ee40sp08hwk1-sol - EE40 Spring 2008 Homework 1 Solution...

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Unformatted text preview: EE40 Spring 2008 Homework 1 Solution Anantharam, Venkat February 1, 2008 NOTE: EACH PROBLEM IS WORTH 10 POINTS Problem 1 Solution (P 1.21) (a) Since we have an active reference configuration, we compute the power as p =- v a i a = 30 W Since this is positive, power is being absorbed by the element. (b) Since we have an active reference configuration, we compute the power as p = v b i b = 30 W Since this is positive, power is being absorbed by the element. (c) We compute the power as p = v DE i DE =- 60 W OR p =- v DE i ED =- 60 W Since this is negative, power is being delivered by the element. Problem 2 Solution The figure from the problem is shown below 1 The solution is shown below A node is a point at which two or more current elements are joined together. All points connected by ideal conductors are considered to be a single node. There are 6 nodes in this circuit as indicated in the figure above....
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ee40sp08hwk1-sol - EE40 Spring 2008 Homework 1 Solution...

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