chap11

chap11 - Chapter 11 Solution 1 v t = 160 cos(50t i t =-20...

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Chapter 11, Solution 1. ) t 50 cos( 160 ) t ( v = ) 90 180 30 t 50 cos( 2 ) 30 t 50 sin( 20 - ) t ( i ° ° + ° = ° = ) 60 t 50 cos( 20 ) t ( i ° + = ) 60 t 50 cos( ) t 50 cos( ) 20 )( 160 ( ) t ( i ) t ( v ) t ( p ° + = = [ ] W ) 60 cos( ) 60 t 100 cos( 1600 ) t ( p ° + ° + = = ) t ( p W ) 60 t 100 cos( 1600 800 ° + + ) 60 cos( ) 20 )( 160 ( 2 1 ) cos( I V 2 1 P i v m m ° = θ θ = = P W 800 Chapter 11, Solution 2. First, transform the circuit to the frequency domain. ° → 0 30 ) t 500 cos( 30 , 500 = ω 150 j L j H 3 . 0 = ω 100 j - ) 10 )( 20 )( 500 ( j - C j 1 F 20 6 - = = ω µ I I 1 I 2 + 30 0 ° V j150 -j100 200 2 . 0 j - 90 2 . 0 150 j 0 30 1 = ° = ° = I ) t 500 sin( 2 . 0 ) 90 t 500 cos( 2 . 0 ) t ( i 1 = ° = 06 . 0 j 12 . 0 56 . 26 1342 . 0 j 2 3 . 0 100 j 200 0 30 2 + = ° = = ° = I

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) 56 . 25 t 500 cos( 1342 . 0 ) t ( i 2 ° + = ° = = + = 49.4 - 1844 . 0 14 . 0 j 12 . 0 2 1 I I I ) 35 t 500 cos( 1844 . 0 ) t ( i ° = For the voltage source, ] ) 35 t 500 cos( 1844 . 0 [ ] ) t 500 cos( 30 [ ) t ( i ) t ( v ) t ( p ° × = = At , s 2 t = ) 35 1000 cos( ) 1000 cos( 532 . 5 p ° = ) 935 . 0 )( 5624 . 0 )( 532 . 5 ( p = = p W 91 . 2 For the inductor, ] ) t 500 sin( 2 . 0 [ ] ) t 500 cos( 30 [ ) t ( i ) t ( v ) t ( p × = = At , s 2 t = ) 1000 sin( ) 1000 cos( 6 p = ) 8269 . 0 )( 5624 . 0 )( 6 ( p = = p W 79 . 2 For the capacitor, ° = = 63.44 - 42 . 13 ) 100 j - ( 2 c I V ) 56 . 25 t 500 cos( 1342 . 0 [ ] ) 44 . 63 500 cos( 42 . 13 [ ) t ( i ) t ( v ) t ( p ° + × ° = = At , s 2 t = ) 56 . 26 1000 cos( ) 44 . 63 1000 cos( 18 p ° + ° = ) 1329 . 0 )( 991 . 0 )( 18 ( p = = p W 37 . 2 For the resistor, ° = = 56 . 25 84 . 26 200 2 R I V ] ) 56 . 26 t 500 cos( 1342 . 0 [ ] ) 56 . 26 t 500 cos( 84 . 26 [ ) t ( i ) t ( v ) t ( p ° + × ° + = = At , s 2 t = ) 56 . 25 1000 ( cos 602 . 3 p 2 ° + = 2 1329 . 0 )( 602 . 3 ( p = = p W 0636 . 0
Chapter 11, Solution 3. 10 , ° → ° + 30 10 ) 30 t 2 cos( 2 = ω 2 j L j H 1 = ω -j2 C j 1 F 25 . 0 = ω 4 2 I I 1 I 2 + 10 30 ° V j2 -j2 2 j 2 2 ) 2 j 2 )( 2 j ( ) 2 j 2 ( || 2 j + = = ° = + + ° = 565 . 11 581 . 1 2 j 2 4 30 10 I ° = = = 565 . 101 581 . 1 j 2 2 j 1 I I I ° = = 565 . 56 236 . 2 2 2 j 2 2 I I For the source, ) 565 . 11 - 581 . 1 )( 30 10 ( 2 1 * ° ° = = I V S 5 . 2 j 5 . 7 18.43 905 . 7 + = ° = S The average power supplied by the source = W 5 . 7 For the 4- resistor, the average power absorbed is = = = ) 4 ( ) 581 . 1 ( 2 1 R 2 1 P 2 2 IW 5 For the inductor, 5 j ) 2 j ( ) 236 . 2 ( 2 1 2 1 2 L 2 2 = = = Z I S The average power absorbed by the inductor = W 0

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For the 2- resistor, the average power absorbed is = = = ) 2 ( ) 581 . 1 ( 2 1 R 2 1 P 2 2 1 IW 5 . 2 For the capacitor, 5 . 2 j - ) 2 j - ( ) 581 . 1 ( 2 1 2 1 2 c 2 1 = = = Z I S The average power absorbed by the capacitor = W 0 Chapter 11, Solution 4. 20 10 I 2 I 1 + -j10 50 V j5 For mesh 1, 2 1 10 j ) 10 j 20 ( 50 I I + = 2 1 j ) j 2 ( 5 I I + = ( 1 ) For mesh 2, 1 2 10 j ) 10 j 5 j 10 ( 0 I I + + = 1 2 2 j ) j 2 ( 0 I I + = ( 2 ) In matrix form, = 2 1 j 2 2 j j j 2 0 5 I I 4 j 5 = , ) j 2 ( 5 1 = , -j10 2 = ° = = = 1 . 12 746 . 1 4 j 5 ) j 2 ( 5 1 1 I ° = = = 66 . 128 562 . 1 j4 - 5 j10 - 2 2 I For the source, ° = = 12.1 - 65 . 43 2 1 * 1 I V S
The average power supplied = ° = ) 1 . 12 cos( 65 . 43 W 68 . 42 For the 20- resistor, = = R 2 1 P 2 1 IW 48 . 30 For the inductor and capacitor, = P W 0 For the 10- resistor, = = R 2 1 P 2 2 2 . 12 Chapter 11, Solution 5. Converting the circuit into the frequency domain, we get: 1 2 + j6 –j2 8 –40 ˚ W 4159 . 1 1 2 6828 . 1 P 38 . 25 6828 . 1 2 j 2 6 j ) 2 j 2 ( 6 j 1 40 8 I 2 1 1 = = ° = + + ° = P 3H = P 0.25F = 0 W 097 . 5 2 2 258 . 2 P 258 . 2 38 . 25 6828 . 1 2 j 2 6 j 6 j I 2 2 2 = = = ° + =

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Chapter 11, Solution 6.
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This document was uploaded on 01/24/2008.

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chap11 - Chapter 11 Solution 1 v t = 160 cos(50t i t =-20...

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