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# Soln9 - Ω(c(a j4 Ω 6 Ω Z th-j2 Ω By placing short...

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EE411 Spring 2008 HW#9 Solution (Q2) 10.46 Let 3 2 1 o v v v v + + = , where 1 v , 2 v , and 3 v are respectively due to the 10-V dc source, the ac current source, and the ac voltage source. For 1 v consider the circuit in Fig. (a). The capacitor is open to dc, while the inductor is a short circuit. Hence, V 10 v 1 = For 2 v , consider the circuit in Fig. (b). 2 = ω 4 j L j H 2 = ω ⎯→ 6 j - ) 12 / 1 )( 2 ( j 1 C j 1 F 12 1 = = ω ⎯→ Applying nodal analysis, 2 2 2 2 4 6 6 1 4 6 - 6 90 4 V V V V + = + + = ° j j j j ° = ° = 44 . 63 45 . 21 5 . 0 1 90 24 2 j V Hence, V ) 56 . 26 t 2 sin( 45 . 21 v 2 ° + = For 3 v , consider the circuit in Fig. (c). (a) + v 1 2 H 10 V 6 Ω 1/12 F + -j6 Ω (b) + V 2 6 Ω 4 ­ 90 ° A j4 Ω

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3 = ω 6 j L j H 2 = ω ⎯→ 4 j - ) 12 / 1 )( 3 ( j 1 C j 1 F 12 1 = = ω ⎯→ At the non-reference node, 6 j 4 j - 6 12 3 3 3 V V V + = ° = + = 56 . 26 - 73 . 10 5 . 0 j 1 12 3 V Hence, V ) 56 . 26 t 3 cos( 73 . 10 v 3 ° = Therefore, = o v 10 + 21.45 sin(2t + 26.56 ° ) + 10.73 cos(3t – 26.56 ° ) V (Q3) 10.56 (a) To find th Z , consider the circuit in Fig. (a). 4 j 6 2 j 4 j ) 2 j - )( 4 j ( 6 ) 2 j - ( || 4 j 6 th N = + = + = = Z Z = 7.211 -33.69 ° Ω j6 Ω 6 Ω 12 0 ° V + -j4
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Unformatted text preview: Ω (c) + (a) j4 Ω 6 Ω Z th-j2 Ω By placing short circuit at terminals a-b, we obtain, = N I 2 ∠ ° A = ° ∠ ° ∠ = = ) 2 ( ) 69 . 33-211 . 7 ( th th th I Z V 14.422 ∠-33.69 ° V (b) To find th Z , consider the circuit in Fig. (b). 20 60 || 30 = 5 j 20 ) 10 j 20 )( 5 j-( ) 10 j 20 ( || 5 j-th N + + = + = = Z Z = 5.423 ∠-77.47 ° Ω To find th V and N I , we transform the voltage source and combine the 30 Ω and 60 Ω resistors. The result is shown in Fig. (c). ) 45 4 )( j 2 ( 5 2 ) 45 4 ( 10 j 20 20 N ° ∠ − = ° ∠ + = I = 3.578 ∠ 18.43 ° A ) 43 . 18 578 . 3 ( ) 47 . 77-423 . 5 ( N th th ° ∠ ° ∠ = = I Z V = 19.4 ∠-59 ° V Z th j10 Ω 60 Ω-j5 Ω (b) 30 Ω 20 Ω (c) j10 Ω-j5 Ω 4 ∠ 45 ° A I N a b...
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