hw 2 - EECE 519 Homework 2(Due Thursday January 31 1 Find...

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Unformatted text preview: EECE 519 Homework 2 (Due Thursday, January 31) 1. Find the current I1 in the circuit. 5A R1 I1 + Vs - -+ 2A 4A 3A I1 = -2A 2. In the circuit shown below V = 8V1 and the power supplied by V is 8mW. Find V1 , V, and R. 2k V 4k + V - 1 R 8k V1 = V = 4 103 I, V = 32 103 I, P = V I = 8 10-3 = (32 103 I)I, 8 -3 I 2 = 810 3 = 1 10-6 , I = 0.5 10-3 A, V = 32 103 I = 16V, V1 = V = 2V 3210 4 8 KVL: V = 16 = i(2 + 4 + R + 8) 10-3 , 16 = 0.5 103 (14 + R) 10-3 , 16 14 + R = 0.5 = 32, R = 32 - 14 = 18k 3. For the circuit shown below find I and V using (a) Mesh-current analysis, and (b) Node-voltage method 8 I3 a 28V + - 6 b 5 c + - I V 12 I1 2 I2 + 32V - (a)Mesh current method: Loop1: 28 - 6(I1 - I3 ) - 2(I1 - I2 ) = 0 8I1 - 2I2 - 6I3 = 28 Loop2: -2(I2 - I1 ) - 5(I2 - I3) - 12I2 - 32 = 0 2I1 - 19I2 + 5I3 = 32 Loop3: -6(I3 - I1 ) - 8I3 - 5(I3 - I2 ) = 0 6I1 + 5I2 - 19I3 = 0 28 -2 -6 32 -19 5 0 5 -19 8 -2 -6 2 -19 5 6 5 -19 8 2 6 8 2 6 28 -6 32 5 0 -19 -2 -6 -19 5 5 -19 I1 = = -(700 + 1216) + (10108 - 960) 7232 = = 4A 2888 - 60 - 60) - (684 + 76 + 200) 1808 I2 = = (4864 + 840) - (-1152 - 1064) -1808 = = -1A 2888 - 60 - 60) - (684 + 76 + 200) 1808 I = I1 - I2 = 4 - (-1) = 5A 8 -2 28 2 -19 32 6 5 0 8 -2 -6 2 -19 5 6 5 -19 I3 = = 1808 = 1A 1808 V = 5(I3 - I2 ) = 5(1 + 1) = 10V (b) Node-voltage method: Node a: Va = 28 Node b: Vb -Va + Vb + Vb -Vc = 0 5Va - 26Vb + 6Vc = 0 6 2 5 -32 Node c: Vc -Vb + Vc -Va + Vc12 = 0 15Va + 24Vb - 49Vc = -320 5 8 substituting for Va in second and third equations, we get -26Vb + 6Vc = -140, 24Vb - 49Vc = -740 Va = 28V Vb = 10V Vc = 20V and I = Vb = 10 = 5A 2 2 V = Vc - Vb = 20 - 10 = 10V ...
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This note was uploaded on 04/07/2008 for the course EECE 519 taught by Professor Chandra during the Spring '08 term at Kansas State University.

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