HW1solutionsWIN10

# HW1solutionsWIN10 - P2.2-5(a(b Using the quadratic formula...

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ECE35 Win’10 Homework 1 Solutions Assignment: Read Chapters 1 and 2, skim Chapter 3, and do at least the following problems: P1.5-4, P1.7-3, P2.2-5, P2.4-5, P2.5-2, P2.6-1, P2.6-5, P2.7-7, P2.9-3, and DP2-2 (design problem) P1.5-4 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). b.) P 1.7-3 Let’s tabulate the power received by each element. We’ll identify each element by its nodes. nodes Power received, W a c a b b c a d b d c d So Total power received = Changing the current reference direction for a particular element will change the total power by twice the power of the particular element. Since the element connected between nodes b and d receives -12 W, changing the reference direction of its current will increase the total power received by 24 W, as required. After that change Total power received = We conclude that the reference direction of the element connected between nodes and b that has been reversed.

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Unformatted text preview: P2.2-5 (a) (b) Using the quadratic formula When v = 0.8 V then . When v = -1.0 V then . (c) Using the quadratic formula So there is no real solution to the equation. P2.4-5 P2.5-2 (a) (b) P2.6-1 (a) (b) The voltage, 12 V, and the current, 0.5 A, of the voltage source obey the passive convention so the power P = 12 (0.5) = 6 W is the power received by the source. The voltage source delivers -6 W. P2.6-5 a.) b.) Element Power supplied voltage source current source resistor total 0 P2.7-7 i d and v d adhere to the passive convention so is the power received by the dependent source. P2.9-3 (a) v = 12 V (b) (c) v = 0 V (d) (the box symbol was supposed to be ~, ‘approximately’) DP2-2 1.) 2.) Therefore 20 < R < 3.75 Ω . These conditions cannot be satisfied simultaneously....
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HW1solutionsWIN10 - P2.2-5(a(b Using the quadratic formula...

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