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Engineering Circuit Analysis, 7th Edition

Chapter Eleven Solutions

10 March 2006

50 (− j80) 106 = − j80 Ω, = 42.40∠ − 32.01°Ω j 500 × 25 50 − j80 ∴ V = 84.80∠ − 32.01° V, I R = 1.696∠ − 32.01° A Zc = I c = 1.0600∠57.99° A ps (π / 2ms) = 84.80 cos (45° − 32.01°) 2 cos 45° = 116.85 W pR = 50 × 1.6962 cos 2 (45° − 32.01°) = 136.55 W pc = 84.80 cos (45° − 32.01°) = 1.060 cos (45° + 57.99°) =−19.69 W

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.

Engineering Circuit Analysis, 7th Edition

Chapter Eleven Solutions

10 March 2006

2. (a)

1 2 1 Li = × 4 (4t 4 − 4t 2 + 1) 2 2 4 2 4 2 ∴ wL =8t − 8t + 2 ∴ wL (3) − wL (1) = 8 × 3 − 8 × 3 + 2 − 8 × 1 + 8 × 1 − 2 = 576 J 4H : i = 2t 2 − 1∴ v = Li′ = 4 (4t ) = 16t , wL =
1 t 2 ⎛2 3 ⎞ ⎛2 3 ⎞ ⎛2 ⎞ 0.2 F : vc = ∫1 (2t − 1) dt + 2 = 5 ⎜ 3 t − t ⎟1 + 2 = 5 ⎜ 3 t − t ⎟ − 5 ⎜ 3 − 1⎟ + 2 0.2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
t

(b)

∴ vc (2) =

10 10 61 61 × 8 − 10 − + 5 + 2 = V ∴ Pc (2) = × 7 = 142.33 W 3 3 3 3

PROPRIETARY MATERIAL. © 2007 The McGraw-HillCompanies, 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

Chapter Eleven Solutions

10 March 2006

3.

vc (0) = −2V, i (0) = 4A, α =

R 1 2 = 2, ω o = = 3, s1,2 = −2 ± 1 = −1, − 3 2L LC

(a)

1 i = Ae− t + Be−3t ∴ A + B= 4; i (0+ ) = vL (0+ ) = (−4 × 4 × +2) = −14 1 ∴−A − 38 = −14 ∴ B = 5, A = −1, i = −e − t + 5e −3t A ∴+vc = 3∫ (−e− t + 5e−3t ) dt − 2 = 3(e− t − 5e −3t ) to − 2 = e −t − 3 − 5e −3t + 5 − 2
o t

∴ vc = 3e − 5e−3t ∴ Pc (0+ ) = (3 − 5) (−1 + 5) = −8 W
−t

(b)

Pc (0.2) = (3e −0.2 − 5e−0.6 ) (−e0.2 + 5e −0.6 ) = −0.5542 W
Pc (0.4) = (3e −0.4 − 5e−1.2 ) (5e−1.2 − e−0.4 ) = 0.4220 W

(c)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

Chapter Eleven Solutions

10 March 2006

4. 2.5

We assume the circuit has already reached sinusoidal steady state by t= 0. kΩ → 2.5 kΩ, 1 H → j1000 Ω, 4 μF → -j250 Ω, 10 kΩ → 10 kΩ Zeq = j1000 || -j250 || 10000 = 11.10 – j333.0 Ω (20∠30)(11.10 − j 333.0) = 2.631∠ − 50.54o V 2500 + 11.10 − j 333.0 Veq Veq I10k = = 0.2631 ∠ - 50.54o mA I1 H = = 2.631 ∠ - 140.5o mA j1000 10000 Veq (20∠30)(2500) I4 μF = = 10.52 ∠39.46o mA V2.5k = = 19.74∠37.55o V − j 250 2500 + 11.10 − j 333.0
Veq =

Thus,

[19.74 cos 37.55 ]P2.5k =

o 2

P1 H = 2.631cos(− 50.54 ) 2.631 × 10-3 cos(−140.5o ) = - 3.395 mW
o

[

2500

=

97.97 mW

][

]

[ ][ [2.631cos(− 50.54 )] = P2.5k =
o 2

P4 μF = 2.631cos(− 50.54o ) 10.52 × 10-3 cos(39.46o ) = 13.58 mW
279.6 μW

]

10000

FREQ IM(V_PRINT1) IP(V_PRINT1) 1.592E+02 7.896E-03 3.755E+01 FREQ VM(R2_5k,\$N_0002)VP(R2_5k,\$N_0002) 1.592E+02 1.974E+01 3.755E+01FREQ IM(V_PRINT2) IP(V_PRINT2) 1.592E+02 2.628E-03 -1.405E+02

FREQ VM(L,0) 1.592E+02 2.629E+00

VP(L,0) -5.054E+01

FREQ IM(V_PRINT11) IP(V_PRINT11) 1.592E+02 1.052E-02 3.946E+01 FREQ IM(V_PRINT12) IP(V_PRINT12) 1.592E+02 2.629E-04 -5.054E+01

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

Engineering Circuit Analysis, 7th Edition

Chapter Eleven Solutions

10 March 2006

5.
is → 5∠0° A, C → − j 4 Ω, Zin = 8 (3 − j 4) = = 3.417∠ − 33.15°∴ Vs = 17.087∠ − 33.15°, vs = 17.087 cos (25t − 33.15°) V ∴

40∠ − 53.13° 11 − j 4

Ps ,abs (0.1) = −17.087 cos (2.5rad − 33.147°) × 5cos 2.5rad = −23.51 W...