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homework01solutions - Chapter 2, Solution 15. (a) Pabs =...

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Chapter 2, Solution 15. (a) P abs = (+3.2 V)(-2 mA) = –6.4 mW (or +6.4 mW supplied ) (b) P abs = (+6 V)(-20 A) = –120 W (or +120 W supplied ) (a) P abs = (+6 V)(2 i x ) = (+6 V)[(2)(5 A)] = +60 W (e) P abs = (4 sin 1000 t V)(-8 cos 1000 t mA) | t = 2 ms = +12.11 W Chapter 2, Solution 22. We are told that V x = 1 V, and from Fig. 2.33 we see that the current flowing through the dependent source (and hence through each element of the circuit) is 5V x = 5 A. We will compute absorbed power by using the current flowing into the positive reference terminal of the appropriate voltage (passive sign convention), and we will compute supplied power by using the current flowing out of the positive reference terminal of the appropriate voltage. (a) The power absorbed by element “A” = (9 V)(5 A) = 45 W (b) The power supplied by the 1-V source = (1 V)(5 A) = 5 W, and the power supplied by the dependent source = (8 V)(5 A) = 40 W (c) The sum of the supplied power = 5 + 40 = 45 W The sum of the absorbed power is 45 W, so
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This note was uploaded on 04/13/2008 for the course ECE 221 taught by Professor N/a during the Fall '05 term at Wisconsin.

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homework01solutions - Chapter 2, Solution 15. (a) Pabs =...

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