chap10

chap10 - Chapter 10 Solution 1 =1 10 cos t 45 10 45 5 sin t...

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Chapter 10, Solution 1. ω 1 = ° → ° 45 - 10 ) 45 t cos( 10 ° → ° + 60 - 5 ) 30 t sin( 5 j L j H 1 = ω → j - C j 1 F 1 = ω → The circuit becomes as shown below. V o 3 + 10 -45 ° V + 2 I o j 5 -60 ° V Applying nodal analysis, j - j ) 60 - 5 ( 3 ) 45 - 10 ( o o o V V V = ° + ° o j 60 - 15 45 - 10 j V = ° + ° ° = ° + ° = 9 . 247 15.73 150 - 15 45 - 10 o V Therefore, = ) t ( v o 15.73 cos(t + 247.9 ° ) V Chapter 10, Solution 2. ω 10 = ° → π 45 - 4 ) 4 t 10 cos( 4 ° → π + 150 - 20 ) 3 t 10 sin( 20 10 j L j H 1 = ω → 5 j - 2 . 0 j 1 C j 1 F 02 . 0 = = ω →
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The circuit becomes that shown below. I o V o 10 j10 -j5 20 -150 ° V + 4 -45 ° A Applying nodal analysis, 5 j - 10 j 45 - 4 10 ) 150 - 20 ( o o o V V V + = ° + ° o ) j 1 ( 1 . 0 45 - 4 150 - 20 V + = ° + ° ° = + ° + ° = = 98 . 150 816 . 2 ) j 1 ( j 45 - 4 150 - 2 10 j o o V I Therefore, = ) t ( i o 2.816 cos(10t + 150.98 ° ) A Chapter 10, Solution 3. ω 4 = ° → 0 2 ) t 4 cos( 2 -j16 90 - 16 ) t 4 sin( 16 = ° → 8 j L j H 2 = ω → 3 j - ) 12 1 )( 4 ( j 1 C j 1 F 12 1 = = ω → The circuit is shown below. V o + 2 0 ° A -j16 V 4 -j3 6 1 j8
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Applying nodal analysis, 8 j 6 1 2 3 j 4 16 j - o o o + + = + V V V o 8 j 6 1 3 j 4 1 1 2 3 j 4 16 j - V + + + = + ° = ° ° = + = 02 . 35 - 835 . 3 88 . 1 2207 . 1 15 . 33 - 682 . 4 04 . 0 j 22 . 1 56 . 2 j 92 . 3 o V Therefore, = ) t ( v o 3.835 cos(4t – 35.02 ° ) V Chapter 10, Solution 4. 16 4 , 10 - 16 ) 10 t 4 sin( = ω ° → ° 4 j L j H 1 = ω → j - ) 4 1 )( 4 ( j 1 C j 1 F 25 . 0 = = ω → j4 1 I x -j 16 -10 ° V + V 1 0.5 I x + V o j 1 2 1 4 j ) 10 - 16 ( 1 x 1 = + ° V I V But 4 j ) 10 - 16 ( 1 x V I ° = So, j 1 8 j ) ) 10 - 16 (( 3 1 1 = ° V V
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4 j 1 - 10 - 48 1 + ° = V Using voltage division, ° = + ° = = 04 . 69 - 232 . 8 ) 4 j 1 j)(- - (1 10 - 48 j 1 1 1 o V V Therefore, = ) t ( v o 8.232 sin(4t – 69.04 ° ) V Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be V x and we get: 0 3 j V 2 j V 4 30 10 I 5 . 0 V x x x x = + + ° x x x x x V 5 . 0 j 2 j V I but 30 30 I 5 . 1 V ) 4 j 6 j 3 ( = = ° = + Combining these equations we get: A 38 . 97 615 . 4 25 . 1 j 3 30 30 5 . 0 j I 25 . 1 j 3 30 30 V or 30 30 V ) 75 . 0 j 2 j 3 ( x x x ° = + ° = + ° = ° = + Chapter 10, Solution 6. Let V o be the voltage across the current source. Using nodal analysis we get: 0 10 j 20 V 3 20 V 4 V o x o = + + where o x V 10 j 20 20 V + = Combining these we get: 30 j 60 V ) 3 5 . 0 j 1 ( 0 10 j 20 V 3 10 j 20 V 4 20 V o o o o + = + = + + + = + = + + = 5 . 0 j 2 ) 3 ( 20 V or 5 . 0 j 2 30 j 60 V x o 29.11 –166 ˚ V .
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Chapter 10, Solution 7. At the main node, + + + = + → + + = + 50 1 30 j 20 j 40 1 V 3 j 196 . 5 20 j 40 058 . 31 j 91 . 115 50 V 30 j V 30 6 20 j 40 V 15 120 o o V 154 08 . 124 0233 . 0 j 04 . 0 7805 . 4 j 1885 . 3 V o = + = Chapter 10, Solution 8. , 200 = ω 20 j 1 . 0 x 200 j L j mH 100 = = ω → 100 j 10 x 50 x 200 j 1 C j 1 F 50 6 = = ω → µ The frequency-domain version of the circuit is shown below.
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