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 = 8 103 = (32 103 I)I, 8 3 I 2 = 810 3 = 1 106 , I = 0.5 103 A, V = 32 103 I = 16V, V1 = V = 2V 3210 4 8 KVL: V = 16 = i(2 + 4 + R + 8) 103 , 16 = 0.5 103 (14 + R) 103 , 16 14 + R = 0.5 = 32, R = 32  14 = 18k 3. For the circuit shown below find I and V using (a) Meshcurrent analysis, and (b) Nodevoltage 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) Nodevoltage 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 ...
View
Full
Document
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.
 Spring '08
 Chandra

Click to edit the document details