2 = −α ± α 2 − ω02 = − 0.069 s −1 and − 0.181 s −1 . So we may write

v(t ) = A1e−0.069t + A2 e −0.181t [1] With iC (0− ) = iC (0+ ) = −8 A and v(0− ) = v(0+ ) = 0 , A1+ A2 = 0 [2] We need to find

diR dt

. We know that t =0

dv = 4 ⎡⎣ −0.069 A1e −0.069t − 0.181A2 e−0.181t ⎤⎦ . So, dt [3] iC (0) = 4 [ −0.069 A1 − 0.181A2 ] = −8

iC (t ) = C

Solving Eqs. [2] and [3] yields A1 = −17.89 V and A2 = 17.89 V So that (b)

v(t ) = −17.89 ⎣⎡ e −0.069t − e−0.181t ⎦⎤ V

dv = 1.236e −0.069t − 8.236e−0.181t . We set this equal to 0 and solve for tm: dt 3.236 e −0.069tm = = e0.112tm , so that tm = 8.61 s. 1.236 e −0.181tm Substituting into our expression for the voltage, the peak value is v(8.61) = –6.1 V (c) The simulation agrees with the analytic results.



16.


10 March 2006

100 = 2A, vc (0) = 100V 50 106 3 × 106+3 α= = 4000, wo2 = = 12 × 106 2 × 50 × 2.5 100 × 2.5 3 16 − 12 × 10 = 200, s1,2 = −4000 ± 2000

iL (0) =

∴ iL (t ) = A1e −2000t + A 2 e −6000t , t > 0 ∴ A1 + A s = 2 −103 × 3 × 100 = −3000 = −2000A1 − 6000A 2 ∴−1.5 = − A1 − 3A 2 ∴ 0.5 = −2A 2 100 ∴ A 2 = −0.25, A1 = 2.25 ∴ iL (t ) = 2.25e −2000t − 0.25e−6000t A, t > 0 iL′ (0+ ) =

t > 0: iL (t ) = 2A ∴ iL (t ) = 2u (−t ) + (2.25e−2000t − 0.25e−6000t ) u (t )A



17.


10 March 2006

12 = 2A, vc (0) = 2V 5 +1 1000 1000 × 45 = 250, ω o2 = = 22500 α= 2 × 1× 2 2

iL (0) =

s1,2 = −250 ± 2502 − 22500 = −50, − 450 s −1 ∴ iL = A1e −50t + A 2 e −450t ∴ A1 + A 2 = 2; iL′ (0+ ) = 45(−2) = −50A1 − 450A 2 ∴ A1 + 9A 2 = 1.8 ∴−8A 2 = 0.2 ∴ A 2 = −0.025, A1 = 2.025(A) ∴ iL (t ) = 2.025e −50t − 0.025e −450t A, t > 0




10 March 2006

18. (a)

1 1440 1440 = = 20, ω o2 = = 144 2RC 72 10 = −20 ± 400 − 144 = −4, − 36: v = A1e −4t + A 2 e −36t

α= s1,2

⎛ 1 18 ⎞ v(0) = 18 = A1 + A2 , v′(0) = 1440 ⎜ − ⎟ = 0 ⎝ 2 36 ⎠ ∴ 0 = −4A1 − 36A 2 = − A1 − 9A 2 = ∴ 18 = −8A 2 , A 2 = −2.25, A1 = 20.25 +

∴ v(t ) = 20.25e −4 − 2.25e −36t V, t > 0

(b)

v 1 v′ = 0.5625e−4t − 0.0625e−36t − 0.05625e−4t + 0.05625e−36t + 36 1440 ∴ i (t ) = 0.50625e−4t − 0.00625e −36t A, t > 0

(c)

vmax at t = 0 ∴ vmax = 18V ∴ 0.18 = 20.25e−4ts − 2.25e −36ts

i (t ) =

Solving using a scientific calculator, we find that ts = 1.181 s.



19. so

Referring to Fig. 9.43, ωo = 1 α=

LC

= 4 rad/s


10 March 2006

L = 1250 mH Since α > ωo, this circuit is over damped.

1 = 5 s −1 2 RC

The capacitor stores 390 J at t = 0−: 1 Wc = C vc2 2 2Wc = 125 V = vc (0+ ) So (vc 01 ) = C The inductor initially stores zero energy, so

iL (0− ) = iL (0+ ) = 0 S1,2 = −α ± α 2 − ωo2 = −5 ± 3 = −8, − 2

Thus, v(t ) = Ae −8t + Be−2t Using the initial conditions, v(0) = 125 = A + B v(0+ ) + ic (0+ ) = 0 iL (0+ ) + iR (0+ ) + ic (0+ ) = 0 + 2 125 v(0+ ) + =− = −62.5 V So (ic 0 ) = − 2 2 dv = 50 ×10−3[−8 Ae −8t − 2 Be −2t ] ic = C dt + [2] ic (0 ) = −62.5 = −50 × 10−3 (8 A + 2 B)

Solving Eqs. [1] and [2],

[1]

A = 150 V B = −25 V

Thus, v(t ) = 166.7e −8t − 41.67e −2t , t > 0



20.


10 March 2006

We want a response v = Ae −4t + Be−6t 1 α= = 5 s −1 2 RC

(a)

S1 = −α + α 2 − ωo2 = −4 = −5 + 25 − ωo2 S2 = −α − α 2 − ωo2 = −6 = −5 − 25 − ωo2 Solving either equation, we obtain ωo = 4.899 rad/s Since (b) with

ωo2 =

1 1 , L = 2 = 833.3 mH LC ωoC

If iR (0+ ) = 10 A and ic (0+ ) = 15 A, find A and B. iR (0+ ) = 10 A, vR (0+ ) = v(0+ ) = vc (0+ ) = 20 V v(0) = A + B = 20 [1] dv ic = C = 50 × 10−3 (−4 Ae −4t − 6 Be−6t ) dt + ic (0 ) = 50 × 10−3 (−4 A − 6 B) = 15 [2] Solving, A = 210 V, B = −190 V v= Thus, 210

e −4t − 190e −6t , t > 0



21. Initial

conditions:


iL (0− ) = iL (0+ ) = 0

(a)

vc (0+ ) = vc (0− ) = 2(25) = 50 V

(b)

ic (0+ ) = −iL (0+ ) − iR (0+ ) = 0 − 2 = −2 A

(c)

t > 0: parallel (source-free) RLC circuit 1 α= = 4000 s −1 2 RC 1 ωo = = 3464 rad/s LC

iR (0+ ) =

10 March 2006

50 =2 A 25

s1,2 = −α ± α 2 − ωo2 = −2000, − 6000

α > ω0, this system is overdamped. Thus,

Since

vc (t ) = Ae −2000t + Be −6000t dv = (5 × 10−6 ) (−2000 Ae −2000t − 6000 Be −6000t ) dt [1] ic (0+ ) = −0.01A − 0.03B = −2 ic = C

and v( c 0+ ) = A + B = 50

[2]

Solving, we find A = −25 and B = 75 so that vc (t ) = −25e −2000t + 75e−6000t , t > 0 (d)

(e)

−25e −2000t + 75e −6000t = 0 ⇒ t = 274.7 μs using a scientific calculator

(f)

vc

So, we

max

= −25 + 75 = 50 V

solving | −25e −2000ts + 75e −6000ts | = 0.5 in view of the graph in part (d), find ts = 1.955 ms using a scientific calculator’s equation solver routine.



22.


10 March 2006

Due to the presence of the inductor, vc (0− ) = 0 . Performing mesh analysis, →i1

→i2 4.444 H

Rearranging, we obtain 2i1 – 2i2 = 0 and

−9 + 2i1 − 2i2 = 0

[1]

2i2 − 2i1 + 3iA + 7i2 = 0

[2]

and i1 − i2 = iA

–4i1 + 6 i2 = 0. Solving, i1 = 13.5 A and i2 = 9 A.

(a)

iA (0− ) = i1 − i2 = 4.5 A and iL (0− ) = i2 = 9 A

(b)

t > 0:

around left mesh:

−vc (0+ ) + 7iA (0+ ) − 3iA (0+ ) + 2iA (0+ ) = 0

4.444 H

so, iA (0+ ) = 0 (c)

vc (0− ) = 0 due to the presence of the inductor.

(d) −vLC + 7 − 3(1) + 2 = 0 1A

(e)

vLC = 6 V ∴ RTH =

1 = 3.333 s −1 2 RC 1 ωo = = 3 rad/s LC

6 = 6Ω 1

α=

S1,2 = −α ± α 2 − ωo2 = −1.881, − 4.785

Thus, iA (t ) = Ae −1.881t + Be−4.785t iA (0+ ) = 0 = A + B

[1]

To find the second equation required to determine the coefficients, we write: iL = −ic − iR = −C

dvc − iA = −25 ×10−3 ⎡ −1.881(6 A)e−1.881t − 4.785(6 B)e−4.785t ⎤ ⎣ ⎦ dt −1.881t −4.785t − Be - Ae

iL (0+ ) = 9 = −25 × 10−3[−1.881(6 A) − 4.785(6 B)] − A − B or 9 = -0.7178A – 0.2822B [2] So

Solving Eqs. [1] and [2], A = −20.66 and B = +20.66 that iA (t ) = 20.66[e −4.785t − e−1.881t ]



23.


Diameter of a dime: approximately 8 mm.

Capacitance L = 4μH ωo =

10 March 2006

Area = π r 2 = 0.5027cm 2

ε r εo A (88) (8.854 ×10−14 F/cm) (0.5027cm 2 ) = = d 0.1cm = 39.17pF

1 = 79.89 Mrad/s LC

For an over damped response, we require α > ωo. Thus,

1 > 79.89 × 106 2 RC 1 R< −12 2(39.17 × 10 ) (79.89 × 106 )

or R < 159.8 Ω *Note: The final answer depends quite strongly on the choice of εr.



24.

(a) For critical damping, R =

α=

(b)


1 L 1 10−3 = = 4.564 Ω . 2 C 2 12 × 10−6

1 1 = = 9.129 ×103 s −1 −6 2 RC 2 ( 4.564 ) 12 × 10

(

)

vC ( t ) = e −9.129×10 t ( A1t + A2 ) 3

Thus,

t = 0, vC (0) = A1 ( 0 ) + A2 = 12

At

10 March 2006

[1] ∴ A2 = 12 V .

Taking the derivative of Eq. [1], dvC ( t ) dt and

= e −9.129×10 t ⎡⎣ −9.129 × 103 A1t + A1 − 9.129 × 103 (12 ) ⎤⎦ 3

also iC = −(iR + iL ) , so dvC dt

Solving,

=− t =0

1 ⎛ vC (0) 1 ⎞ ⎛ 12 ⎞ + 0⎟ = − + 0 ⎟ A1 − 9.129 × 103 (12 ) ⎜ −6 ⎜ C⎝ R 12 × 10 ⎝ 4.565 ⎠ ⎠

A1 = −109.6 × 103 V , so we may write

(

)

vC ( t ) = e −9.129×10 t −109.6 ×103 t + 12 . 3

(c) We see from plotting both the analytic result in Probe and the simulated voltage, the two are in excellent agreement (the curves lie on top of one another).



25. (b)

(a) For critical damping, R =

α=

10 March 2006

1 L 1 10−8 = = 1.581 mΩ . 2 C 2 10−3

1 1 = = 3.162 ×105 s −1 −3 −3 2 RC 2 1.581× 10 10

(

)(

)

iL ( t ) = e−3.162×10 t ( A1t + A2 ) 5

Thus, At


[1]

t = 0, iL (0) = A1 ( 0 ) + A2 = 10

∴ A2 = 10 A .

Taking the derivative of Eq. [1], diL ( t ) dt and

= e −3.162×10 t ⎡⎣ −3.162 × 105 A1t + A1 − 3.162 × 105 (10 ) ⎤⎦ [2]

also L

5

diL dt

= vC (0) = 0 [3], so t =0

Solving Eqs. [2] and [3], A1 = 3.162 × 105 (10 ) = 3.162 × 106 V , so we may write

(

(

)

)

iL ( t ) = e −3.162×10 t 3.162 ×106 t + 10 . 5

(c) We see from plotting both the analytic result in Probe and the simulated voltage, the two are in reasonable agreement (some numerical error is evident).



26.


10 March 2006

It is unlikely to observe a critically damped response in real-life circuits, as it would be virtually impossible to obtain the exact values required for R, L and C. However, using carefully chosen components, it is possible to obtain a response which is for all intents and purposes very close to a critically damped response.



27.

crit. damp. (b)


10 March 2006

L

(a)

L = 4R 2 C = 4 × 1× 2 × 10−3 = 8mH

1 1000 = = 250 ∴ iL = e −250t (A1t + A 2 ) 2RC 2 ×1× 2 iL (0) = 2A, vc (0) = 2V ∴ iL = e−250t (A1t + 2)

α = ωo

Then 8 × 10−3 iL′ (0+ ) = −2 = 8 × 10−3 (A1 − 500), = e −1.25 (1.25 + 2) = 0.9311A

(c)

iL max : (250tm + 2) = 0, 1 = 250tm + 2, tm < 0 No! ∴ tm = 0, iL max = 2A ∴ 0.02 = e −250ts (250ts + 2); SOLVE: ts = 23.96ms




10 March 2006

28.

R

crit. damp.

(b)

(a)

L = 4R 2 C =

100 × 10−3 = 4R 2 ×10−6 ∴ R = 57.74Ω 3

1 × 2.5 = 3464 s −1 30 ∴ vc (t ) = e −3464t (A1t + A 2 ) vc (0) = 100V

ω o = α = 103 /

100 = 1.7321A ∴100 = A 2 57.74 106 ⎛ 100 ⎞ 5 vc′ (0+ ) = ⎜ 1.7321 − ⎟ = 0 = A1 − 3464A 2 ∴ A1 = 3.464 × 10 2.5 ⎝ 57.74 ⎠ t 3464 ∴ vc (t ) = e − (3.464 × 105 t + 100) V, t > 0 iL (0) =



29.


10 March 2006

Diameter of a dime is approximately 8 mm. The area, therefore, is πr2 = 0.5027 cm2. ε r εo A (88) (8.854 × 10−14 ) (0.5027) = The capacitance is d 0.1 = 39.17 pF

w

ith L = 4μH, ωo =

1 = 79.89 Mrad/s LC

For critical damping, we require or R =

1 = 159.8 Ω 2ωoC

1 = ωo 2 RC




30.

L = 5mH, C = 10−8 F, crit. damp. v(0) = −400V, i (0) = 0.1A

(a)

L = 4R 2 C = 5 × 10−3 = 4R 2 10−8 ∴ R = 353.6Ω

(b)

10 March 2006

108 α= = 141, 420 ∴ i = e −141,420t (A1t + A 2 ) 2 × 353.6 ∴ A 2 = 0.1∴= e −141,421t (A1t + 0.1), 5 × 10−3 (A1 − 141, 420 × 0.1) = −400 ∴ A1 = −65,860 ∴ i = e −141,421t (−65,860t + 0.1). i′ = 0 ∴ e −α t (+65860) + 141, 420e −α t (−65,860tm + 0.1) = 0 ∴ tm = 8.590 μ s ∴ i (tm ) = e −141,420×8.590×10

−6

(−65,860 × 8.590 ×10−6 + 0.1) = −0.13821A ∴ i

(c)

max

= i (tm ) = 0.13821A

∴imax = i (0) = 0.1A



31.


10 March 2006

Critically damped parallel RLC with α = 10−3 s −1 , R = 1MΩ .

1 103 = 10−3 , so 500 C= = 2 RC 2 ×106 1 Since α = ωo, ωo = = 10−3 LC 1 or 10 = −6 LC so L = 2 GH (!) We know

μF

μN 2 A = 2 × 109 L= S 2

⎡⎛ 50 turns ⎞ ⎤ ⎛ 1m ⎞ 2 (4π× 10 H/m) ⎢⎜ ⎟ ⎟ . s ⎥ (0.5cm) .π . ⎜ ⎝ 100 cm ⎠ ⎣⎝ cm ⎠ ⎦ If So s = 2 ×109 −7

(4π2 ×10−9 ) (50) 2 (0.5) 2 s = 2 × 109 So 8.106 s=

×1013 cm




10 March 2006

32. 1 4 1 4 × 13 = = 1, ω o2 = = = 26, ω d = 26 − 1 = 5 2RC 2 × 2 LC 2 ∴ vc (t ) = e − t ( B1 cos 5t + B2 sin 5t )

α=

(a)

iL (0+ ) = iL (0) = 4A

(b)

vc (0+ ) = vc (0) = 0

(c)

iL′ (0+ ) =

1 vc (0+ ) = 0 L

(d)

vc′ (0+ ) =

⎡ vc (0+ ) ⎤ 1 + + [−iL (0 ) − iR (0 )] = 4 ⎢ −4 − ⎥ = 4 (−4 + 0) = −16 V/s c 2 ⎦ ⎣

(e)

∴ (e) 0 = 1(B1 )∴ B1 = 0, vc (t ) = B2 e − t sin 5t , vc′ (0 + ) = B2 (5) = −16 ∴ B2 = −3.2, vc (t ) = −3.2e − t sin 5t V, t > 0

(f)



33.

α=


10 March 2006

1 106 1 106+3 = = 4000, ω o2 = = = 2 × 107 2RC 100 × 2.5 LC 50

ω d = 20 × 106 − 16 × 106 = 2000 ∴ ic = e −4000t (B1 cos 2000t + B2 sin 2000t ) iL (0) = 2A, vc (0) = 0 ∴ ic (0+ ) = −2A; ic′ (0+ ) = −iL′ (0+ ) − iR′ (0+ ) 1 1 1 2 ×106 vc (0) − vc′ (0+ ) = 0 − ic (0+ ) = L R RC 125 6 2 ×10 ∴ B1 = −2A, = 16, 000 = 2000B2 + (−2) (−4000) ∴ B2 = 4 125 ∴ ic (t ) = e −4000t (−2 cos 2000t + 4sin 2000t )A, t > 0 ∴ ic′ (0+ ) = −




10 March 2006

34. (a)

(b)

1 100 1 100 2 = = 8, ωo2 = = , ωd = 36 = ωo2 − 64 2RC 12.5 LC L 100 ∴ωo2 = 100 = ∴ L = 1H L α=

t < 0: iL (t ) = 4A; t > 0: iL (t ) = e −8t (B1 cos 6t + B2 sin 6t ) iL (0) = 4A ∴ B1 = 4A, iL = e −8t (4 cos 6t + B2 sin 6t ) vc (0) = 0 iL′ (0+ ) = t vc (0+ ) = 0 ∴ 6B2 − 8(4) = 0, B2 = 16 / 3 ∴ iL (t ) = 4u (−t ) + e −8t (4 cos 6t + 5.333sin 6t ) u (t ) A

(c)




10 March 2006

35. (a)

α

1 109−3 1 109 = = 5000, ωo2 = = = 1.25 × 108 2RC 2 × 20 × 5 LC 1.6 × 5

ωd = ωo2 − α 2 = 125 × 106 − 25 × 106 = 10, 000 ∴ vc (t ) = e −5000t (B1 cos104 t + B2 sin104 t ) vc (0) = 200V, iL (0) = 10mA ∴ vc (t ) = e −5000t (200 cos104 t + B2 sin104 t ) 1 109 vc′ (0+ ) = ic (0+ ) = 5 c =

vc (0) ⎤ ⎡ ⎢iL (0) − 20, 000 ⎥ ⎣ ⎦

109 ⎛ −2 200 ⎞ = 0 = 104 B2 − 200 (5000) 10 − ⎜ ⎟ 5 ⎝ 20, 000 ⎠

∴ B2 = 100V ∴ vc (t ) = e −5000t (200 cos104 t + 100sin104 t ) V, t > 0 (b)

isw = 10−2 − iL , iL =

1 vc + Cvc′ R

vc′ = e −5000t [104 (−200sin + 100 cos] − 5000 (200 cos + 100sin)] = e −500t [106 (−2sin − 0.5cos)] = −2.5 × 106 e−5000t sin104 t v / s ⎡ 1 ⎤ (200 cos + 100sin) − 5 × 10−9 × 2.5 × 106 e−5000t sin104 t ⎥ ∴ iL = e −5000t ⎢ ⎣ 20, 000 ⎦ 4 4 −5000 t =e (0.01cos10 t − 0.0075sin10 t ) A ∴ isw = 10 − e −5000t (10 cos104 t − 7.5sin104 t ) mA, t > 0




10 March 2006

36. (a)

α=

1 106 1 1.01× 106 = = 20, ωo2 = = = 40, 400 2RC 2000 × 25 LC 25

ωd = ωo2 − α 2 = 40, 400 − 400 = 200 ∴ v = e −20t (A1 cos 200t + A 2 sin 200t ) v(0) = 10V, iL (0) = 9mA ∴ A1 = 10V ∴ v = e −20t (10 cos 200t + A 2 sin 200t ) V, t > 0 v′(0+ ) = 200A 2 − 20 ×10 = 200 (A 2 − 1) =

1 io (0+ ) C

106 (−10−3 ) = −40 ∴ A 2 = 1 − 0.2 = 0.8 25 ∴ v(t ) = e −20t (10 cos 200t + 0.8sin 200t ) V, t > 0 =

(b)

v = 10.032e−20t cos (200t − 4.574°)V T=

2π = 3.42ms 200



37.


10 March 2006

1 106−3 1 = = 100 s −1 , ωo2 = = 1.01× 106 2RC 2 × 5 LC 60 ∴ωd = 101× 104 − 104 = 100; iL (0) = = 6mA 10 vc (0) = 0 ∴ vc (t ) = e −100t (A1 cos1000t + A 2 sin1000t ), t > 0 α=

∴ A1 = 0, vc (t ) = A 2 e −100t sin1000t 1 1 ic (0+ ) = 106 [−i1 (0+ ) − vc (0+ )] = 106 C 5000 (−6 ×10−3 ) = −6000 = 1000 A 2 ∴ A 2 = −6

vc′ (0+ ) =

∴ vc (t ) = −6e −100t sin1000tV, t > 0 ∴ i1 (t ) = −

1 104

vc (t ) = −10−4 (−6) e −100t sin1000tA ∴ i1 (t ) = 0.6e −100t sin1000t mA, t > 0



38.


10 March 2006

We replace the 25-Ω resistor to obtain an underdamped response:

α =

1 2RC

and

1 10 × 10− 6 R

Thus,

1 ; we require α < ω0. LC

ω0 =

< 3464

or

R > 34.64 mΩ.

For R = 34.64 Ω (1000× the minimum required value), the response is: v(t) = e-αt (A cos ωdt + B sin ωdt) where α = 2887 s-1 and ωd = 1914 rad/s. iL(0+) = iL(0-) = 0 and vC(0+) = vC(0-) = (2)(25) = 50 V = A. dvL dv = L C dt dt −αt = L e (− Aω d t sin ω d t + Bω d t cos ω d t ) - αe −αt ( A cos ω d t + Bsinω d t )

iL(t) = L

[

]

50 × 10−3 iL(0 ) = 0 = [B ω d - αA ], so that B = 75.42 V. 3 +

Thus,

v(t) = e-2887t (50 cos 1914t + 75.42 sin 1914t) V. Sketch of v(t).

PSpice schematic for t > 0 circuit.

From PSpice the settling time using R = 34.64 Ω is approximately 1.6 ms.



39.


10 March 2006

v(0) = 0; i (0) = 10A v = eαt (A cos ωd t + Bsin ωd t ) ∴ A = 0, v = Be −αt sin ωd t v′ = e −αt [−α Bsin ωd t + ωd B cos ωd t ] = 0 ∴ tan ωd t = tm 2 = tm1 +

ωd 1 ω , tm1 = tan −1 d α ωd α

1 π Td = tm1 + ; 2 ωd

vm1 = Be −αtm1 sin ωd tm1 vm 2 = − Be −αtm1 −απ / ωd sin ωd tm1 ∴

vm 2 v 1 = −e −απ / ωd ; let m 2 = Vm1 vm1 100

ωd 1 21 ln 100; α = = , 2RC R π 1 21 ln100 6R 2 − 441 ω02 = = 6 ∴ωd = 6 − 441/ R 2 ∴ LC R πR

∴ eαπ / ωd = 100, α =

2 ⎡ ⎛ 21π ⎞ ⎤ ∴ R = 1/ 6 ⎢ 441 + ⎜ ⎟ ⎥ = 10.3781Ω To keep ⎝ 100 ⎠ ⎥⎦ ⎢⎣

vm 2 < 0.01, chose R = 10.3780Ω v′(0+ ) = ωd vm1 2

0 ⎛ 21 ⎞ ⎛ ⎞ B = B 6−⎜ ⎟ = 4R ⎜10 + ⎟ ∴ B = 1.380363 10.3780 ⎠ ⎝ 10.378 ⎠ ⎝ 2

α=

21 ⎛ 21 ⎞ = 2.02351; ωd = 6 − ⎜ ⎟ = 1.380363 10.378 ⎝ 10.378 ⎠

∴ v = 304.268e −2.02351t sin 1.380363t v tm1 = 0.434 s, vm1 = 71.2926v Computed values show ts = 2.145sec; vm 2 = 0.7126 < 0.01vm1



40. (a)


10 March 2006

For t < 0 s, we see from the circuit that the capacitor and the resistor are shorted by the presence of the inductor. Hence, iL(0-) = 4 A and vC(0-) = 0 V. When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that α = 1/2RC = 0.4 s -1 and ω0 = 5.099 rad/s. Since α < ω0, the response will be underdamped with ωd = 5.083 rad/s. Assume the form iL(t) = e-αt (C cos ωdt + D sin ωdt) for the response.

and

ith iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that di vC(t) = vL(t) = L L dt so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)]

W

ith vC(0+) = 0 = (2/13) (5.083D – 0.4C), we obtain D = 0.3148 A.

W

Thus, (b)

iL(t) = e-0.4t (4 cos 5.083t + 0.3148 sin 5.083t) A and iL(2.5) = 1.473 A.

α = 1/2RC = 4 s-1 and ω0 = 5.099 rad/s. Since α < ω0, the new response will still be underdamped, but with ωd = 3.162 rad/s. We still may write vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)] and so with vC(0+) = 0 = (2/13) (3.162D – 4C), we obtain D = 5.06 A.

Thus, (c)

iL(t) = e-4t (4 cos 3.162t + 5.06 sin 3.162t) A and iL(.25) = 2.358 A. We see from the simulation result below that our hand calculations are correct; the slight disagreement is due to numerical inaccuracy. Changing the step ceiling from the 10-ms value employed to a smaller value will improve the accuracy.




10 March 2006

41. (a,b) For t < 0 s, we see from the circuit below that the capacitor and the resistor are shorted by the presence of the inductor. Hence, iL(0-) = 4 A and vC(0-) = 0 V.

When the 4-A source turns off at t = 0 s, we are left with a parallel RLC circuit such that α = 1/2RC = 1 s-1 and ω0 = 5.099 rad/s. Since α < ω0, the response will be underdamped with ωd = 5 rad/s. Assume the form iL(t) = e-αt (C cos ωdt + D sin ωdt) for the response.

and

ith iL(0+) = iL(0-) = 4 A, we find C = 4 A. To find D, we first note that di vC(t) = vL(t) = L L dt so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)]

W Thus,

ith vC(0+) = 0 = (2/13) (5D – 4), we obtain D = 0.8 A. iL(t) = e-t (4 cos 5t + 0.8 sin 5t) A

W

We see that the simulation result confirms our hand analysis; there is only a slight difference due to numerical error between the simulation result and our exact expression.

(c)

Using the cursor tool, the settling time is approximately 4.65 s.



42. vc (0) = 50 + 80 × 2 = 210 V, iL (0) = 0, α =


10 March 2006

R 80 = = 20 2L 4

100 = 500 : ω d = 500 − 202 = 10 2 ∴ vc (t ) = e −20t (A1 cos10t + A 2 sin10t ) ∴ A1 = 210 V

ω o2 =

1 ic (0+ ) = 0 C −20 t ∴ 0 = 10A 2 − 20 (210), A 2 = 420 ∴ vc (t ) = e (210 cos10t + 420sin10t )

∴ vc (t ) = e −20t (210 cos10t + A 2 sin10t ); vc′ (0+ ) =

∴ vc (40ms) = e −0.8 (210 cos 0.4 + 420sin 0.4) = 160.40 V Also, i(L = e −20t B1 cos10t + B2 sin10t ), iL (0+ ) =

1 1 1 vL (0+ ) = [0 − vc (0+ )] = × 210 L 2 2

∴ iL′ (0+ ) = −105 = 10B2 ∴ B2 = 10.5 ∴ iL (t ) = −10.5e −20t sin10t A, t > 0 ∴ vR (t ) = 80iL = 840e −20t sin 10tV ∴ vR (40ms) = −840e −0.8 sin 0.4 = −146.98 V vL (t ) = −vc (t ) − vc (t ) − vR (t ) ∴ vL (40ms) = −160.40 + 146.98 = −13.420 V [check: vL = e −20t (−210 cos− 420sin + 840sin) = e −20t (−210 cos10t + 420sin10t ) V, t > 0 ∴ vL (40ms) = e −0.8 (−210 cos− 420sin + 840 sin) = e −20t (−210 cos10t + 420sin10t )V, t > 0 ∴ VL (40ms) = e −0.8 (420sin 0.4 − 210 cos 0.4) = −13.420 V Checks]



43. Series:


10 March 2006

R 2 1 4 = = 4, ω o2 = = = 20, ω d = 20 − 16 = 2 2L 1/ 2 LC 0.2 ∴ iL = e −4t (A1 cos 2t + A 2 sin 2t ); iL (0) = 10A, vc (0) = 20V

α=

∴ A1 = 10; iL′ (0+ ) =

1 vL (0+ ) = 4 (20 − 20) = 0 L

∴ iL′ (0+ ) = 2A 2 − 4 × 10 ∴ A 2 = 20 ∴ iL (t ) = e −4t (10 cos 2t + 20sin 2t )A, t > 0



44. (a)


10 March 2006

R2 1 1 = ω o2 = ∴ L = R 2C 2 4L LC 4 1 200 ∴ L = × 4 × 104 −6 = 0.01H, α = = 104 = ω o 4 0.02 ∴ vc (t ) = e −10000t (A1t + A 2 ); vc (0) = −10V, iL (0) = −0.15A crit. damp; α 2 =

∴ A 2 = −10, vc (t ) = e −10000t (A1t − 0); vc′ (0+ ) = −

1 C

iL (0) = −106 (−0.15) = 150, 000 Now, (vc′ 0+ ) = A1 + 105 = 150, 000 ∴ A1 = 50, 000 ∴ vc (t ) = e −10,000t (50, 000t − 10) V, t > 0 (b)

vc′ (t ) = e −10,000t [50, 000 − 10, 000 (50, 000t − 10)] =∴ 5 = 50, 000tm − 10 ∴ tm =

15 = 0.3ms 50, 000

∴ vc (tm ) = e −3 (15 − 10) = 5e −3 = 0.2489V vc (0) = −10V ∴ vc

(c)

max

= 10V

vc ,max = 0.2489V



45.


10 March 2006

“Obtain an expression for vc(t) in the circuit of Fig. 9.8 (dual) that is valid for all t′′. μF Ω

α=

mF

A

R 0.02 ×106 106 × 3 = = 4000, ω o2 = = 1.2 × 107 2L 2 × 2.5 2.5 × 10

∴ s1,2 = −4000 ± 16 ×106 − 12 × 106 = −2000, − 6000 1 × 100 = 2V 50 1 iL (0) = 100A ∴ 2 = A1 + A 2 , vc′ (0+ ) = C 3 (−iL (0)) = − × 103 × 100 = −3000v / s 100 ∴−3000 = −200A1 − 600A 2 , − 1.5 = − A1 − 3A 2 ∴ vc (t ) = A1e −2000t + A 2 e−6000t ; vc (0) =

∴ 0.5 = −2A 2 , = −0.25, A1 = 2.25 ∴ vc (t ) = (2.25e −200t − 0.25e−6000t ) u (t ) + 2u (−t ) V (checks)



46. (a)


10 March 2006

R 2 1 = = 1, ω o2 = = 5, ω d = ω o2 − α 2 = 2 2L 2 LC −t ∴ iL = e (B1 cos 2t + B2 sin 2t ), iL (0) = 0, vc (0) = 10V

α=

∴ B1 = 0, iL = B2 e− t sin 2t 1 i1 (0) = vL (0+ ) = vR (0+ ) − Vc (0+ ) = 0 − 10 = 2B2 1 ∴ B2 = 5 ∴ iL = −5e− t sin 2tA, t > 0 (b)

iL′ = −5[e − t (2 cos 2t − sin 2t )] = 0 ∴ 2 cos 2t = sin 2t , tan 2t = 2 ∴ t1 = 0.5536 s, iL (t1 ) = −2.571A 2t2 = 2 × 0.5536 + π , t2 = 2.124, iL (t2 ) = 0.5345 ∴ iL

max

= 2.571A

and i0.5345A L max =



47. (a)

α=


10 March 2006

R 250 1 106 = = 25, ω o2 = = = 400 2L 10 LC 2500

s1,2 = −α ± α 2 − ω o2 = −25 ± 15 = −10, −40 ∴ iL = A1 e −10t + A 2 e−40t , iL (0) = 0.5A, vc (0) = 100V 1 1 ∴ 0.5 = A1 + A 2 , iL′ (0+ ) = vL (0+ ) = 5 5 (100 − 25 − 100) = −5 A / s = −10A1 − 40A 2 ∴ 5 = 10 A1 + 40 (0.5 − A1 ) = 10A1 − 40 A1 + 20 ∴−30A1 = −15, A1 = 0.5, A 2 = 0 ∴ iL (t ) = 0.5e−10t A, t > 0

(b)

vc = A 3e −10t + A 4 e−40t ∴100 = A 3 + A 4 ; 1 106 vc′ = ic′ (0+ ) (−0.5) = −1000 c 500 ∴−10A 3 − 40A 4 = −1000 ∴−3A 4 = 0, A 4 = 0, A 3 = 100 ∴ vc (t ) = 100e−10t V t > 0



48.


10 March 2006

Considering the circuit as it exists for t < 0, we conclude that vC(0-) = 0 and iL(0-) = 9/4 = 2.25 A. For t > 0, we are left with a parallel RLC circuit having α = 1/2RC = 0.25 s-1 and ωo = 1/ LC = 0.3333 rad/s. Thus, we expect an underdam ped response with ωd = 0.2205 rad/s: iL(t) = e-αt (A cos ωdt + B sin ωdt) iL(0+) = iL(0-) = 2.25 = A so iL(t) = e–0.25t (2.25 cos 0.2205t + B sin 0.2205t) In order to determine B, we must invoke the remaining boundary condition. Noting that di vC(t) = vL(t) = L L dt = (9)(-0.25)e-0.25t (2.25 cos 0.2205t + B sin 0.2205t) + (9) e-0.25t [-2.25(0.2205) sin 0.2205t + 0.2205B cos 0.2205t] vC(0+) = vC(0-) = 0 = (9)(-0.25)(2.25) + (9)(0.2205B) so B = 2.551 and iL(t) = e-0.25t [2.25 cos 0.2205t + 2.551 sin 0.2205t] A Thus, iL(2) = 1.895 A This answer is borne out by PSpice simulation:



49.


10 March 2006

We are presented with a series RLC circuit having α = R/2L = 4700 s-1 and ωo = 1/ LC = 447.2 rad/s; therefore we expect an overdamped response with s1 = -21.32 s-1 and s2 = -9379 s-1. From the circuit as it exists for t < 0, it is evident that iL(0-) = 0 and vC(0-) = 4.7 kV

Thus,

vL(t) = A e–21.32t + B e-9379t

[1]

With iL(0+) = iL(0-) = 0 and iR(0+) = 0 we conclude that vR(0+) = 0; this leads to vL(0+) = -vC(0-) = -4.7 kV and hence A + B = -4700 [2] di Since vL = L , we may integrate Eq. [1] to find an expression for the inductor current: dt B − 9379t ⎤ 1 ⎡ A − 21.32t iL(t) = e e ⎥ L ⎢⎣ 21.32 9379 ⎦ 1 B ⎤ ⎡ A t = 0+, iL = 0 so we have = 0 [3] At -3 ⎢ 500 × 10 ⎣ 21.32 9379 ⎥⎦ Simultaneous solution of Eqs. [2] and [3] yields A = 10.71 and B = -4711. Thus,

vL(t) = 10.71e-21.32t - 4711 e-9379t V,

t>0

and the peak inductor voltage magnitude is 4700 V.



50.


10 March 2006

With the 144 mJ originally stored via a 12-V battery, we know that the capacitor has a value of 2 mF. The initial inductor current is zero, and the initial capacitor voltage is 12 V. We begin by seeking a (painful) current response of the form

ibear = Aes1t + Bes2t Using our first initial condition, ibear(0+) = iL(0+) = iL(0-) = 0 = A + B

di/dt = As1 es1t + Bs2 es2t vL = Ldi/dt = ALs1 es1t + BLs2 es2t vL(0+) = ALs1 + BLs2 = vC(0+) = vC(0-) = 12 What else is known? We know that the bear stops reacting at t = 18 μs, meaning that the current flowing through its fur coat has dropped just below 100 mA by then (not a long shock). Thus, A exp[(18×10-6)s1] + B exp[(18×10-6)s2] = 100×10-3 Iterating, we find that Rbear = 119.9775 Ω. This corresponds to A = 100 mA, B = -100 mA, s1 = -4.167 s-1 and s2 = -24×106 s-1



51.


10 March 2006

Considering the circuit at t < 0, we note that iL(0-) = 9/4 = 2.25 A and vC(0-) = 0. 1 1 = , which, with For a critically damped circuit, we require α = ωo, or 2RC LC L = 9 H and C = 1 F, leads to the requirement that R = 1.5 Ω (so α = 0.3333 s-1). The inductor energy is given by wL = ½ L [iL(t)]2, so we seek an expression for iL(t):

iL(t) = e-αt (At + B) that iL(0+) = iL(0-) = 2.25, we see that B = 2.25 and hence

Noting

iL(t) = e-0.3333t (At + 2.25) Invoking the remaining initial condition requires consideration of the voltage across the capacitor, which is equal in this case to the inductor voltage, given by:

vC(t) = vL(t) = L

diL = 9(-0.3333) e-0.3333t (At + 2.25) + 9A e-0.3333t dt

vC(0+) = vC(0-) = 0 = 9(-0.333)(2.25) + 9A so A = 0.7499 amperes and iL(t) = e-0.3333t (0.7499t + 2.25) A Thus,

iL(100 ms) = 2.249 A and so wL(100 ms) = 22.76 J




10 March 2006

v ⎛ 50 ⎞ to t = 0, we find that v = (10 + i1 ) ⎜ ⎟ and i1 = 5 ⎝ 15 ⎠ ⎛ 10 ⎞ 500 so v = 100 V . v ⎜1 − ⎟ = ⎝ 15 ⎠ 15

52. Prior Thus,

Therefore, vC (0+ ) = vC (0− ) = 100 V, and iL (0+ ) = iL (0− ) = 0. The circuit for t > 0 may be reduced to a simple series circuit consisting of a 2 mH inductor, 20 nF capacitor, and a 10 Ω resistor; the dependent source delivers exactly the current to the 5 Ω that is required.

α=

Thus,

ω0 =

and

R 10 = = 2.5 × 103 s −1 2 L 2 2 ×10−3

(

1 = LC

)

1

( 2 ×10 )( 20 ×10 ) −3

−9

= 1.581× 105 rad/s

ith α < ω0 we find the circuit is underdamped, with

W

ωd = ω02 − α 2 = 1.581× 105 rad/s We may therefore write the response as iL (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t ) At

t = 0, iL = 0 ∴ B1 = 0 .

Noting

that L

Finally,

diL dt

diL d −α t = ( e B2 sin ωd t ) = B2 e −α t ( −α sin ωd t + ωd cos ωd t ) and dt dt

= −100 we find that B2 = -0.316 A. B

t =0

iL (t ) = −316e−2500t sin1.581× 105 t mA



53. Prior


10 March 2006

to t = 0, we find that vC = 100 V, since 10 A flows through the 10 Ω resistor.

Therefore, vC (0+ ) = vC (0− ) = 100 V, and iL (0+ ) = iL (0− ) = 0. The circuit for t > 0 may be reduced to a simple series circuit consisting of a 2 mH inductor, 20 nF capacitor, and a 10 Ω resistor; the dependent source delivers exactly the current to the 5 Ω that is required to maintain its current.

α=

Thus,

ω0 =

and

R 10 = = 2.5 × 103 s −1 −3 2 L 2 2 ×10

(

1 = LC

)

1

( 2 ×10 )( 20 ×10 ) −3

−9

= 1.581× 105 rad/s

ith α < ω0 we find the circuit is underdamped, with

W

ωd = ω02 − α 2 = 1.581× 105 rad/s We may therefore write the response as vC (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t ) At

t = 0, vC = 100 ∴ B1 = 100 V .

Noting

dvC = iL and dt

that C

d −α t ⎡e (100 cos ωd t + B2 sin ωd t ) ⎤⎦ dt ⎣ = e −α t ⎡⎣ −α (100 cos ωd t + B2 sin ωd t ) − 100ωd sin ωd t + B2ωd cos ωd t ⎤⎦

which is equal to zero at t = 0 (since iL = 0) we find that B2 = 1.581 V . B

Finally,

vC (t ) = e−2500t ⎡⎣100 cos (1.581×105 t ) + 1.581sin (1.581×105 t ) ⎤⎦ V



54. Prior Thus, After =


10 March 2006

to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V. vC(0+) = vC(0–) = 7.5 + 5 = 12.5 V and iL = 0

iL . We may replace the 4 dependent current source with a 0.5 Ω resistor. Thus, we have a series RLC circuit with R 1.25 Ω, C = 1 F, and L = 3 H. t = 0 we are left with a series RLC circuit where i1 = −

R 1.25 = = 0.208 s −1 2L 6 1 1 ω0 = = = 577 mrad/s 3 LC

α=

Thus, and

ith α < ω0 we find the circuit is underdamped, so that

W

ωd = ω02 − α 2 = 538 mrad/s We may therefore write the response as iL (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t ) At

t = 0, iL = 0 ∴ B1 = 0 A .

Noting

,

diL dt

that L

= −vC (0) and t =0

v (t ) −12.5 diL d −α t = ⎣⎡ e ( B2 sin ωd t ) ⎦⎤ = B2 e−α t [ −α sin ωd t + ωd cos ωd t ] = C = (t = 0) dt dt L 3 we find that B2 = –7.738 V. B

Finally,

iL (t ) = 1.935e −0.208t sin 0.538t A for t > 0 and 2.5 A, t < 0



55. Prior Thus, After =


10 March 2006

to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V. vC(0+) = vC(0–) = 12.5 V and iL = 0

iL . We may replace the 4 dependent current source with a 0.5 Ω resistor. Thus, we have a series RLC circuit with R 1.25 Ω, C = 1 mF, and L = 3 H. t = 0 we are left with a series RLC circuit where i1 = −

R 1.25 = = 0.208 s −1 2L 6 1 1 ω0 = = = 18.26 rad/s −3 LC 3 10 × ( )

α=

Thus, and

ith α < ω0 we find the circuit is underdamped, so that

W

ωd = ω02 − α 2 = 18.26 rad/s We may therefore write the response as vC (t ) = e −α t ( B1 cos ωd t + B2 sin ωd t ) At

t = 0, vC = 12.5 ∴ B1 = 12.5 V .

Noting

that dvC d −α t = ⎡⎣ e ( B1 cos ωd t + B2 sin ωd t ) ⎤⎦ dt dt −α t = −α e [12.5cos ωd t + B2 sin ωd t ] + e−α t [ −12.5ωd sin ωd t + ωd B2 cos ωd t ] and this expression is equal to 0 at t = 0, we find that B2 = 0.143 V. B

Finally,

vC (t ) = e −0.208t [12.5cos18.26t + 0.143sin18.26t ] V for t > 0 and 12.5 V, t < 0



56. (a)

Series, driven: α =


10 March 2006

R 100 = = 500, 2L 0.2

1 10 × 106 = = 250, 000 LC 40 ∴ Crit. damp iL ( f ) = 3(1 − 2) = −3,

ω o2 =

iL (0) = 3, vc (0) = 300V ∴ iL = −3 + e −500t (A1t + A 2 ) ∴ 3 = −3 + A 2 , A 2 = 6A 1 [vc (0) − vR (0+ )] = 0 L −5000 t ∴ A1 = 3000 e ∴ iL (t ) = −3 + e −500t iL (0+ ) = A1 − 300 = (3000t + 6), t > 0 ∴ iL (t ) = 3u (−t ) + [−3 + e −500t (3000t + 6)] u (t )A

(b)

e−500to (3000to + 6) = 3; by SOLVE, to = 3.357ms



57. vc (0) = 0, iL (0) = 0, α =


10 March 2006

R 2 1 = = 4, ω o2 = = 4 × 5 = 20 2L 0.5 LC

∴ω d = 20 − 16 = 2 ∴ iL (t ) = e −4t (A1 cos 2t + A 2 sin 2t ) + iL , f iL , f = 10A ∴ iL (t ) = 10 + e −4t (A1 cos 2t + A 2 sin 2t ) ∴ 0 = 10 + A1 , A1 = −10, iL (t ) = 10 + e −4t (A 2 sin 2t − 10 cos 2t ) iL (0+ ) =

1 vL (0+ ) = 4 × 0 = 0 ∴ iL (0+ ) = 0 = 2A 2 + 40, A 2 = −20 L

iL(t) = 10 - e-4t (20 sin 2t + 10 cos 2t) A, t > 0




10 March 2006

58.

R 250 1 106 = = 25, ωo2 = = = 400 2L 10 LC 2500 s1,2 = −25 ± 625 − 400 = −10, −40

α=

iL (0) = 0.5A, vc (0) = 100V, iL , f = −0.5A ∴ iL (t ) = −0.5 + A1e −10t + A 2 e −40t A t = 0+ : vL (0+ ) = 100 − 50 × 1 − 200 × 0.5 = −50V ∴−50 = 5iL′ (0+ ) ∴ iL′ (0+ ) = −10 ∴−10 = −10A1 − 40A 2 , 0.5 = −0.5 + A1 + A 2 ∴ A1 + A 2 = 1∴−10 = −10A 2 − 40 (−1+A1 ) = −50A1 + 40, A1 = 1, A 2 = 0 ∴ iL (t ) = −0.5 + 1e −10t A, t > 0; iL (t ) = 0.5A, t > 0



59. α=


10 March 2006

1 106 1 106+3 = = 4000, ωo2 = = = 20 × 106 2RC 100 × 2.5 LC 50

∴ωd = ωo2 − α 2 = 2000, iL (0) = 2A, vc (0) = 0 ic , f = 0, (vc , f = 0) ∴ ic = e −400t (A1 cos 2000t + A 2 sin 2000t ) work with vc : vc (t ) = e −4000t (B1 cos 2000t + B2 sin 2000t ) ∴ B1 = 0 1 106 + (2 × 1) = 8 × 105 ∴ vc = B2 e sin 2000t , vc′ (0 ) = ic (0 ) = C 2.5 −4000 t 5 ∴ 8 × 10 = 2000B2 , B2 = 400, vc = 400e sin 2000t −4000 t

+

∴ ic (t ) = Cvc′ = 2.5 × 10−6 × 400e −4000t (−4000sin 200t + 2000 cos 200t ) = 10−6+3+3 e −4000t (−4sin 2000t + 2 cos 2000t ) = e −4000t (2 cos 2000t − 4sin 2000t ) A, t > 0



60. (a)

α=


10 March 2006

1 8 × 106 8 ×106 × 13 2 = = 1000, ω = = 26 × 106 o 3 2RC 2 × 4 ×10 4

∴ωd = 26 − 1 × 103 = 5000, vc (0) = 8V iL (0) = 8mA, vc , f = 0 ∴ vc = e−1000t (A1 cos1000t + A 2 sin 5000t ) 1 8 − 0.008) = 0 ic (0+ ) = 8 × 106 (0.01 − C 4000 ∴ 5000A 2 − 1000 × 8 = 0, A 2 = 1.6 ∴ A1 = 8; vc′ (0+ ) =

So vc(t) = e-1000t (8 cos 1000t + 1.6 sin 1000t) V, t > 0 (b)




10 March 2006

61.

R 1 1 = = 1, ωo2 = = 1∴ crit. damp 2L 1 LC 5 vc (0) = × 12 = 10V, iL (0) = 2A, vc , f = 12V 6 1 1 ∴ vc (t ) = 12 + e − t (A1t − 2); vc′ (0+ ) = ic (0+ ) = × iL (0+ ) = 1 C 2 −t ∴1 = A1 + 2; A1 = −1∴ vc (t ) = 12 − e (t + 2) V, t > 0 α=



62. (a)


10 March 2006

1 106 vs = 10u (−t ) V : α = = = 1000 2RC 2000 × 0.5 1 2 ×106 × 3 ωo2 = = = 0.75 × 106 ∴ s1,2 = −500, − 1500 LC 8 ∴ vc = A1e −500t + A 2 e −1500t , vo (0) = 10V, iL (0) = 10mA ∴ A1 + A 2 = 10, vc′ (0+ ) = 2 ×106 [iL (0) − iR (0+ )] = 2 × 106 10 ⎞ ⎛ ⎜ 0.01 − ⎟ = 0 ∴−500A1 − 1500 A 2 = 0, 1000 ⎠ ⎝ − A1 − 3A 2 = 0; add: − 2 A 2 = 10, A 2 = −5, A1 = 15 ∴ vc (t ) = 15e −500t − 5e−1500t V t > 0 ∴ iR (t ) = 15e −500t − 5e−1500t mA, t > 0

(b)

vs = 10u (t ) V, vc , f = 10, vc = 10 + A 3e−500t + A 4 e−1500t , vc (0) = 0, iL (0) = 0 ∴ A 3 + A 4 = −10V, vc′ (0+ ) = 2 × 106 [iL (0) − iR (0+ )] = 2 × 106 (0 − 0) = 0 = −500A 3 − 1500A 4 ∴− A 3 − 3A 4 = 0, add: − 2A 4 = −10, A 4 = 5 ∴ A 3 = −15 ∴ vc (t ) = 10 − 15e −500t + 5e −1500t V, t > 0 ∴ iR (t ) = 10 − 15e −500t + 5e1500t mA, t > 0



63. (a)


10 March 2006

1 106 = = 1000 2RC 1000 1 106 × 3 3 ωo2 = = ∴ s1,2 = −1000 ± 106 − × 106 = −500, − 1500 LC 4 4 −500 t −1500 t vc , f = 0 ∴ vc = A1e + A2e , vc (0) = 10V, iL (0) = 0 vs (t ) = 10u (−t ) V: α =

10 ⎤ ⎡ ∴10 = A1 + A 2 , vc′ = 106 ic (0+ ) = 106 ⎢0 − = −2 × 104 ⎥ ⎣ 500 ⎦ 4 ∴−2 × 10 = −500A1 − 1500A ∴ 40 = A1 + 3A 2 ∴ 30 = 2A 2 , A 2 = 15, A1 = −5 ∴ vc = −5e −500t + 15e −1500t V, t > 0 ∴ is = ic = Cvc′ ∴ is = 10−6 (2500e −500t − 22,500e −1500t ) = 2.5e −500t − 22.5e −1500t mA, t > 0

(b)

vs (t ) = 10u (t ) V ∴ vc , f = 10V, vc (0) = 0, iL (0) = 0 ∴ vc = 10 + A3 e −500t + A 4 e−1500t ∴ A 3 + A 4 = −10 10 ⎞ ⎛ 4 vc′ (0+ ) = 106 ic (0+ ) = 106 ⎜ 0 + ⎟ = 2 × 10 = −500 A 3 − 1500 A 4 500 ⎝ ⎠ ∴− A 3 − 3A 4 = 40, add: 2− A 4 = 30, A 4 = −15, A 3 = 5, vc = 10 + 5e −500t − 15e−1500t V, is = ic = 10−6 (−2500e −500t + 22,500e −1500t ) = 25e −500t + 22.5e−1500t mA, t > 0



64.


10 March 2006

Considering the circuit at t < 0, we see that iL(0-) = 15 A and vC(0-) = 0. The circuit is a series RLC with α = R/2L = 0.375 s -1 and ω0 = 1.768 rad/s. W e therefore of the expect an underdam ped response with ωd = 1.728 rad/s. The general form response will be vC(t) = e-αt (A cos ωdt + B sin ωdt) + 0

(vC(∞) = 0)

vC(0+) = vC(0-) = 0 = A and we may therefore write vC(t) = Be-0.375t sin (1.728t) V iC(t) = -iL(t) = C At Thus,

dvC = (80×10-3)(-0.375B e-0.375t sin 1.728t dt

t = 0+, iC = 15 + 7 – iL(0+) = 7 = (80×10-3)(1.728B) so that B = 50.64 V. vC(t) = 50.64 e–0.375t sin 1.807t V and vC(t = 200 ms) = 16.61 V. The energy stored in the capacitor at that instant is ½ CvC2 = 11.04 J




65. (a)

vS(0-) = vC(0-) = 2(15) = 30 V

(b) Thus,

iL(0+) = iL(0-) = 15 A iC(0+) = 22 – 15 = 7 A and vS(0+) = 3(7) + vC(0+) = 51 V

10 March 2006

(c) As t → ∞, the current through the inductor approaches 22 A, so vS(t→ ∞,) = 44 A. (d) We are presented with a s eries RLC circuit having α = 5/2 = 2.5 s -1 and ωo = 3.536 rad/s. The natural response will therefore be underdamped with ωd = 2.501 rad/s. iL(t) = 22 + e-αt (A cos ωdt + B sin ωdt) iL(0+) = iL(0-) = 15 = 22 + A so A = -7 amperes Thus, iL(t) = 22 + e-2.5t (-7 cos 2.501t + B sin 2.501t) di di vS(t) = 2 iL(t) + L L = 2iL + L = 44 + 2e-2.5t (-7cos 2.501t + Bsin 2.501t) dt dt -2.5t – 2.5e (-7cos 2.501t + Bsin 2.501t) + e-2.5t [7(2.501) sin 2.501t + 2.501B cos 2.501t)] vS(t) = 51 = 44 + 2(-7) – 2.5(-7) + 2.501B so B = 1.399 amperes and hence vS(t) = 44 + 2e-2.5t (-7cos 2.501t + 1.399sin 2.501t) -2.5e-2.5t (-7cos 2.501t + 1.399sin 2.501t) + e-2.5t [17.51sin 2.501t + 3.499cos 2.501t)] and vS(t) at t = 3.4 s = 44.002 V. This is borne out by PSpice simulation:



66. For


10 March 2006

t < 0, we have 15 A dc flowing, so that iL = 15 A, vC = 30 V, v3Ω = 0 and vS = 30 V. This is a series RLC circuit with α = R/2L = 2.5 s-1 and ω0 = 3.536 rad/s. We therefore expect an underdamped response with ωd = 2.501 rad/s. 0 0, vC(t) = A e-2000t + B e-8000t - 50 vC(0+) = vC(0-) = 75 = A + B – 50 so A + B = 125 [1] dvC = -2000 Ae-2000t – 8000 Be-8000t dt iC(0+) = C

dvC dt

= 3 – 5 – iL(0-) = -5 = -25×10-6 (2000A + 8000B) t =0+

Thus, 2000A + 8000B = 5/25×10-6

[2]

Solving Eqs. [1] and [2], we find that A = 133.3 V and B = -8.333 V. Thus,

vC(t) = 133.3 e-2000t – 8.333 e-8000t – 50 and vC(1 ms) = -31.96 V. This is confirmed by the PSpice simulation shown below.



71.


10 March 2006

α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞) ωo2 = 0.05 therefore ωd = 0.223 rad/s. We anticipate a response of the form: v(t) = A cos 0.2236t + B sin 0.2236t v(0+) = v(0-) = 0 = A therefore v(t) = B sin 0.2236t dv/dt = 0.2236B cos 0.2236t;

iC(t) = Cdv/dt = 0.4472B cos 0.2236t

iC(0+) = 0.4472B = -iL(0+) = -iL(0-) = -1×10-3 so B = -2.236×10-3 and thus v(t) = -2.236 sin 0.2236t mV In designing the op amp stage, we first write the differential equation: dv 1 t v dt ′ + 10-3 + 2 = 0 (iC + iL = 0) ∫ 0 dt 10 and then take the derivative of both sides: d 2v 1 = - v 2 dt 20 dv With = (0.2236)(−2.236 × 10− 3 ) = −5 × 10 − 4 , one possible solution is: dt t = 0 +

PSpice sim ulations ar e very sens itive to par ameter v alues; bette r re sults were obta ined using LF411 instead of 741s (both were compared to the simple LC circuit simulation.)

Simulation using 741 op amps

Simulation using LF411 op amps



72.


10 March 2006

α = 0 (this is a series RLC with R = 0, or a parallel RLC with R = ∞) ωo2 = 50 therefore ωd = 7.071 rad/s. We anticipate a response of the form: v(t) = A cos 7.071t + B sin 7.071t, knowing that iL(0-) = 2 A and v(0-) = 0. v(0+) = v(0-) = 0 = A therefore v(t) = B sin 7.071t dv/dt = 7.071B cos 7.071t;

iC(t) = Cdv/dt = 0.007071B cos 7.071t

iC(0+) = 0.007071B = -iL(0+) = -iL(0-) = -2 so B = -282.8 and thus v(t) = -282.8 sin 7.071t V In designing the op amp stage, we first write the differential equation: dv 1 t v dt ′ + 2 + 10-3 = 0 (iC + iL = 0) ∫ 0 dt 20 and then take the derivative of both sides: d 2v = - 50v dt 2 dv With = (7.071)(−282.8) = −2178 , one possible solution is: dt t = 0 +




10 March 2006

73.

(a)

v dv + 3.3 × 10- 3 1000 dt or

= 0

dv 1 = v dt 3.3 (b) One possible solution:



74.


10 March 2006

We see either a series RLC with R = 0 or a parallel RLC with R = ∞; either way, α = 0. ωd = 0.5477 rad/s (com bining the tw o inductors in parallel for the calculation). We expect a response of the form i(t) = A cos ωdt + B sin ωdt.

ω02 = 0.3 so

i(0+) = i(0-) = A = 1×10-3 di/dt = -Aωd sin ωdt + Bωd cos ωdt vL = 10di/dt = -10Aωd sin ωdt + 10Bωd cos ωdt vL(0+) = vC(0+) = vC(0-) = 0 = 10B(0.5477) so that B = 0 and hence i(t) = 10-3 cos 0.5477t A The differential equation for this circuit is

and

1 1 dv vdt ′ + 10-3 + ∫ vdt ′ + 2 ∫ 10 0 20 dt t

di dt

=0 t =0+

t

or d 2v = − 0.3v dt 2

= 0

↓ i 1Ω

One possible solution is:



75. (a)

vR = vL

20(-

iL) = 5

diL dt


10 March 2006

diL = - 4iL dt

or

(b) We expect a response of the form iL(t) = A e-t/ τ where τ = L/R = 0.25. We know that iL(0-) = 2 amperes, so A = 2 and iL(t) = 2 e-4t diL = -4(2) = -8 A/s. dt t = 0 + One possible solution, then, is

8V 1 μF

1 MΩ

i ↓ 1Ω

4 kΩ

1 kΩ



Chapter Ten Solutions

10 March 2006

1. (a)

2π103 = 290.9t rad/s 21.6 ∴ f (t ) = 8.5sin (290.9t + Φ ) ∴ 0 = 8.5sin (290.9 × 2.1× 10−3 + Φ ) T = 4 (7.5 − 2.1)10−3 = 21.6 × 10−3 , ω =

∴Φ = −0.6109rad + 2π = 5.672rad or 325.0 ° ∴ f (t ) = 8.5sin (290.9t + 325.0°) (b)

8.5sin (290.9t + 325.0°) = 8.5 cos(290.9t + 235°) = 8.5 cos (290.9t − 125°)

(c)

8.5 cos (−125°) cos ωt + 8.5sin125° sin ωt = −4.875+ cos 290.9t + 6.963sin 290.9t




10 March 2006

2. (a)

−10 cos ωt + 4sin ωt + ACos ( wt + Φ ), A > 0, − 180° < Φ ≤ 180° A = 116 = 10.770, A cos Φ = −10, A sin Φ = −4 ∴ tan Φ = 0.4, 3d quad ∴Φ = 21.80° = 201.8°, too large ∴Φ = 201.8° − 360° = −158.20°

(b)

200 cos (5t + 130°) = Fcos 5t + G sin 5t ∴ F = 200cos130° = −128.6 G = −200sin130° = −153.2

(c)

(d)

i (t ) = 5cos10t − 3sin10t = 0, 0 ≤ t ≤ 1 s sin10t 5 ∴ = , 10t = 1.0304, cos10t 3 t = 0.10304 s; also, 10t = 1.0304 + π, t = 0.4172 s; 10t = 1.0304 + 2π, t = 0.7314 s 0 < t < 10ms, 10 cos100πt ≥ 12sin100πt ; let 10cos100πt =12sin100πt 10 ∴ tan100πt = , 100πt = 0.6947 ∴ t = 2.211 ms ∴ 0 < t < 2.211 ms 12




10 March 2006

3. ⎛ ⎛ −B ⎞ ⎞ that A cos x + B sin x = A2 + B 2 cos ⎜ x + tan −1 ⎜ ⎟ ⎟ . For f(t), the angle is in the ⎝ A ⎠⎠ ⎝ second quadrant; most calculators will return –30.96o, which is off by 180o.

(a) Note

f (t ) = −50 cos ωt − 30sin ωt = 58.31cos (ωt + 149.04°) g (t ) = 55cos ωt − 15sin ωt = 57.01cos (ωt + 15.255°) ∴ ampl. of f (t ) = 58.31, ampl. of g (t ) = 57.01

(b)

f (t ) leads g (t ) by 149.04° − 15.255° = 133.8°



4.


10 March 2006

i (t ) = A cos (ωt − θ), and L(di / dt ) + Ri = Vm cos ωt [−ωA sin (ω L t − θ)] + RA cos (ωt − θ) = Vm cos ωt

∴

−ωLA sin ωt cos θ + ωLA cos ωt sin θ + RA cos ωt cos θ + RA sin ωt sin θ = Vm cos ωt ∴ ωLA cos θ = RA sin θ and ωLA sin θ + RA cos θ = Vm ωL R ωL

Thus, tan θ = and ωLA

R 2 + ω2 L2

* + RA

R R 2 + ω2 L2

= Vm

⎛ ⎞ R2 ω2 L2 so that ⎜ + ⎟ A = Vm 2 2 2 R 2 + ω2 L2 ⎠ ⎝ R +ω L Thus,

(

)

R 2 + ω2 L2 A = ( Vm ) and therefore we may write A =

Vm R 2 + ω2 L2

*



5.


f = 13.56 MHz so ω = 2πf = 85.20 Mrad/s. Delivering 300 W (peak) to a 5-Ω load implies that

Finally, Since

10 March 2006

Vm2 = 300 so Vm = 38.73 V. 5

(85.2×106)(21.15×10–3) + φ = nπ, n = 1, 3, 5, … (85.2×106)(21.15×10–3) = 1801980, which is 573588π, we find that

φ = 573589π - (85.2×106)(21.15×10–3) = 573589π - 573588π = π



6.

(a)


10 March 2006

-33 sin(8t – 9o) → -33∠(-9-90)o = 33∠81o 12 cos (8t – 1o) → 12∠-1o 33∠81o

-33 sin(8t – 9o) leads 12 cos (8t – 1o) by 81 – (-1) = 82o.

12∠-1o

(b)

15 cos (1000t + 66o) -2 cos (1000t + 450o)

→ 15 ∠ 66o → -2 ∠ 450o = -2 ∠90o = 2 ∠ 270o 15 cos (1000t + 66o) leads -2 cos (1000t + 450o) by 66 – -90 = 156o.

15∠66o

2∠270o

(c)

sin (t – 13o) cos (t – 90o)

→ →1

1∠-103o ∠ -90o cos (t – 90o) leads sin (t – 13o) by 66 – -90 = 156o.

1∠-103o

(d)

1 ∠ -90o

sin t cos (t – 90o)

→1 →1

∠ -90o ∠ -90o

These two waveforms are in phase. Neither leads the other.



7.

(a)

6 cos (2π60t – 9o) -6 cos (2π60t + 9o)


10 March 2006

→ 6∠-9o → 6∠189o -6 cos (2π60t + 9o) lags 6 cos (2π60t – 9o) by 360 – 9 – 189 = 162o.

6∠-9o

6∠189o

cos (t - 100o) -cos (t - 100o)

(b)

→ 1 ∠ -100o → -1 ∠ -100o = 1 ∠80o -cos (t - 100o) lags cos (t - 100o) by 180o.

1∠80o

1∠-100o (c)

-sin t sin t

∠-90o = 1∠90o 1∠ -90o

→ -1 →

-sin t lags sin t by 180o. 1∠90o 1 ∠ -90o (d)

7000 cos (t – π) 9 cos (t – 3.14o)

→ 7000 ∠ -π = 7000 ∠ -180o →9 ∠ -3.14o 7000 cos (t – π) lags 9 cos (t – 3.14o) by 180 – 3.14 = 176.9o.

7000 ∠ -180o 9 ∠ -3.14o



8.


v(t) = V1 cos ωt - V2 sin ωt

10 March 2006

[1]

We assume this can be written as a single cosine such that v(t) = Vm cos (ωt + φ) = Vm cos ωt cos φ - Vm sin ωt sin φ [2] Equating terms on the right hand sides of Eqs. [1] and [2], V1 cos ωt – V2 sin ωt = (Vm cos φ) cos ωt – (Vm sin φ) sin ωt yields

V1 = Vm cos φ and V2 = Vm sin φ Dividing, we find that

V2 V sin φ = m = tan φ and φ = tan-1(V2/ V1) V1 Vm cos φ

V12 + V22

V2

φ

V1 Next, we see from the above sketch that we may write Vm = V1/ cos φ or Vm =

V1 V1

2 1

2 2

V +V

Thus, we can write v(t) = Vm cos (ωt + φ) =

=

V12 + V22 V12 + V22 cos [ ωt + tan-1(V2/ V1)].



9.


10 March 2006

(a) In the range 0 ≤ t ≤ 0.5, v(t) = t/0.5 V. Thus, v(0.4) = 0.4/0.5 = 0.8 V. (b) Remembering to set the calculator to radians, 0.7709 V. (c) 0.8141 V.

(d)

0.8046 V.



10. (a)

Vrms

⎡ V2 = ⎢ m ⎣T ⎡ V2 = ⎢ m ⎣T ⎡ V2 = ⎢ m ⎣ 2T ⎡ V2 = ⎢ m ⎣ 2T

∫

T

0

∫

T

∫

T

∫

T

0

0

0


⎤ cos 2 ωt dt ⎥ ⎦

1

2πt ⎤ cos dt ⎥ T ⎦

2

1

2

2

4πt ⎞ ⎤ ⎛ ⎜1 + cos ⎟ dt ⎥ T ⎠ ⎦ ⎝ V2 dt + m 2T

∫

T

0

10 March 2006

1

2

4πt ⎤ cos dt ⎥ T ⎦

⎡ V2 V2 4π ⎤ = ⎢ m T + m cos u 0 ⎥ 8π ⎣ 2T ⎦ V = m 2

1

1

2

2

*

(b)

Vm = 110 2 = 155.6 V , 115 2 = 162.6 V , 120 2 = 169.7 V



11.


10 March 2006

We begin by defining a clockwise current i. Then, KVL yields –2×10–3cos5t + 10i + vC = 0.

Since

i = iC = C

dvC , we may rewrite our KVL equation as dt dv 30 C + vC = 2 × 10−3 cos 5t dt

[1]

dvC = −5 A sin(5t + θ ) , dt we now may write Eq. [1] as –150Asin(5t + θ) + Acos(5t + θ) = 2×10–3 cos5t. Using a common trigonometric identity, we may combine the two terms on the left hand side into a single cosine function:

We anticipate a response of the form vC(t) = Acos(5t + θ). Since

(150 A)

2

150 A ⎞ ⎛ −3 + A2 cos ⎜ 5t + θta+ n −1 ⎟ = 2 × 10 cos 5t A ⎝ ⎠

Equating terms, we find that A = 13.33 μV and θ = –tan–1 150 = –89.62o. Thus, vC(t) = 13.33 cos (5t – 89.62o) μV.



12. KVL


10 March 2006

yields

–6cos400t + 100i + vL = 0. di di Since vL = L = 2 , we may rewrite our KVL equation as dt dt di 2 + 100i = 6 cos 400t dt We anticipate a response of the form i(t) = Acos(400t + θ). Since di = −400 A sin(400t + θ ) , dt we now may write Eq. [1] as

[1]

–800Asin(400t + θ) + 100Acos(400t + θ) = 6 cos400t. Using a common trigonometric identity, we may combine the two terms on the left hand side into a single cosine function:

(800 A) + (100 A) 2

2

800 A ⎞ ⎛ cos ⎜ 400t + θt + an −1 ⎟ = 6 cos 400t 100 A ⎠ ⎝

Equating terms, we find that A = 7.442 mA and θ = –tan–1 8 = –82.88o. Thus, i(t) = 7.442 cos (400t – 82.88o) mV, so vL = L

di di = 2 = 5.954cos (400t + 7.12o) dt dt



13. 20cos500


10 March 2006

t V → 20∠0o V. 20 mH → j10 Ω.

Performing a quick source transformation, we replace the voltage source/20-Ω resistor series combination with a 1∠0o A current source in parallel with a 20-Ω resistor. 20 || 60k = 19.99 Ω. By current division, then, IL =

19.99 = 0.7427∠-21.81o A. Thus, iL(t) = 742.7 cos (500t – 21.81o) mA. 19.99 + 5 + j 10




10 March 2006

14. At x − x : R th = 80 20 = 16Ω 80 cos 500t 85 ∴ voc = 4.8cos 500t V voc = −0.4 (15 85)

4.8

10 ⎞ ⎛ cos ⎜ 500t − tan −1 ⎟ 15 ⎠ ⎝ 162 + 102 = 0.2544 cos (500t − 32.01°) A

(a)

iL =

(b)

vL = LiL′ = 0.02 × 0.02544 (−500) sin (500t − 32.01°) = − 2.544sin (500t − 32.01°) V ∴ vL = 2.544 cos (500t + 57.99°) V, ix = 31.80 cos (500t + 57.99°) mA




10 March 2006

15.

100

800 ⎞ ⎛ 5 cos ⎜105 t − ⎟ = 0.10600 cos (10 t − 57.99°) A 2 2 500 ⎠ ⎝ 500 + 800 π 57.99° pR = 0 when i = 0 ∴105 t − π = , t = 25.83μs 180 2

(a)

i=

(b)

± vL = Li′ = 8 × 10−3 × 0.10600 (−105 ) sin (105 t − 57.99°) ∴ vL = −84.80sin (105 t − 57.99°) ∴ pL = vL i = −8.989sin (105 t − 57.99°) cos (105 t − 57.99°) = −4.494 sin (2 × 165 t − 115.989°) ∴ pL = 0 when 2 × 105 t − 115.989° = 0°, 180°, ∴ t = 10.121 or 25.83μs

(c)

ps = vs iL = 10.600 cos105 t cos (105 t − 57.99°) ∴ ps = 0 when 105 t =

π , t = 15.708μs and also t = 25.83μs 2



16.


10 March 2006

vs = 3cos105 t V, is = 0.1cos105 t A vs in series with 30Ω → 0.1cos105 t A 30Ω Add, getting 0.2 cos105 t A 30 Ω change to 6 cos 105 t V in series with 30Ω; 30Ω + 20Ω = 50Ω 10 ⎞ ⎛ cos ⎜105 t − tan −1 ⎟ = 0.11767 cos (105 t − 11.310°) A 50 ⎠ ⎝ 502 + 102 At 10 t = μs, 105 t = 1∴ iL = 0.1167 cos (1rad − 11.310°) = 81.76mA

∴ iL =

6

∴ vL = 0.11767 × 10 cos (1rad − 11.30° + 90°) = −0.8462V



17. cos500


10 March 2006

t V → 1∠0o V. 0.3 mH → j0.15 Ω.

Performing a quick source transformation, we replace the voltage source-resistor series combination with at 0.01∠0o A current source in parallel with a 100-Ω resistor. Current division then leads to 100 = IL ( 0.01 + 0.2I L ) 100 + j 0.15 1 + 20IL = (100 + j0.15) IL Solving, we find that IL = 0.0125∠-0.1074o A, so that iL(t) = 12.5cos(500t – 0.1074o) mA.



18.


10 March 2006

vs1 = Vs 2 = 120 cos120 π t V 120 120 = 2A, = 1A, 2 + 1 = 3A, 60 120 = 40Ω 60 12 3 × 40 = 120 V, ωL = 12π = 37.70Ω 37.70 ⎞ ⎛ cos ⎜120 πt − tan −1 ⎟ 40 ⎠ ⎝ 402 + 37.702 = 2.183cos (120 π t − 43.30°) A

∴ iL =

(a)

(b)

120

1 ∴ ωL = × 0.1× 2.1832 cos 2 (120π t − 43.30°) 2 = 0.2383cos 2 (120π t − 43.30°) J ωL , av =

1 × 0.2383 = 0.11916 J 2




10 March 2006

19. vs1 = 120 cos 400t V, vs 2 = 180 cos 200t V Performing two quick source transformations, 120 180 = 2 A, = 1.5 A, and noting that 60 120 = 40 Ω, 60 120 results in two current sources (with different frequencies) in parallel, and also in parallel with a 40 Ω resistor and the 100 mH inductor. Next we employ superposition. Open-circuiting the 200 rad/s source first, we perform a source transformation to obtain a voltage source having magnitude 2 × 40 = 80 V. Applying Eqn. 10.4, 80 400(0.1) cos (400t − tan −1 ) iL′ = 40 402 + 4002 (0.1) 2 Next , we open-circuit the 400 rad/s current source, and perform a source transformation to obtain a voltage source with magnitude 1.5 × 40 = 60 V. Its contribution to the inductor current is 60 200(0.1) cos (200t − tan −1 )A iL′′ = 40 402 + 2002 (0.1) 2 so that iL = 1.414 cos (400t − 45°) + 1.342 cos (200t − 26.57°) A



20.

R i = ∞, R o = 0, A = ∞, ideal, iupper = −

v Vm cos ω t , ilower = out R R1

∴ ic1 = iupper + ilower =


= C1 R

1

10 March 2006

L R

i ′ (vout − Vm cos ω t ) = −C1vout R1

L ′ vout R d ⎛v ⎞ For RL circuit, Vm cos ω t = vr + L ⎜ R ⎟ dt ⎝ R ⎠ L ∴ Vm cos ω t = vR + v′R R By comparison, vR = vout ′ = vout + ∴ Vm cos ω t = vout + R1C1vout

*




21. (a)

1 idt (ignore I.C) C∫ 1 ∴−ω Vm sin ω t = Ri′ + i C

(b)

Assume iA=

10 March 2006

Vm cos ω t = Ri +

cos (ω t + Φ )

∴−ω Vm sin ω t = −Rω A sin (ω t + Φ ) +

A cos (ω t + Φ ) C

∴−ω Vm sin ω t = − Rω A cos Φ sin ω t-Rω A sin Φ cos ω t +

A A cos ω t cos Φ − sin ω t sin Φ C C

Equating terms on the left and right side, A 1 [1] Rω A sin Φ = cos Φ∴ tan Φ = so tΦ = an −1 (1 ω CR ) , and ω CR C

ω CR

[2] −ω Vm = −Rω A ∴ω Vm = ∴i =

1+ ω C R 2

2

A ⎡ R ω 2C2 + 1 ⎢ C ⎣ 1 + ω 2 C2 R 2 2

2

−

A 1 C 1 + ω 2C2 R 2

⎤ A ω CVm 1 + ω 2 C2 R 2 ∴ A = ⎥= 1 + ω 2C2 R 2 ⎦ C

ω CVm

⎛ 1 ⎞ cos ⎜ ω t + tan −1 ⎟ ω CR ⎠ 1 + ω 2 C2 R 2 ⎝



22. (a)


10 March 2006

7 ∠ -90o = -j 7 (b) 3 + j + 7 ∠ -17o = 3 + j + 6.694 – j 2.047 = 9.694 – j 1.047 o

(c)

14ej15 = 14 ∠ 15o = 14 cos 15o + j 14 sin 15o = 13.52 + j 3. 623

(d)

1 ∠ 0o = 1 (e) –2 (1 + j 9) = -2 – j 18 = 18.11 ∠ - 96.34o (f) 3 = 3 ∠ 0o



23. (b)


10 March 2006

(a) 3 + 15 ∠ -23o = 3 + 13.81 – j 5.861 = 16.81 – j 5.861 (j 12)(17 ∠ 180o) = (12 ∠ 90o)(17 ∠ 180o) = 204 ∠ 270o = –j 204 (c) 5 – 16(9 – j 5)/ (33 ∠ -9o) = 5 – (164 ∠ -29.05o)/ (33 ∠ -9o) = 5 – 4.992 ∠ -20.05o = 5 – 4.689 – j 1.712 = 0.3109 + j 1.712



24.


10 March 2006

(a) 5 ∠ 9o – 9 ∠ -17o = 4.938 + j 0.7822 – 8.607 + j 2.631 = -3.668 + j 3.414 = 5.011∠ 137.1o (b) (8 – j 15)(4 + j 16) – j = 272 + j 68 – j = 272 + j 67 = 280.1 ∠ 13.84o (c) (14 – j 9)/ (2 – j 8) + 5 ∠ -30o = (16.64 ∠-32.74o)/ (8.246 ∠ - 75.96o) + 4.330 – j 2.5 = 1.471 + j 1.382 + 4.330 – j 2.5 = 5.801 – j 1.118 = 5.908 ∠ -10.91o

(d)

17 ∠ -33o + 6 ∠-21o + j 3 = 14.26 – j 9.259 + 5.601 – j 2.150 + j 3 = 19.86 – j 8.409 = 21.57 ∠ -22.95o



25.


10 March 2006

o

(a) ej14 + 9 ∠ 3o – (8 – j 6)/ j2 = 1 ∠ 14o + 9 ∠ 3o – (8 – j 6)/ (-1) = 0.9703 + j 0.2419 + 8.988 + j 0.4710 + 8 – j 6 = 17.96 – j 5.287 = 18.72 ∠ -16.40o

(b) =

o

(5 ∠ 30o)/ (2 ∠ -15o) + 2 e j5 / (2 – j 2) 2.5 ∠ 45o + (2 ∠ 5o)/ (2.828 ∠ -45o) = 1.768 + j 1.768 + 0.7072 ∠ 50o = 1.768 + j 1.768 + 0.4546 + j 0.5418 = 2.224 + j 2.310 = 3.207 ∠ 46.09o




10 March 2006

26. (a)

5∠ − 110° = −1.7101 − j 4.698

(b)

6e j160° = −5.638 + j 2.052

(c)

(3 + j 6) (2∠50°) = −5.336 + j12.310

(d)

−100 − j 40 = 107.70∠ − 158.20°

(e)

2∠50° + 3∠ − 120° = 1.0873∠ − 101.37°




10 March 2006

27. (a)

40∠ − 50° − 18∠25° = 39.39∠ − 76.20°

(b)

3+

(c)

(2.1∠25°)3 = 9.261∠75° = 2.397 + j8.945+

(d)

0.7e j 0.3 = 0.7∠0.3rad = 0.6687 + j 0.2069

2 2 − j5 + = 4.050− ∠ − 69.78° j 1+ j2




10 March 2006

28. ic = 20e(40t +30°) A ∴ vc = 100∫ 20e j (40t +30°) dt vc = − j 50e j (40t +30°) , iR = − j10e j (40t + 30°) A ∴ iL = (20 − j10) e j (40t +30°) , vL = j 40 × 0.08(20 − j10) e j (40t + 30°) ∴ vL = (32 + j 64) e j (40t +30°) V ∴ vs = (32 + j 64 − j 50) e j (40t + 30°) ∴ vs = 34.93e j (40t −53.63° ) V




10 March 2006

29. iL = 20e j (10t + 25°) A vL = 0.2

d [20e j (10t + 25°) ] = j 40e(10t = 25° ) dt

vR = 80e j (10t + 25°) vs = (80 + j 40) e j (10t + 25°) , ic = 0.08(80 + j 40) j10e j (10t + 25°) ∴ ic = (−32 + j 64) e j (10t + 25°) ∴ is = (−12 + j 64) e j (10t + 25°) ∴ is = 65.12e j (10t +125.62°) A




30.

80 cos(500t − 20°) V → 5 cos (500t + 12°) A

(a)

vs = 40 cos (500t + 10°) ∴ iout = 2.5cos (500t + 42°) A

(b)

vs = 40sin (500t + 10°) = 40 cos (500t − 80°)

10 March 2006

∴ iout = 2.5cos (500t − 48°) A (c)

vs = 40e j (500t +10°) = 40 cos (500t + 10°) + j 40sin (500t + 10°) ∴ iout = 2.5e j (500t + 42°) A

(d)

vs = (50 + j 20) e j 500t = 53.85+ e j 21.80°+ j 500t ∴ iout = 3.366e j (500t +53.80° ) A




10 March 2006

31. (a)

12 sin (400t + 110°) A → 12∠20°A

(b)

−7 sin 800t − 3cos 800t → j 7 − 3 = −3 + j 7 = 7.616∠113.20° A

(c)

4 cos (200t − 30°) − 5cos (200t + 20°) → 4∠ − 30° − 5∠20° = 3.910∠ − 108.40° A

(d)

ω = 600, t = 5ms : 70∠30° V → 70 cos (600 × 5 × 10−3rad + 30°) = −64.95 V

(e)

ω = 600, t = 5ms : 60 + j 40 V = 72.11∠146.3° → 72.11cos (3rad + 146.31°) = 53.75 V



32.

ω = 4000, t = 1ms

(a)

I x = 5∠ − 80° A


10 March 2006

∴ ix = 5cos (4rad − 80°) = −4.294 A (b)

I x = −4 + j1.5 = 4.272∠159.44° A ∴ ix = 4.272 cos (4rad + 159.44°) = 3.750− A

(c)

vx (t ) = 50sin (250t − 40°) = 50 cos (250t − 130°) → Vx = 50∠ − 130° V

(d)

vx = 20 cos108t − 30sin108t → 20 + j 30 = 36.06∠56.31° V

(e)

vx = 33cos (80t − 50°) + 41cos (80t − 75°) → 33∠ − 50° + 41∠ − 75° = 72.27∠ − 63.87° V




10 March 2006

33. V1 = 10∠90° mV, ω = 500; V2 = 8∠90° mV, ω = 1200, M by − 5, t = 0.5ms vout = (−5) [10 cos (500 × 0.5 ×10−3rad + 90°) + 8cos (1.2 × 0.5 + 90°)] = 50sin 0.25rad + 40sin 0.6rad = 34.96mV



34. (2.5


10 March 2006

Begin with the inductor: ∠40o) (j500) (20×10-3) = 25∠130o V across the inductor and the 25-Ω resistor. The current through the 25-Ω resistor is then (25∠130o) / 25 = 1∠130o A.

The current through the unknown element is therefore 2.5∠40o + 1∠130o = 2.693 ∠61.80o A; this is the same current through the 10-Ω resistor as well. Armed with this information, KVL provides that Vs = 10(26.93∠61.8o) + (25∠ -30o) + (25∠130o) = 35.47 ∠58.93o

and

so vs(t) = 35.47 cos (500t + 58.93o) V.




10 March 2006

35.

ω = 5000 rad/s.

(a)

The inductor voltage = 48∠ 30o = jωL IL = j(5000)(1.2×10-3) IL So IL = 8∠-60o and the total current flowing through the capacitor is ∠ 0o - IL = 9.165∠49.11o A and the voltage V1 across the capacitor is

10

V1 = (1/jωC)(9.165∠49.11o) = -j2 (9.165∠49.11o) = 18.33∠-40.89o V.

Thus,

v1(t) = 18.33 cos (5000t – 40.89o) V.

(b)

V2 = V1 + 5(9.165∠49.11o) + 60∠120o = 75.88∠79.48o V ∴ v2 (t ) = 75.88cos (5000t + 79.48°) V

(c)

V3 = V2 – 48∠30o = 75.88 ∠79.48o – 48∠30o = 57.70∠118.7o V ∴ v3 (t ) = 57.70 cos (5000t + 118.70°) V



36.


10 March 2006

VR = 1∠0o V, Vseries = (1 + jω –j/ω)(1∠0o)

VR = 1 and Vseries = 1 + (ω - 1/ω )

2

We desire the frequency w at which Vseries = 2VR or Vseries = 2 2 Thus, we need to solve the equation 1 + (ω - 1/ω ) = 4 or ω 2 - 3 ω - 1 = 0 Solving, we find that ω = 2.189 rad/s.



37.


10 March 2006

With an operating frequency of ω = 400 rad/s, the impedance of the 10-mH inductor is jωL = j4 Ω, and the impedance of the 1-mF capacitor is –j/ωC = -j2.5 Ω. ∴ Vc = 2∠40° (− j 2.5) = 5∠ − 50° A ∴ I L = 3 − 2∠40° = 1.9513∠ − 41.211° A ∴ VL = 4 × 1.9513∠90° − 4.211° = 7.805+ ∠48.79° V ∴ Vx = VL − Vc = 7.805+ ∠48.79° − 5∠ − 50° ∴ Vx = 9.892∠78.76° V, vx = 9.892 cos (400t + 78.76°) V




10 March 2006

38. If I si = 2∠20° A, I s 2 = 3∠ − 30° A → Vout = 80∠10° V I s1 = I s 2 = 4∠40° A → Vout = 90 − j 30 V Now let I s1 = 2.5∠ − 60° A and I s 2 = 2.5∠60° A Let Vout = AI s1 + BI s 2 ∴ 80∠10° = A(2∠20°) + B(3∠ − 30°) and 90 − j 30 = (A + B) (4∠40°) ∴ A + B =

90 − j 30 = 12.415+ − j 20.21 4∠40°

80∠10° 3∠ − 30° =A+B ∴ A = 40∠ − 10° − B(1.5∠ − 50°) 2∠20° 2∠20 ∴12.415+ − j 20.21 − B = 40∠ − 10° − B(1.5∠ − 50°) ∴

∴12.415+ − j 20.21 − 40∠ − 10° = B (1 − 1.5∠ − 50°) = B (1.1496∠ + 88.21°) 30.06∠ − 153.82° = 26.148∠117.97° 1.1496∠ + 88.21° ∴ A = 12.415+ − j 20.21 − 10.800 + j 23.81 ∴B =

= 49.842∠ − 60.32° Vout = (49.842∠ − 60.32°) (2.5∠ − 60°) + (26.15∠117.97°) (2.5∠60°) = 165.90∠ − 140.63°V



39.


10 March 2006

We begin by noting that the series connection of capacitors can be replaced by a single 1 equivalent capacitance of value C = = 545.5 F μ . Noting ω = 2πf, 1+ 1 + 1 2 3 − j106 = − j 291.8 Ω . 2π ( 545.5 )

(a)

ω = 2π rad/s, therefore ZC = –j/ωC =

(b)

ω = 200π rad/s, therefore ZC = –j/ωC =

(c)

− j106 ω = 2000π rad/s, therefore ZC = –j/ωC = = − j 291.8 mΩ . 2000π ( 545.5 )

(d)

ω = 2×109π rad/s, therefore ZC = –j/ωC =

− j106 = − j 2.918 Ω . 200π ( 545.5 )

− j106 = − j 291.8 nΩ . 2 × 109 π ( 545.5 )



40.


10 March 2006

We begin by noting that the parallel connection of inductors can be replaced by a single 1 5 equivalent inductance of value L = = nH . In terms of impedance, then, we have 1+ 5 6

Noting

ω = 2πf,

5 ⎛ ⎞ 5 ⎜ jω × 10−9 ⎟ 6 ⎠ Z= ⎝ 5 −9 5 + jω ×10 6

(a)

ω = 2π rad/s, therefore Z = j5.236×10–9 Ω (the real part is essentially zero).

(b)

ω = 2×103π rad/s, therefore Z = 5.483×10–12 + j5.236×10–6 Ω.

(c)

ω = 2×106π rad/s, therefore Z = 5.483×10–6 + j5.236×10–6 Ω.

(d)

ω = 2×109π rad/s, therefore Z = 2.615 + j2.497 Ω.

(e)

ω = 2×1012π rad/s, therefore Z = 5 + j4.775×10–3 Ω .



41. (a)


10 March 2006

ω = 800 : 2μF → − j 625, 0.6H → j 480 300(− j 625) 600( j 480) + 300 − j 625 600 + j 480 = 478.0 + j175.65Ω

∴ Zin =

(b)

ω = 1600 : Zin = +

300(− j 312.5) 300 − j 312.5

600( j 960) = 587.6 + j119.79Ω 600 + j 960




10 March 2006

42.

(a)

At ω = 100 rad/s, 2 mF → - j 5 Ω; 0.1 H → j10 Ω. 50 − j 50 10 − j10 2 − j1 (10 + j10) (− j 5) = = 10 + j 5 2 + j1 2 − j1 = 2 − j 6 Ω∴ Zin = 20 + 2 − j 6 = 22 − j 6 Ω

(b)

SCa, b : 20 10 = 6.667, (6.667 − j 5) j10 50 + j 66.67 150 + j 200 30 + j 40 4 − j 3 = = × 6.667 + j 5 20 + j15 4 + j3 4 − j3 = Z in ∴ Z in (1.2 + j1.6) (4 − j 3) = 9.6 + j 2.8 Ω =




10 March 2006

43. ω = 800 : 2μF → − j 625, 0.6H → j 480 300(− j 625) 600( j 480) + 300 − j 625 600 + j 480 = 478.0 + j175.65Ω

∴ Zin =

− j 625 120 × 478.0 + j175.65 300 − j 625 or I = 0.2124∠ − 45.82° A ∴I =

Thus, i(t) = 212.4 cos (800t – 45.82o) mA.




44. (a)

3 Ω + 2mH : V = (3∠ − 20°) (3 + j 4) = 15∠33.13° V

(b)

3 Ω + 125μF : V = (3∠ − 20°) (3 − j 4) = 15∠ − 73.3° V

(c)

3 Ω 2mH 125μF : V = (3∠ − 20°) 3 = 9∠ − 20° V

(d)

same: ω 4 = 000 ∴ V = (3∠ − 20°) (3 + j8 − j 2) ∴ V = (3∠ − 20°) (3 + j 6) = 20.12∠43.43° V

10 March 2006




10 March 2006

45. (a)

C = 20μF, ω = 100 1

Zin =

=

1 0.005 − j 0.01 + j 0.002

1 1 + + j1000 × 20 × 10−6 200 j1000 1 ∴ Zin = = 196.12∠ − 11.310°Ω 0.005 + j 0.001

(b)

ω = 100 rad/s ∴ Z in = Z in = 125 =

(c)

1 0.005 − j 0.001 + j100C 1

0.0052 + (100C − 0.001)

2

or

64 × 10-6 = 0.0052 + (100C − 0.001)

so or

6.245 ×10-3 = 39 × 10-6 = 100C − 0.001 C = 72.45 μF

C = 20μF ∴ Zin =

2

1 1 = 100∠ = −5 0.01∠ 0.0005 − j 0.1/ ω + j 2 ×10 ω 2

2

0.1 ⎞ 0.1 ⎞ ⎛ ⎛ −5 −5 ∴ 0.0052 + ⎜ 2 ×10−5 ω − ⎟ = 0.0001, ⎜ 2 × 10 − ⎟ = 7.5 × 10 ω ⎠ ω ⎠ ⎝ ⎝ 0.01 ∴ 2 ×10−5 − ∓ 866.0 × 10−5 = 0 ∴ 2 × 10−5 ω2 ∓ 866.0 × 10−5 ω − 0.1 = 0 ω use − sign: ω =

866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6 = 444.3 and < 0 4 × 10−5

−866.0 × 10−5 ± 7.5 × 10−5 + 8 × 10−6 = 11.254 and 0) = 1200C

X=



49. At


10 March 2006

ω = 4 rad/s, the 1/8-F capacitor has an impedance of –j/ωC = -j2 Ω, and the 4-H inductor has an impedance of jωL = j16 Ω.

(a) Term

inals ab open circuited: Zin = 8 + j16 || (2 – j2) = 10.56 – j1.92 Ω

(b) Term

inals ab short-circuited: Zin = 8 + j16 || 2 = 9.969 + j0.2462 Ω



50. 2 3.2 1 1 20

f = 1 MHz, ω = 2πf = 6.283 Mrad/s μF → -j0.07958 Ω μH → j20.11 Ω μF → -j0.1592 Ω μH → j6.283 Ω μH → j125.7 Ω 200 pF → -j795.8 Ω


10 March 2006

= Z1 = Z2 = Z3 = Z4 = Z5 = Z6

The three impedances at the upper right, Z3, 700 kΩ, and Z3 reduce to –j0.01592 Ω Then we form Z2 in series with Zeq: Z2 + Zeq = j20.09 Ω. Next we see 106 || (Z2 + Zeq) = j20.09 Ω. Finally,

Zin = Z1 + Z4 + j20.09 = j26.29 Ω.



51. f


10 March 2006

As in any true design problem, there is more than one possible solution. Model answers ollow: (a) Using at least 1 inductor, ω = 1 rad/s. Z = 1 + j4 Ω. Construct this using a single 1 Ω resistor in series with a 4 H inductor.

(b)

Force jL = j/C, so that C = 1/L. Then we construct the network using a single 5 Ω resistor, a 2 H inductor, and a 0.5 F capacitor, all in series (any values for these last two will suffice, provided they satisfy the C = 1/L requirement).

(c)

Z = 7∠80o Ω. R = Re{Z} = 7cos80o= 1.216 Ω, and X = Im{Z} = 7sin80o = 6.894 Ω. We can obtain this impedance at 100 rad/s by placing a resistor of value 1.216 Ω in series with an inductor having a value of L = 6.894/ω = 68.94 mH. (d) A single resistor having value R = 5 Ω is the simplest solution.



52. f


10 March 2006

As in any true design problem, there is more than one possible solution. Model answers ollow: (a) 1 + j4 kΩ at ω = 230 rad/s may be constructed using a 1 kΩ resistor in series with an inductor L and a capacitor C such that j230L – j/(230C) = 4000. Selecting arbitrarily C = 1 F yields a required inductance value of L = 17.39 H. Thus, one design is a 1 kΩ resistor in series with 17.39 H in series with 1 F. (b) To obtain a purely real impedance, the reactance of the inductor must cancel the reactance of the capacitor, In a series string, this is obtained by meeting the criterion ωL 1/ωC, or L = 1/ω2C = 1/100C.

=

Select a 5 MΩ resistor in series with 1 F in series with 100 mH. If Z = 80∠–22o Ω is constructed using a series combination of a single resistor R and single capacitor C, R = Re{Z} = 80cos(–22o) = 74.17 Ω. X = –1/ωC = Im{Z} = o 80sin(–22 ) = –29.97 Ω. Thus, C = 667.3 μF. (c)

(d) The simplest solution, independent of frequency, is a single 300 Ω resistor.




10 March 2006

53. Note that we may replace the three capacitors in parallel with a single capacitor having value 10 −3 + 2 × 10−3 + 4 × 10−3 = 7 mF . (a)

ω = 4π rad/s.

Y = j4πC = j87.96 mS

(b)

ω = 400π rad/s.

Y = j400πC = j8.796 S

(c)

ω = 4π×103 rad/s. Y = j4π×103C = j879.6 S

(d)

ω = 4π×1011 rad/s. Y = j4π×1011C = j8.796×109 S



54.


10 March 2006

(a) Susceptance is 0 (b) B = ωC = 100 S

(c)

Z = 1 + j100 Ω, so Y =

1 1 − j100 =G + jB , where B = –9.999 mS. = 1 + j100 1+1002




10 March 2006

55. 2 H → j 2, 1F → − j1 Let I1∈ = ∠0° A ∴ VL = j 2V ∴ I c = I in + 0.5 VL = 1 + j1 ∴ Vin = j 2 + (1 + j1) (− j1) = 1 + j1 ∴ Vin =

1∠0° 1 1 − j1 = = 0.5 − j 0.5 1 + j1 1 − j1 Vin

Now 0.5 S → 2 Ω, − j 0.5 S =

1 →2H j2



56. (a)


10 March 2006

ω = 500, ZinRLC = 5 + j10 − j1 = 5 + j 9 1 5 − j9 9 = ∴ Yc = = 500C 5 + j9 106 106 9 ∴C = = 169.8 μ F 53, 000 ∴ YinRLC =

106 = 21.2 Ω 5

(b)

R in , ab =

(c)

ω = 1000 rad/s ∴ Z S = 5 + j 2 − j 5 = 5 − j 3 = 5.831∠ − 30.96o Ω and Z C = − j 58.89 Ω. Thus, Yin ,ab =

1 1 + = 0.1808∠35.58o S Z S ZC

= 147.1 + j105.2 mS



57. (a)


10 March 2006

j 0.1ω 100 + j 0.001ω 50, 000 + j 0.6ω 100 − j 0.001ω ∴ Zin = × 100 + j 0.001ω 100 − j 0.001ω R in = 550 Ω : Zin = 500 +

5 × 106 + 0.0006ω2 + j (60ω − 50ω) 104 + 10−6 ω2 5 × 106 + 0.006ω2 ∴ R in = = 550 ∴ 5.5 ×106 4 −6 2 10 + 10 ω −4 2 + 5.5 × 10 ω = 5 ×106 × 10−4 ω2 ∴ Zin =

∴ 0.5 × 10−4 ω2 = 0.5 ×106 , ω2 = 1010 , ω = 105 rad/s (b)

10ω = 0.5 × 106 + 0.5 × 10−4 ω2 − 10ω −6 2 10 + 10 ω 2 = 0, ω − 2 × 105 ω + 1010 = 0 X in = 50 Ω =

∴ω =

(c)

4

2 × 105 ± 4 × 1010 − 4 × 1010 = 105 ∴ω = 105 rad/s 2

G in = 1.8 × 10−3 : Yin =

100 + j 0.001ω 50, 000 − j 0.6ω × 50, 000 + j 0.6ω 50, 000 − j 0.6ω

5 × 106 + 6 × 10−4 ω2 + j (50ω − 6ω) 25 × 108 + 0.36ω2 5 × 106 + 6 × 10−4 ω2 ∴1.8 × 103 = 25 × 108 + 0.36ω2 ∴ 5 × 106 + 6 ×10−4 ω2 = 4.5 × 106 + 648 × 10−6 ω2 =

∴ 0.5 ×106 = 48 × 10−6 ω2 ∴ω = 102.06 krad/s (d)

−10ω 25 ×108 + 0.36ω2 ∴10ω = 37.5 × 104 + 54 × 10−6 ω2 Bin = 1.5 × 10−4 =

∴ 54 × 10−6 ω2 − 10ω + 37.5 × 104 = 0, ω = 10 ±

100 − 81 = 52.23 and 133.95 krad/s 108 × 10−6




10 March 2006

58. I1 0.1∠30° = = 20∠ − 23.13°∴ V1 = 20 V Y1 (3 + j 4)10−3

(a)

V1 =

(b)

V2 = V1 ∴ V2 = 20V

(c)

I 2 = Y2 V2 = (5 + j 2)10−3 × 20∠ − 23.13° = 0.10770∠ − 1.3286° A ∴ I3 = I1 + I 2 = 0.1∠30° + 0.10770∠ − 1.3286° = 0.2∠13.740° A ∴ V3 =

(d)

I3 0.2∠13.740° = = 44.72∠77.18° V ∴ V3 = 44.72V Y3 (2 − j 4)10−3

Vin = V1 + V3 + 20∠ − 23.13° + 44.72∠77.18° = 45.60∠51.62° ∴ Vin = 45.60V




10 March 2006

59. (a)

50 μ F → − j 20 Ω∴ Yin = 0.1 + j 0.05 1 1000 1 ∴ R1 − j = = 8 − j4 1000 C 0.1 + j 0.05 R1 − j C 1 ∴ R 1 = 8 Ω and C1 = = 250 μ F 4ω Yin =

(b)

ω = 2000 : 50μ F → − j10 Ω ∴ Yin = 0.1 + j 0.1 =

∴ R1 − j

1 R1 − j

500 C1

500 = 5 − j 5 ∴ R1 = 5 Ω, C1 = 100μ F C1




10 March 2006

60. (a)

10 10 + jω = jω jω jω 10 − jω ∴ Yin × 10 + jω 10 − jω Zin = 1 +

ω 2 + j10ω ω 2 + 100 10ω ω2 G in = 2 , Bin = 2 ω + 100 ω + 100

∴ Yin =

ω

Gin

Bin

0 1 2 5 10 20 ∞

0 0.0099 0.0385 0.2 0.5 0.8 1

0 0.0099 0.1923 0.4 0.5 0.4 0

B




10 March 2006

61. f

As in any true design problem , there is m ore than one possible solution. Model answers ollow:

(a)

Y = 1 – j4 S at ω = 1 rad/s. Construct this using a 1 S conductance in = 4, or L = 250 mH.

(b) 1/

Y = 200 m S (purely real at ω = 1 rad/s). This can be constructed using a 200 mS conductance (R = 5 Ω), in pa rallel with an ind uctor L and capacitor C such th at ωC – ωL = 0. Arbitrarily selecting L = 1 H, we find that C = 1 F.

H

One solution therefore is a 5 Ω resistor in parallel with a 1 F capacitor in parallel with a 1 inductor.

(c) mF.

parallel with an induc tance L such that 1/ ωL

Y = 7∠80o μS = G + jB at ω = 100 rad/s. G = Re{ Y} = 7cos80 o = 1.216 S (an 822.7 mΩ resistor). B = Im {Y} = 7sin8 0o = 6.894 S. W e m ay rea lize th is suscep tance by placing a capacito r C in parallel with the resis tor such th at jωC = j6.894, or C = 68.94 One solution therefore is an 822.7 mΩ resistor in parallel with a 68.94 mF. (d) The simplest solution is a single conductance G = 200 mS (a 5 Ω resistor).



62. f (a)


10 March 2006

As in any true design problem , there is m ore than one possible solution. Model answers ollow: Y = 1 – j4 pS at ω = 30 rad/s. Construct this using a 1 pS conductance (a 1 T Ω resistor) in parall el with an inductor L such that –j4×10–12 = –j/ωL, or L = 8.333 GH.

200

(b) We may realise a pu rely real admittance of 5 μS by pla cing a 5 μS conductance (a kΩ resistor) in parallel with a capa citor C and inducta nce L such that ωC – 1/ ωL = 0. Arbitrarily selecting a value of L = 2 H, we find a value of C = 1.594 μF.

One possible solution, then, is a 20 0 k Ω resistor in parallel w ith a 2 H inductor and a 1.594 μF capacitor. (c)

Y = 4 ∠–10o nS = G + jB at ω = 50 rad/s. G = Re{ Y} = 4×10 –9cos(–10o) = 3.939 nS (an 253.9 MΩ resistor). B = Im{Y} = 4×10 –9sin(–10o) = –6.946×10 –10 S. We may r ealize this susceptance by placing an inductor L in parallel with the resistor such that –j/ωL = –j6.946×10–10, or L = 28.78 μH. One possible solution, then, is a 253.9 MΩ resistor in parallel with a 28.78 μH inductor. (d) The simplest possible solution is a 60 nS resistor (a 16.67 MΩ resistor).



63.


10 March 2006

v1 V1 − V2 v1 − V2 , − j 75 = 5V1 + j 3V1 − j 3V2 − j 5V1 + j 5V2 + + 3 − j5 j3 ∴ (5 − j 2) V1 + j 2V2 = − j 75 (1)

− j5 =

v2 − V1 V2 − V1 V2 + + = 10 6 − j5 j3 − j10V2 + j10V1 + j 6V2 − j 6V1 + 5V2 = 300 ∴ j 4V1 + (5 − j 4) V2 = 300

(2)

5 − j 2 − j 75 j4 300 1500 − j 600 − 300 1200 − j 600 = 34.36∠23.63° V = ∴ V2 = = j2 5 − j2 25 − j 30 17 − j30 + 8 j4 5 − j4




10 March 2006

64. j 3I B − j 5(I B − I D ) = 0 ∴−2I B + j 5I D = 0 3(I D + j 5) − j 5(I D − I B ) + 6 (I D + 10) = 0 ∴ j 5I B + (9 − j 5) I D = −60 − j15 0 j5 −60 − j15 9 − j 5 −75 + j 300 = IB = − j2 j5 15 − j18 j5 9 − j5 = 13.198∠154.23° A



65.


10 March 2006

vs1 = 20 cos1000t V, vs 2 = 20sin1000t V ∴ Vs1 = 20∠0° V, Vs 2 = − j 20V

0.01H → j10 Ω, 0.1mF → − j10 Ω vx − 20 vx vx + j 20 + + = 0, 0.04vx + j 2 − 2 = 0, − j10 j10 25 Vx = 25(2 − j 2) = 70.71∠ − 45° V ∴

∴ vx (t ) = 70.71cos(1000t − 45°) V



66. (a)


10 March 2006

Assume V3 = 1V ∴ V2 = 1 − j 0.5V, I 2 = 1 − j 0.5 mA ∴ V1 = 1 − j 0.5 + (2 − j 0.5) (− j 0.5) = 0.75 − j1.5V ∴ I1 = 0.75 − j1.5 mA, ∴ Iin = 0.75 − j1.5 + 2 − j 0.5 = 2.75 − j 2 mA ∴ Vin = 0.75 − j1.5 − j1.5 + (2.75 − j 2) (− j 0.5) = −0.25 − j 2.875 V ∴ V3 =

(b)

100 = 34.65+ ∠94.97°V − j 0.25 − j 2.875

− j 0.5 → − jx Assume 1V3 = V ∴ I 3 = 1A, V2 = 1 − jX, I 2 = 1 − jX, → I12 = 2 − jX ∴ V1 = 1 − jX + (2 − jX) (− jX) = 1 − X 2 − j 3X, I1 = 1 − X 2 − j 3X, I in = 3 − X 2 − j 4 X ∴ Vin = 1 − X 2 − j 3X − 4X 2 + jX 3 − j3X = 1 − 5X 2 + j (X 3 − 6X) ∴ X 3 − 6X = 0 ∴ X 2 = 6, X = 6, Z c = − j 2.449 kΩ



67. m


10 March 2006

Define three clockwise mesh currents i1, i2, i3 with i1 in the left mesh, i2 in the top right esh, and i3 in the bottom right mesh. Mesh 1: -10∠0o + (1 + 1 – j0.25)I1 – I2 – (-j0.25)I3 = 0 Mesh 2: – I1 + (1 + 1 + j4)I2 – I3 = 0 Mesh 3: (-j0.25 + 1 + 1)I3 – I2 – (-j0.25I1) = 0 −1 10 2 − j 0.25 −1 2 + j4 0 −1 j 0.25 0 Ix = 2 − j 0.25 −1 j 0.25 −1

2 + j4

−1

j 0.25

−1

2 − j 0.25

10 (1 + 1 − j 0.5) j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1) 20 − j 5 = ∴ I x = 1.217∠ − 75.96° A, ix (t ) = 1.2127 cos (100t − 75.96°) A 8 + j15

∴Ix =



68.


10 March 2006

V1 − 10 − j 0.25V1 + j 0.25Vx + V1 − V2 = 0 ∴ (2 − j 0.25) V1 − V2 + j 0.25 Vx = 10 V2 − V1 + V2 − Vx + j 4V2 = 0 − V1 + (2 + j 4) V2 − Vx = 0 − j 0.25Vx + j 0.25V1 + Vx + Vx − V2 ∴ j 0.25V1 − V2 + (2 − j 0.25) Vx = 0 −1 10 2 − j 0.25 −1 2 + j4 0 −1 j 0.25 0 Vx = j 0.25 − j 0.25 −1 −1 −1 2 + j4 j 0.25 −1 2 − j 0.25 10 (1 + 1 − j 0.5) j 0.25(2 − j 0.5) + (−2 + j 0.25 + j 0.25) + (2 − j 0.25) (4 + 1 − j 0.5 + j8 − 1) 20 − j 5 = = 1.2127∠ − 75.96° V 8 + j15 ∴ vx = 1.2127 cos(100t − 75.96°) V =



69. (a)


10 March 2006

R1 = ∞, R o = 0, A = −Vo / Vi >> 0 I=

V1 + AVi = jω C1 (Vs − Vi ) Rf

∴ Vi (1 + A + jω C1R f ) = jω C1R f Vs Vo (1 + A + jω C1R f ) = jω C1R f Vs A jω C1R f A V V ∴ o =− As A → ∞, o → − jω C1R f Vs 1 + A + jω C1R f Vs Vo = − AVi ∴−

(b)

R f Cf =

I=

1 jω C f +

1 Rf

=

Rf 1 + jω C f R f

(V1 + AVi ) (1 + jω C f R f ) = (Vs − Vi ) jω C1 , Vo = − AVi Rf

∴ Vi (1 + A) (1 + jω C f R f ) = Vs jω C1R f − jω C1R f Vi , Vi [(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs Vo [(1 + A) (1 + jω C f R f ) + jω C1R f ] = jω C1R f Vs A − jω C1R f − jω C1R f A V V ∴ o = As A → ∞, o → Vs 1 + jω C f R f Vs (1 + A) (1 + jω C f R f ) + jω C1R f ∴−



70.


10 March 2006

Define the nodal voltage v1(t) at the junction between the two dependent sources. The voltage source may be replaced by a 3∠-3o V source, the 600-μF capacitor by a –j/ 0.6 Ω impedance, the 500-μF capacitor by a –j2 Ω impedance, and the inductor by a j2 Ω impedance. V1 - 3∠ - 3o (V - V ) 5V2 + 3V2 = + 1 2 [1] 100 − j / 0.6 - j2

-5V2 =

( V2 − V1 ) − j2

+

V2 j2

[2]

Solving, we find that V2 = 9.81 ∠ – 13.36o mV. Converting back to the time domain, v2(t) = 9.81 cos (103t – 13.36o) mV



71. m

and Mesh


10 March 2006

Define three clockwise mesh currents: i1(t) in the left-most mesh, i2(t) in the bottom right esh, and i3(t) in the top right mesh. The 15-μF capacitor is replaced with a –j/ 0.15 Ω impedance, the inductor is replaced by a j20 Ω impedance, the 74 μF capacitor is replaced by a –j1.351 Ω impedance, the current source is replaced by a 2∠0o mA source, and the voltage source is replaced with a 5∠0o V source. Around the 1, 2 supermesh: (1 + j20) I1 + (13 – j1.351) I2 – 5 I3 = 0 –I1 + I2 = 2×10–3 3:

5

∠0o – 5 I2 + (5 – j6.667) I3 = 0

Solving, we find that I1 = 148.0∠179.6o mA. Converting to the time domain, i1(t) = 148.0cos (104t + 179.6o) μA Thus, P1Ω = [i1(1 ms)]2 • 1 = (16.15×10–3)(1) W = 16.15 mW.



72.


10 March 2006

We define an additional clockwise mesh current i4(t) flowing in the upper right-hand mesh. The inductor is replaced by a j0.004 Ω impedance, the 750 μF capacitor is replaced by a –j/ 0.0015 Ω impedance, and the 1000 μF capacitor is replaced by a –j/ 2 Ω impedance. We replace the left voltage source with a a 6 ∠ -13o V source, and the right voltage source with a 6 ∠ 0o V source. = 6 ∠ –13o [1]

(1 – j/ 0.0015) I1 + j/0.0015I2 – I3 (0.005 + j/ 0.0015) I1 + (j0.004 – j/0.0015) I2 –I1

+ (1 – j500) I3

– j0.004 I4 = 0 +

[2]

j500 I4 = –6 ∠ 0o [3]

–j0.004 I2 + j500I3 + (j0.004 – j500) I4 = 0

[4]

Solving, we find that I1 = 2.002∠ –6.613o mA, I2 = 2.038 ∠ –6.500o mA, and I3 = 5.998 ∠ 179.8o A.

Converting to the time domain, i1(t) = 1.44 cos (2t – 6.613o) mA i2(t) = 2.038 cos (2t – 6.500o) mA i3(t) = 5.998 cos (2t + 179.8o) A



73. Defi


We replace the voltage source with a 115 2 ∠0o V source, the capacitor with a –j/ 2πC1 Ω impedance, and the inductor with a j0.03142 Ω impedance. ne Z such that Z-1 = 2πC1 - j/0.03142 + 1/20 By voltage division, we can write that 6.014 ∠85.76o = 115 2

Thus,

10 March 2006

Z Z + 20

Z = 0.7411 ∠ 87.88o Ω. This allows us to solve for C1:

2πC1 – 1/0.03142 = -1.348 so that C1 = 4.85 F.



74.


10 March 2006

Defining a clockwise mesh current i1(t), we replace the voltage source with a 115 2 ∠0o V source, the inductor with a j2πL Ω impedance, and the capacitor with a –j1.592 Ω impedance. Ohm’s law then yields I1 = Thus, 20 =

115 2 = 8.132∠0o 20 + j (2πL − 1.592 )

202 + (2πL − 1.592)

2

and we find that L = 253.4 mH.



75.


10 March 2006

(a) By nodal analysis: 0 = (Vπ – 1)/ Rs + Vπ / RB + Vπ / rπ + jωCπ Vπ + (Vπ – Vout) jωCμ [1] -gmVπ = (Vout – Vπ) jωCμ + Vout / RC + Vout / RL

[2]

Simplify and collect terms: ⎡⎛ 1 ⎤ 1 1⎞ 1 [1] + + ⎟⎟ + jω (Cπ + C μ )⎥ Vπ - jωC μ Vout = ⎢⎜⎜ RS ⎣⎢⎝ R S R B rπ ⎠ ⎦⎥

(-gm + jωCμ) Vπ - (jωCμ + 1/RC + 1/RL) Vout = 0 Defi

ne

1

′ RS

1 1 1 + + R S R B rπ

′ and R L = RC || RL

⎛ C + Cπ C μ ⎞⎟ -1 + ω 2 (2C2μ + C μ Cπ ) - jω ⎜ g mC μ + μ + ′ ′ ′ ′ ⎜ RS R L R R S ⎟⎠ L ⎝

Then

Δ =

And

Vout =

Therefore,

=

[2]

g m R S − jω C μ R S

⎛ C μ + Cπ C μ ⎞⎟ -1 + ω 2 2C2μ + C μ Cπ - jω ⎜ g m C μ + + ′ ′ ′ ′ ⎜ RS R L RL R S ⎟⎠ ⎝ ⎛ ⎛ ⎞⎞ ⎜ − ω ⎜ g C + C μ + Cπ + C μ ⎟ ⎟ ⎜ ′ ′ ⎟ ⎜ m μ − jωCμ ⎞ RL R S ⎟⎠ ⎟ −1 ⎛ ⎝ −1 ⎜ ⎟ tan ang(Vout) = tan ⎜⎜ 2 ⎟ ⎜ -1 ⎟ 2 2 ⎝ g m RS ⎠ ⎜ ′ ′ + ω (2C μ + C μ Cπ ) ⎟ ⎜ RS R L ⎟ ⎝ ⎠

(

)

(b)

(c) The output is ~180o out of phase with the input for f < 105 Hz; only for f = 0 is it exactly 180o out of phase with the input. PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


76. OC : −


10 March 2006

Vx 100 − Vx + − 0.02Vx = 0 − j10 20

j10 = (0.05 + j 0.1 + 0.02) Vx , Vx =

j10 0.07 + j 0.1

∴ Vx = 67.11 + j 46.98 ∴ Vab ,oc = 100 − Vx = 32.89 − j 46.98 = 57.35∠ − 55.01° V SC :Vx = 100 ∴↓ I SC = 0.02 ×100 + ∴ Zth =

100 = 7A 20

57.35∠ − 55.01° = 4.698 − j 6.711Ω 7



77.


10 March 2006

Let I in = 1∠0. Then VL = j 2ω I in = j 2ω ∴ 0.5VL = jω ∴ Vin = (1 + jω )

1 + j 2ω jω

1 + j 2ω jω ω V 1 ∴ Zin = in = 1 + + j 2ω so Yin = 1 jω ω + j (2ω 2 − 1) At ω = 1, Zin = 1 − j1 + j 2 = 1 + j = 1+

∴ Yin =

1 = 0.5 − j 0.5 1 + j1

R = 1/0.5 = 2 Ω

and

L = 1/0.5 = 2 H.




10 March 2006

78. (a)

(b) I

−15 (1 − j1)1 2 + j1 3 − j1 × = ∴ V1 = × 0.6 − j 0.2 2 − j1 2 + j1 5 j 2 + 0.6 − j 0.2 ∴ V1 = 5∠90°∴ v1 (t ) = 5cos (1000t + 90°) V Vs :

s:

j2 1− j2 = 0.8 + j 0.4 ∴ V1 1+ j2 1− j2 −10 + j 20 0.8 + j 0.4 = j 25 = = 11.785+ ∠135° V 1 − j1 + 0.8 + j 0.4 1.8 − j 0.6 j2 1=

so

v1(t) = 11.79 cos (1000t + 135o) V.




10 March 2006

79. OC :VL = 0 ∴ Vab ,oc = 1∠0° V SC : ↓ I N ∴ VL = j 2I N ∴1∠0° = − j1[0.25( j 2I N ) + I N ] + j 2I N ∴1 = (0.5 − j + j 2) I N = (0.5 + j1) I N I 1 = 0.4 − j 0.8 ∴ YN = N = 0.4 − j 0.8 0.5 + j1 1∠0° 1 1 1 1 ∴RN = = 2.5 Ω, = = − j 0.8, L N = = 1.25H 0.4 0.8 jω L N jL N ∴ IN =

I N = 0.4 − j 0.8 = 0.8944∠ − 63.43° A



80.

=


10 March 2006

To solve this problem , we e mploy super position in order to separate sources having different frequencies. First cons idering the sources operating at w = 200 rad/s, we opencircuit the 100 rad/s current source. This leads to VL′ = (j)(2∠0) = j2 V. Therefore, vL′ ( t ) 2cos(200t + 90o) V. For the 100 rad/s source, we find j VL′′ = (1∠0 ) , vL′′ = 0.5cos (100t + 90°) V 2 ∴ vL ( t ) = 2 cos (200t + 90°) + 0.5cos (100t + 90°) V




10 March 2006

81.

j100 j100 − j 300 − j 300 = −50∠0° V Right: Vab = j100 = j150 V − j 300 + j100 ∴ Vth = −50 + j150 = 158.11∠108.43° V Use superposition. Left: Vab = 100

Zth = j100 − j 300 =

30, 000 = j150 Ω − j 200



82.

Then


10 March 2006

This problem is easily solved if we first perform two source transformations to yield a circuit containing only voltage sources and impedances:

5∠17o + 0.240∠-90o − 2.920∠-45o 73 + 10 + j13 − j 4 = (4.264∠ 50.42o)/ (83.49 ∠ 6.189o) = 51.07 ∠ 44.23 mA I =

Converting back to the time domain, we find that i(t) = 51.07 cos (103t + 43.23o) mA




10 March 2006

83.

(a) There are a number of possible approaches: Thévenizing everything to the left of the capacitor is one of them. VTH = 6(j2)/ (5 + j2) = 2.228 ∠ 68.2o V ZTH = 5 || j2 = j10/ (5 + j2) = 1.857 ∠ 68.2o Ω

Then, by simple voltage division, we find that − j /3 VC = (2.228 ∠ 68.2o) 1.857∠68.2o - j / 3 + j 7 = 88.21 ∠-107.1o mV Converting back to the time domain, vC(t) = 88.21 cos (t – 107.1o) mV. (b)

PSpice verification.

Running an ac sweep at the frequency f = 1/2π = 0.1592 Hz, we obtain a phasor magnitude of 88.23 mV, and a phasor angle of –107.1o, in agreement with our calculated result (the slight disagreement is a combination of round-off error in the hand calculations and the rounding due to expressing 1 rad/s in Hz.



84.


10 March 2006

(a) Performing nodal analysis on the circuit, Node 1:

Node

2:

1 = V1/ 5 + V1/ (-j10) + (V1 – V2)/ (-j5) + (V1 – V2)/ j10 j0.5 = V2/ 10 + (V2 – V1)/ (-j5) + (V2 – V1)/ j10

[1]

[2]

Simplifying and collecting terms, (0.2 + j0.2) V1 – j0.1 V2 = 1

[1]

-j V1 + (1 + j) V2 = j5

[2]

Solving, we find that V2 = VTH = 5.423 ∠ 40.60o V ZTH = 10 || [(j10 || -j5) + (5 || -j10)] = 10 || (-j10 + 4 – j2) = 5.882 – j3.529 Ω.

(b)

FREQ

VM($N 0002,0)

VP($N 0002,0)

1.592E+01

4.474E+00

1.165E+02

FREQ

VM($N_0005,0)

VP($N_0005,0)

1.592E+01

4.473E+00

1.165E+02



85.


10 March 2006

Consider the circuit below: Vout Vin

1 jω C

Using voltage division, we may write: Vout = Vin

V out 1 / jωC , or R + 1 / jωC V in

=

1 1 + j ω RC

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is 1 Vout = 2 Vin 1 + (ωRC ) As

ω → 0, this magnitude → 1, its maximum value.

As

ω → ∞, this magnitude → 0; the capacitor is acting as a short circuit to the ac signal.

=

Thus, low frequency signals are transferred from the input to the output relatively unaffected by this circuit, but high frequency signals are attenuated, or “filtered out.” This is readily apparent if we plot the magnitude as a function of frequency (assuming R 1 Ω and C = 1 F for convenience):



86.


10 March 2006

Consider the circuit below:

Vout

1/jωC

Vin

R

Using voltage division, we may write:

Vout = Vin

R jωRC V , or out = R + 1 / jωC Vin 1 + jωRC

The magnitude of this ratio (consider, for example, an input with unity magnitude and zero phase) is ωRC Vout = 2 Vin 1 + (ωRC ) As

ω → ∞, this magnitude → 1, its maximum value.

As

ω → 0, this magnitude → 0; the capacitor is acting as an open circuit to the ac signal.

=

Thus, high frequency signals are transferred from the input to the output relatively unaffected by this circuit, but low frequency signals are attenuated, or “filtered out.” This is readily apparent if we plot the magnitude as a function of frequency (assuming R 1 Ω and C = 1 F for convenience):



87.


10 March 2006

(a) Removing the capacitor temporarily, we easily find the Thévenin equivalent:

Vth = (405/505) VS and Rth = 100 || (330 + 75) = 80.2 Ω

80.2 Ω 405 VS 505

31.57 fF

(b)

Vout =

405 1/jωC VS 505 80.2 + 1 / jωC

and

hence

Vout = VS

so

+ Vout -

Vout 1 ⎛ 405 ⎞ = ⎜ ⎟ −12 VS ⎝ 505 ⎠ 1 + j 2.532 × 10 ω

0.802 1 + 6.411 × 10− 24 ω 2

(c)

Both the MATLAB plot of the frequency response and the PSpice simulation show essentially the same behavior; at a frequency of approximately 20 MHz, there is a sharp roll-off in the transfer function magnitude.



88.


10 March 2006

From the derivation, we see that

- g m (R C || R L ) + jω (R C || R L )C μ Vout = Vin 1 + jω (R C || R L )Cμ so

that 1

2 ⎡ 2 ⎛ R R ⎞2 ⎤ 2 2 2 ⎛ R CR L ⎞ C L ⎟ Cμ ⎥ ⎟ + ω ⎜⎜ ⎢ g m ⎜⎜ R C + R L ⎟⎠ R C + R L ⎟⎠ Vout ⎢ ⎥ ⎝ ⎝ = ⎢ 2 ⎥ Vin 2 2 ⎛ R CR L ⎞ ⎢ ⎥ ⎟⎟ C μ 1 + ω ⎜⎜ ⎢ ⎥ ⎝ RC + RL ⎠ ⎣ ⎦ This function has a maximum value of gm (RC || RL) at ω = 0. Thus, the capacitors reduce the gain at high frequencies; this is the frequency regime at which they begin to act as short circuits. Therefore, the maximum gain is obtained at frequencies at which the capacitors may be treated as open circuits. If we do this, we may analyze the circuit of Fig. 10.25b without the capacitors, which leads to

Vout VS

low frequency

⎛ R R ⎞ (rπ || R B ) ⎛ R R ⎞ rπ R B = - g m ⎜⎜ C L ⎟⎟ = - g m ⎜⎜ C L ⎟⎟ ⎝ R C + R L ⎠ R S + rπ || R B ⎝ R C + R L ⎠ R S (rπ + R B ) + rπ R B

The resistor network comprised of rπ, RS, and RB acts as a voltage divider, leading to a reduction in the gain of the amplifier. In the situation where rπ || RB >> RS, then it has minimal effect and the gain will equal its “maximum” value of –gm (RC || RL). (b) If we set RS = 100 Ω, RL = 8 Ω, RC | max = 10 kΩ and rπgm = 300, then we find that

Vou t rπ || R B = - g m (7.994) VS 100 + rπ || R B of

We seek to maximize this term within the stated constraints. This requires a large value gm, but also a large value of rπ || RB. This parallel combination will be less than the smaller of the two terms, so even if we allow RB → ∞, we are left with Vou t - 2398 g r ≈ - (7.994) m π = VS 100 + rπ 100 + rπ

Considering this simpler expression, it is clear that if we select rπ to be small, (i.e. rπ

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