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Engineering Circuit Analysis, 7th Edition

1.

Chapter Nine Solutions

10 March 2006

Parallel RLC circuit: (a) α =

1 1 1 = = = 175 × 103 s −1 −6 2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (2.857)(10−6 )

(b) ω0 =

1 = LC

1

( 2 ×10 )(10 ) −3

−6

=

22.4 krad/s

(c) The circuit is overdamped since α > ω0 .

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.

Engineering Circuit Analysis, 7th Edition

2.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

3.

Chapter Nine Solutions

10 March 2006

Parallel RLC circuit: (a) α =

1 1 1 = = = −6 2 RC ( 2 ) (4 ||10)(10 ) ( 2 ) (1)(10−9 )

5 × 108 s −1

ω0 =

1 = LC

31.6 Trad/s

1

(10 )(10 ) −12

−9

= 3.16 × 1013 rad/s =

(b) 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 .

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Engineering Circuit Analysis, 7th Edition

4.

Chapter Nine Solutions

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.

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Engineering Circuit Analysis, 7th Edition

5.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

6.

(a)

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

7. (a)

(b)

Chapter Nine Solutions

10 March 2006

Parallel RLC with ωo = 70.71 × 1012 rad/s. L = 2 pH. 1 = (70.71× 1012 ) 2 LC 1 So C = = 100.0 aF 12 2 (70.71×10 ) (2 ×10−12 ) ωo2 =

1 = 5 × 109 s −1 2 RC 1 So R = 10 =1 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

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Engineering Circuit Analysis, 7th Edition

Given: L = 4 R 2C , α =

8.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

9.

Chapter Nine Solutions

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%

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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 =

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Engineering Circuit Analysis, 7th Edition

11.

1 L 1 10−3 1 10 = = = 2 C 2 10−4 2 Therefore (a) RC =

Chapter Nine Solutions

10 March 2006

1.581 Ω

R = 0.1RC = 158.1 mΩ (b) α =

1 = 3.162 ×104 s −1 and ω0 = 2RC

1 = 3.162 × 103 rad/s LC

Thus, s1,2 = −α ± α 2 − ω02 = − 158.5 s −1 and − 6.31× 104 s −1 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

[1]

Noting v(0+ ) = L

(

di = 10 dt t =0

)

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

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Engineering Circuit Analysis, 7th Edition

12.

(a) α =

Chapter Nine Solutions

1 = 500 s −1 and ω0 = 2RC

10 March 2006

1 = 100 rad/s LC

Thus, s1,2 = −α ± α 2 − ω02 = − 10.10 s −1 and − 989.9 s −1 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

Thus, iC (0+ ) = C

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

Therefore, 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

)

mA (c) We see that the simulation agrees.

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Engineering Circuit Analysis, 7th Edition

13.

i (0) = 40A, v(0) = 40V, L =

(a)

α=

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

14.

(a) α =

Chapter Nine Solutions

1 = 6.667 × 108 s −1 and ω0 = 2RC

10 March 2006

1 = 105 rad/s LC

Thus, s1,2 = −α ± α 2 − ω02 = − 7.5 s −1 and − 1.333 × 109 s −1 . So we may write 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 9

A1+ A2 = –0.133×106 We need to find

C

diR dt

. We know that L t =0

[2] di dt

di 2 = = 106 . Also, dt t =0 2 ×10−6

= 2 so t =0

(

)

9 i di di 1 dv 1 dv so R = C = = iC and R = A1e−7.5t + A2 e−1.333×10 t . dt dt R dt dt CR CR

Using

di di diR diC + + = 0 so C dt dt dt dt

= −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

(

iC (t ) = − 0.75 ×10−3 e −7.5t + 0.133 × 106 e−1.333×10

9

t

)

A

(b)

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Engineering Circuit Analysis, 7th Edition

15.

(a) α =

1 = 0.125 s −1 and ω0 = 2RC

Chapter Nine Solutions

10 March 2006

1 = 0.112 rad/s LC

Thus, 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.

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Engineering Circuit Analysis, 7th Edition

16.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

17.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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.

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Engineering Circuit Analysis, 7th Edition

19.

Chapter Nine Solutions

10 March 2006

Referring to Fig. 9.43, L = 1250 mH

so

ωo = 1 α=

LC

= 4 rad/s

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

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Engineering Circuit Analysis, 7th Edition

20.

Chapter Nine Solutions

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 ωo2 = (b)

1 1 , L = 2 = 833.3 mH ωoC LC

If iR (0+ ) = 10 A and ic (0+ ) = 15 A, find A and B. with 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 Thus, v = 210e −4t − 190e −6t , t > 0

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

iL (0− ) = iL (0+ ) = 0

21.

Initial conditions:

(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 =2A 25

s1,2 = −α ± α 2 − ωo2 = −2000, − 6000

Since α > ω0, this system is overdamped. Thus, 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 vc (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

max

= −25 + 75 = 50 V

So, solving | −25e −2000ts + 75e −6000ts | = 0.5 in view of the graph in part (d), we find ts = 1.955 ms using a scientific calculator’s equation solver routine. 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.

Engineering Circuit Analysis, 7th Edition

22.

Chapter Nine Solutions

10 March 2006

Due to the presence of the inductor, vc (0− ) = 0 . Performing mesh analysis, →i1

→i2 4.444 H

−9 + 2i1 − 2i2 = 0

[1]

2i2 − 2i1 + 3iA + 7i2 = 0

[2]

and i1 − i2 = iA Rearranging, we obtain 2i1 – 2i2 = 0 and –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] Solving Eqs. [1] and [2], A = −20.66 and B = +20.66 So that iA (t ) = 20.66[e −4.785t − e−1.881t ] 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.

Engineering Circuit Analysis, 7th Edition

23.

Chapter Nine Solutions

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.

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Engineering Circuit Analysis, 7th Edition

24.

(a) For critical damping, R =

(b) α =

Chapter Nine Solutions

10 March 2006

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

(

)

Thus, vC ( t ) = e −9.129×10 t ( A1t + A2 ) 3

At t = 0, vC (0) = A1 ( 0 ) + A2 = 12

[1] ∴ A2 = 12 V .

Taking the derivative of Eq. [1], dvC ( t ) dt

= e −9.129×10 t ⎡⎣ −9.129 × 103 A1t + A1 − 9.129 × 103 (12 ) ⎤⎦ 3

and 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).

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Engineering Circuit Analysis, 7th Edition

25.

(a) For critical damping, R =

(b) α =

Chapter Nine Solutions

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

(

)(

)

Thus, iL ( t ) = e−3.162×10 t ( A1t + A2 ) 5

[1]

At t = 0, iL (0) = A1 ( 0 ) + A2 = 10

∴ A2 = 10 A .

Taking the derivative of Eq. [1], diL ( t ) dt

= e −3.162×10 t ⎡⎣ −3.162 × 105 A1t + A1 − 3.162 × 105 (10 ) ⎤⎦ [2]

and 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).

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Engineering Circuit Analysis, 7th Edition

26.

Chapter Nine Solutions

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.

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

10 March 2006

27. L

crit. damp.

(b)

(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

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

10 March 2006

28.

R

crit. damp.

(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 /

(b)

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) =

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Engineering Circuit Analysis, 7th Edition

29.

Chapter Nine Solutions

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

with L = 4μH, ωo =

1 = 79.89 Mrad/s LC

For critical damping, we require or R =

1 = ωo 2 RC

1 = 159.8 Ω 2ωoC

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

31.

Chapter Nine Solutions

10 March 2006

Critically damped parallel RLC with α = 10−3 s −1 , R = 1MΩ .

1 103 = 10−3 , so C = = 500 μF 2 RC 2 ×106 1 Since α = ωo, ωo = = 10−3 LC 1 = 10−6 or LC so L = 2 GH (!) We know

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 s = 8.106 ×1013 cm

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Chapter Nine Solutions

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)

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Engineering Circuit Analysis, 7th Edition

33.

α=

Chapter Nine Solutions

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+ ) = −

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Chapter Nine Solutions

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)

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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

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Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

37.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

38.

Chapter Nine Solutions

10 March 2006

We replace the 25-Ω resistor to obtain an underdamped response:

α = Thus,

1 2RC

and

1 10 × 10− 6 R

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.

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Engineering Circuit Analysis, 7th Edition

39.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

40.

Chapter Nine Solutions

10 March 2006

(a) 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. With 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 and so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)] With vC(0+) = 0 = (2/13) (5.083D – 0.4C), we obtain D = 0.3148 A. Thus, iL(t) = e-0.4t (4 cos 5.083t + 0.3148 sin 5.083t) A and iL(2.5) = 1.473 A. (b) 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, iL(t) = e-4t (4 cos 3.162t + 5.06 sin 3.162t) A and iL(.25) = 2.358 A. (c)

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.

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Engineering Circuit Analysis, 7th Edition

41.

Chapter Nine Solutions

10 March 2006

(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. With 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 and so vC(t) = (2/13) [e-αt (-Cωd sin ωdt + Dωd cos ωdt) - αe-αt (C cos ωdt + D sin ωdt)] With vC(0+) = 0 = (2/13) (5D – 4), we obtain D = 0.8 A. Thus, iL(t) = e-t (4 cos 5t + 0.8 sin 5t) A

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.

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

10 March 2006

42. vc (0) = 50 + 80 × 2 = 210 V, iL (0) = 0, α =

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, iL = 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]

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Engineering Circuit Analysis, 7th Edition

43.

Series:

Chapter Nine Solutions

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

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44.

(a)

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

45.

Chapter Nine Solutions

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)

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Engineering Circuit Analysis, 7th Edition

46.

(a)

Chapter Nine Solutions

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 iL max = 0.5345A

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Engineering Circuit Analysis, 7th Edition

47.

(a)

α=

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

48.

Chapter Nine Solutions

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 underdamped 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:

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Engineering Circuit Analysis, 7th Edition

49.

Chapter Nine Solutions

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 At t = 0+, iL = 0 so we have = 0 [3] -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.

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Engineering Circuit Analysis, 7th Edition

50.

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

51.

Chapter Nine Solutions

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 For a critically damped circuit, we require α = ωo, or , which, with = 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) Noting that iL(0+) = iL(0-) = 2.25, we see that B = 2.25 and hence

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

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Engineering Circuit Analysis, 7th Edition

52.

Chapter Nine Solutions

10 March 2006

v ⎛ 50 ⎞ Prior to t = 0, we find that v = (10 + i1 ) ⎜ ⎟ and i1 = 5 ⎝ 15 ⎠ ⎛ 10 ⎞ 500 so v = 100 V . Thus, v ⎜1 − ⎟ = ⎝ 15 ⎠ 15 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. Thus, α =

R 10 = = 2.5 × 103 s −1 2 L 2 2 ×10−3

and ω0 =

1 = LC

(

)

1

( 2 ×10 )( 20 ×10 ) −3

−9

= 1.581× 105 rad/s

With α < ω0 we find the circuit is underdamped, with

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

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

Finally, iL (t ) = −316e−2500t sin1.581× 105 t mA

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Engineering Circuit Analysis, 7th Edition

53.

Chapter Nine Solutions

10 March 2006

Prior 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. Thus, α =

R 10 = = 2.5 × 103 s −1 −3 2 L 2 2 ×10

and ω0 =

1 = LC

(

)

1

( 2 ×10 )( 20 ×10 ) −3

−9

= 1.581× 105 rad/s

With α < ω0 we find the circuit is underdamped, with

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 .

dvC = iL and dt

Noting 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

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Engineering Circuit Analysis, 7th Edition

54.

Chapter Nine Solutions

10 March 2006

Prior to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V. Thus, 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.

After t = 0 we are left with a series RLC circuit where i1 = −

R 1.25 = = 0.208 s −1 2L 6 1 1 and ω0 = = = 577 mrad/s 3 LC

Thus, α =

With α < ω0 we find the circuit is underdamped, so that

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 . diL dt

Noting 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

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Engineering Circuit Analysis, 7th Edition

55.

Chapter Nine Solutions

10 March 2006

Prior to t = 0, i1 = 10/4 = 2.5 A, v = 7.5 V, and vg = –5 V. Thus, 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.

After t = 0 we are left with a series RLC circuit where i1 = −

R 1.25 = = 0.208 s −1 2L 6 1 1 and ω0 = = = 18.26 rad/s −3 LC 3 10 × ( )

Thus, α =

With α < ω0 we find the circuit is underdamped, so that

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

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Engineering Circuit Analysis, 7th Edition

56.

(a)

Series, driven: α =

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

10 March 2006

57. vc (0) = 0, iL (0) = 0, α =

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

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

10 March 2006

59. α=

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

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Engineering Circuit Analysis, 7th Edition

60.

(a)

α=

Chapter Nine Solutions

10 March 2006

1 8 × 106 8 ×106 × 13 2 = = ω = = 26 × 106 1000, 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)

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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=

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Engineering Circuit Analysis, 7th Edition

62.

(a)

Chapter Nine Solutions

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

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Engineering Circuit Analysis, 7th Edition

63.

(a)

Chapter Nine Solutions

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: − 2A 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

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Engineering Circuit Analysis, 7th Edition

64.

Chapter Nine Solutions

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. We therefore expect an underdamped response with ωd = 1.728 rad/s. The general form of the 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

dvC = (80×10-3)(-0.375B e-0.375t sin 1.728t dt

At t = 0+, iC = 15 + 7 – iL(0+) = 7 = (80×10-3)(1.728B) so that B = 50.64 V. Thus, 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

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Engineering Circuit Analysis, 7th Edition

65.

Chapter Nine Solutions

10 March 2006

(a) vS(0-) = vC(0-) = 2(15) = 30 V (b) iL(0+) = iL(0-) = 15 A Thus, iC(0+) = 22 – 15 = 7 A and vS(0+) = 3(7) + vC(0+) = 51 V (c) As t → ∞, the current through the inductor approaches 22 A, so vS(t→ ∞,) = 44 A. (d) We are presented with a series 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:

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Engineering Circuit Analysis, 7th Edition

66.

Chapter Nine Solutions

10 March 2006

For 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. vC(t) = e-αt (A cos ωdt + B sin ωdt)

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.

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Engineering Circuit Analysis, 7th Edition

71.

Chapter Nine Solutions

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 simulations are very sensitive to parameter values; better results were obtained using LF411 instead of 741s (both were compared to the simple LC circuit simulation.)

Simulation using 741 op amps

Simulation using LF411 op amps

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Engineering Circuit Analysis, 7th Edition

72.

Chapter Nine Solutions

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 +

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Engineering Circuit Analysis, 7th Edition

Chapter Nine Solutions

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:

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Engineering Circuit Analysis, 7th Edition

74.

Chapter Nine Solutions

10 March 2006

We see either a series RLC with R = 0 or a parallel RLC with R = ∞; either way, α = 0. (combining the two 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 ωd = 0.5477 rad/s

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

t

di dt

=0 t =0+

t

1 1 dv vdt ′ + 10-3 + ∫ vdt ′ + 2 ∫ 10 0 20 dt or d 2v = − 0.3v dt 2

= 0

↓ i 1Ω

One possible solution is:

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Engineering Circuit Analysis, 7th Edition

75.

Chapter Nine Solutions

10 March 2006

(a) vR = vL 20(-iL) = 5

diL dt

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

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