solutions_chap09 Engineering Circuit Analysis, 6 th Edition.pdf PDF

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CHAPTER NINE SOLUTIONS 1. 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 1 ∴−6 = −7 + 49 − ω o2 ∴ω o2 = 48 , 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 Engineering Circuit Analysis, 6...

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CHAPTER NINE SOLUTIONS 1.

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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 2. (a)

ic = 40e −100t − 30e −200t mA, C = 1mF, v(0) = −0.25V t 1 t − = i dt 0.25 (40e−100t − 30e−200t ) dt − 0.25 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 = 150 s − 1 1 500 ∴150 + = 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−100 t + 0.045e−200 t 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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 3. (a)

(b)

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 =1 MΩ So R = 10 (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 S2 = −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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS Given: L = 4 R 2C , α =

4.

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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 5.

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  12 in  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%

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 6.

L = 5H, R = 8Ω, C = 12.5mF, v (0+ ) = 40V

(a)

i (0+ ) = 8A: α = ωo = 4 s1,2

1 1000 1 = = 5, ωo2 = = 16, 2RC 2 × 8 × 12.5 LC = −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 =

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS

7.

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

(a)

α=

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 ; v(0)  1  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 −10 t − 600e −40 t − 0.2(-20)(-10)e -10t − (0.2)(60)(-40)e −40t dt −10 t = 160e − 120e −40t A

i(t) = – v/ R – C

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 8.

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−6000 t , 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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CHAPTER NINE SOLUTIONS 9.

12 = 2A, vc (0) = 2V 5 +1 1000 1000 × 45 α= = 250, ω o2 = = 22500 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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 10. (a)

1 1440 1440 = = 20, ω o2 = = 144 2RC 72 10 s1,2 = −20 ± 400 − 144 = −4, − 36: v = A1e−4t + A 2 e−36t

α=

 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.

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 11. so

L = 1250 mH ωo = 1 = 4 rad/s Since α > ωo, this circuit is over damped. LC 1 α= = 5 s −1 2 RC

The capacitor stores 390 J at t = 0−: 1 Wc = C vc2 2 2Wc So vc (01 ) = = 125 V = vc (0+ ) 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+ ) + + + iL (0 ) + iR (0 ) + ic (0 ) = 0 + + ic (0+ ) = 0 2 + v(0 ) 125 So ic (0 + ) = − =− = −62.5 V 2 2 dv ic = C = 50 ×10−3[−8 Ae −8t − 2 Be −2t ] dt ic (0+ ) = −62.5 = −50 × 10−3 (8 A + 2 B) [2] Solving Eqs. [1] and [2],

[1]

A = 150 V B = −25 V

Thus, v(t ) = 166.7e−8t − 41.67 e−2 t , t > 0

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 12.

We want a response v = Ae−4t + Be−6t 1 α= = 5 s −1 2 RC

(a)

S1 = −α + α 2 − ω2o = −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 LC ωo C

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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CHAPTER NINE SOLUTIONS iL (0− ) = iL (0+ ) = 0

13.

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+ ) =

50 =2A 25

s1,2 = −α ± α 2 − ωo2 = −2000, − 6000

Since α > ω0, this system is overdamped. Thus, vc (t ) = Ae −2000t + Be −6000 t dv = (5 ×10−6 ) (−2000 Ae−2000t − 6000 Be −6000t ) dt ic (0+ ) = −0.01A − 0.03B = −2 [1] ic = C

and vc (0+ ) = A + B = 50

[2]

Solving, we find A = −25 and B = 75 so that vc (t ) = −25e−2000t + 75e−6000 t , t > 0 (d)

(e)

−25e −2000t + 75e −6000 t = 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.

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 14.

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 + ) + 2i A (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 vLC = 6 V ∴ RTH =

1A

(e)

1 = 3.333 s −1 2 RC 1 ωo = = 3 rad/s LC

6 = 6Ω 1

α=

Thus, iA (t ) = Ae−1.881t + Be−4.785t

S1,2 = −α ± α 2 − ωo2 = −1.881, − 4.785

iA (0+ ) = 0 = A + B

To find the second equation required to determine the coefficients, we write: iL = −ic − iR = −C

dvc − i A = −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 ]

Engineering Circuit Analysis, 6th Edition

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[1]

CHAPTER NINE SOLUTIONS Diameter of a dime: approximately 8 mm. Area = π r 2 = 0.5027cm 2

15.

Capacitance L = 4µH ωo =

εr εo A (88) (8.854 ×10−14 F/cm) (0.5027cm2 ) = = 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.

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 16.

R

crit. damp.

(a)

L = 4R 2 C =

100 × 10−3 = 4R 2 × 10−6 ∴ R = 57.74Ω 3

1 × 2.5 = 3464s −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  ∴ vc (t ) = e −3464t (3.464 × 105 t + 100) V, t > 0 iL (0) =

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 17. 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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 18.

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)

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 −6

∴ tm = 8.590 µ s ∴ i (tm ) = e−141,420×8.590×10

(−65,860 × 8.590 × 10−6 + 0.1) = −0.13821A ∴ i (c)

max

= i (tm ) = 0.13821A

∴imax = i (0) = 0.1A

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 19.

Diameter of a dime is approximately 8 mm. The area, therefore, is πr2 = 0.5027 cm2.

ε r ε o 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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 20.

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 L= = 2 × 109 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

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 21.

α=

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+ ) = −

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 22. 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+ ) = B 2 (5) = −16 ∴ B2 = −3.2, vc (t ) = −3.2e −t sin 5t V, t > 0

(f)

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 23. (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 cos10 4 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+ ) = c 5 =

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 [10 4 (−200sin + 100 cos] − 5000 (200 cos + 100sin)] = e−500 t [106 (−2 sin − 0.5cos)] = −2.5 × 106 e−5000t sin104 t v / s  1  ∴ iL = e−5000t  (200 cos + 100sin) − 5 × 10−9 × 2.5 × 106 e−5000t sin104 t   20,000  4 4 −5000 t (0.01cos10 t − 0.0075sin10 t ) A =e ∴ isw = 10 − e−5000t (10 cos104 t − 7.5sin104 t ) mA, t > 0

Engineering Circuit Analysis, 6th Edition

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CHAPTER NINE SOLUTIONS 24. (a)

(b)

1 100 1 100 2 , ωd = 36 = ωo2 − 64 = = 8, ωo2 = = 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)

Engineering Circuit Analysis, 6th Edition

Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

CHAPTER NINE SOLUTIONS 25.

1 106−3 1 = = 100s −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

Engineering Circuit Analysis, 6th Edition

Copyright 2002 McGraw-Hill, Inc. All Rights Reserved

CHAPTER NINE SOLUTIONS 26. (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....