W08 314 HW04 solutions

W08 314 HW04 solutions - Part 1 & 2 2A 1A b 50V...

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Unformatted text preview: Part 1 & 2 2A 1A b 50V 48V 5.2 2.5 5 8 2 2 a = = 2A b 50V 2V 5.2 2.5 5 8 2 2 a 6A 10A b 50V 2V 2.5 5 1.6 7.2 a 4A = b 6.4V 2V 2.5 5 1.6 7.2 a 10A = b 4.4V 2.5 5 8.8 a 10A 0.5A = b 2.5 5 8.8 a 10.5A = b 2.5 3.19 a 33.5V = b 2.5 3.19 a 33.5V = b 5.69 a 5.89A = b 5.69 a EECS 314 Winter 2008 HW 04 Solutions Problem 1 Part 1 There are multiple ways to solve this problem; here, calculate based on a Thevenin circuit equivalent. Assume f xed V T and R s and construct two equations based on the load values: V T R s 1 10A V T R s 7 5A V T 60V R s 5 R L ,max = R s = 5 . Convert to the Norton equivalent current by dividing, and calculate maximum power using the provided equation: I N V T R s 60V 5 12A P max V 2 T 4 R s 180W Graph the volt-amp characteristic of the source. The open-circuit voltage is simply equal to the Thevenin equivalent voltage, and the short-circuit current equal to the Norton equivalent current: 12 60 I (A) V (V) EECS 314 Winter 2008 HW 04 Solutions Problem 2 Part 2 Again, there are multiple ways to solve this problem; here, calculate based on a Norton circuit equivalent. Assume f xed V T and R s and construct two equations based on the load values: I N R s R s R L 2 R L P I N R s R s 4 2 4 64W I N R s R s 64 2 64 64W I N 5A R s 16 Again, R L ,max = R s = 5 . Convert to the Thevenin equivalent voltage by multiplying, and calculate maximum power using the provided equation: V T I N R s 80V P max V 2 T 4 R s 100W Graph the volt-amp characteristic of the source: 5 80 I (A) V (V) EECS 314 Winter 2008 HW 04 Solutions Problem 2 Problem 3 Part 1 In this problem, we are given that when = 4 L R and A I S 15 = , the source transfers 100 Watts to the load. In order to find the max power transfer to the load, we need to find the value of . S R From current division, we know that the current through the load is: ) 4 ( 15 ) ( + = + = S S L S S S L R R R R R I I We also know that . 100 4 2 2 = = = L L L L I R I P Substituting in , we get: L I 4 ) 4 ( 15 100 2 + = S S R R Solving this for , we find that: S R = 2 S R Since this circuit is already in its Norton configuration, 15 = N I A....
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W08 314 HW04 solutions - Part 1 & 2 2A 1A b 50V...

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