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HW9%20solution

# HW9%20solution - I 1-I s 1 R 1-r m I x-R 2 I 1-I 2 = 0 I x...

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ECE 201 Spring 2010 Homework 9 Solutions Problem 37 (a) The following loop equation can be written, - ( I 1 - 0 . 75)200 - 300 I 1 - ( I 1 +0 . 1)500 = 0 I 1 =0 . 1 A P = VI P s 1 = 200(0 . 75 - 0 . 1)0 . 75 = 97 . 5 W P s 2 = 500(0 . 1 + 0 . 1)0 . 1 = 10 W (b) Again, writing the loop equation and comparing the equation obtained with that given in the problem, - ( I 1 - 0 . 4) R 1 - 600 I 1 - ( I 1 +0 . 1) R 2 =0 ( R 1 + R 2 + 600) I 1 =0 . 4 R 1 - 0 . 1 R 2 R 1 + R 2 = 1400 4 R 1 - R 2 = 600 R 1 = 400 Ω R 2 = 1000 Ω 1

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Problem 42 (a) The following loop equations can be written, 21 - 20 I 1 - 80( I 1 - I 2 )=0 24 + 80 I 2 - 80( I 1 - I 2 )=0 I 1 =0 . 15 A I 2 = - 0 . 075 A V R 3 =( I 1 - I 2 ) R 3 =0 . 225 × 80 = 18 V P R 3 = 18(0 . 225) = 4 . 05 W (b) Again, writing the loop equations, 21 - ( I 1 + I 2 )20 - 80 I 2 - 24 = 0 21 - ( I 1 + I 2 )20 - 80 I 1 =0 I 1 =0 . 225 A I 2 = - 0 . 075 A V R 3 = I 1 R 3 =0 . 225 × 80 = 18 V P R 3 = 18(0 . 225) = 4 . 05 W Problem 52 (a) Assume clockwise loop currents I 1 and I 2 in the two rightmost loops respec- tively. The following equations can then be written for the two loops,
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Unformatted text preview: -( I 1-I s 1 ) R 1-r m I x-R 2 ( I 1-I 2 ) = 0 I x = I s 1-I 1-R 2 ( I 2-I 1 )-R 3 I 2-V s 2 = 0 2 Putting in matrix form, r m-R 1-R 2 R 2 R 2-( R 2 + R 3 ) I 1 I 2 = ( r m-R 1 ) I s 1 V s 2 (b) Putting the respective literal values, -80 40 40-120 I 1 I 2 = -40 40 ⇒ I 1 = 0 . 4 A I 2 =-. 2 A (c) V A = (1-. 4)100 = 60 V V B = 0 . 6(40) = 24 V (d) P s 1 = 60 × 1 = 60 W P s 2 = 40 × . 2 = 8 W (e) P dep = 60 × . 6 ×-. 4 =-14 . 4 W 3...
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HW9%20solution - I 1-I s 1 R 1-r m I x-R 2 I 1-I 2 = 0 I x...

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