HW1_Solutions

HW1_Solutions - Homework Solutions HW 1 ECE302 Spring 2007 1.20 V = 10 V R1 = 22 k R2= 47 k and R3 = 180 k V 1 I2 R R 1 2 V I 3 V 2 R 3 V1 = 10V V2

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Homework Solutions ECE302 Spring 2007 HW 1 1-1 6/9/06 1.20 V = 10 V, R 1 = 22 k , R 2 = 47 k and R 3 = 180 k . V + - V 1 V 2 + - R 1 R 2 R 3 I 3 I 2 V 1 = 10 V 22 k 22 k Ω+ 47 k 180 k ( 29 = 10 V 22 k 22 k Ω+ 37.3 k = 3.71 V V 2 = 10 V 37.3 k 22 k Ω+ 37.3 k = 6.29 V Checking: 6.29+ 3.71= 10.0 V I 2 = I 1 180 k 47 k Ω+ 180 k = 10 V 22 k Ω+ 37.3 k 180 k 47 k Ω+ 180 k = 134 m A I 3 = I 1 47 k 47 k Ω+ 180 k = 10 V 22 k Ω+ 37.3 k 47 k 47 k Ω+ 180 k = 34.9 m A Checking: I 1 = 10 V 22 k Ω+ 37.3 k = 169 m A and I 1 = I 2 + I 3 1.23 I 2 = 250 m A 150 k 150 k Ω+ 150 k = 125 m A I 3 = 250 m A 150 k 150 k Ω+ 150 k = 125 m A V 3 = 250 m A 150 k 150 k ( 29 82 k 68 k Ω+ 82 k = 10.3 V Checking: I 1 + I 2 = 250 m A and I 2 R 2 = 125 m A 82 k ( 29 = 10.3 V
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1-2 ©R. C. Jaeger & T. N. Blalock 6/9/06 1.26 (a) R 1 R 2 β i v s i + - v th V th = V oc =- b i R 2 but i = - v s R 1 and V th = b v s R 2 R 1 = 120 v s 39 k 100 k = 46.8 v s R 1 R 2 β i i R th v x i x R th = v x i x ; i x = v x R 2 + b i but i = 0 since V R 1 = 0. R th = R 2 = 39 k . Thévenin equivalent circuit: 58.5v s 39 k (b) R 1 R 2 β i i s i + - v th V th = V oc =- b i R 2 where i + bi + i s = 0 and V th =- b - i s b + 1 R 2 = 38700 i
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Homework Solutions ECE302 Spring 2007
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This note was uploaded on 03/30/2008 for the course ECE 302 taught by Professor Barlage during the Spring '07 term at N.C. State.

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HW1_Solutions - Homework Solutions HW 1 ECE302 Spring 2007 1.20 V = 10 V R1 = 22 k R2= 47 k and R3 = 180 k V 1 I2 R R 1 2 V I 3 V 2 R 3 V1 = 10V V2

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