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HW1solutions - ECE35 Win09 Homework 1 Solutions P1.5-4 a.)...

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ECE35 Win’09 Homework 1 Solutions P1.5-4 a.) Assuming no more energy is delivered to the battery after 5 hours (battery is fully charged). 5 3600 5 3600 2 0 0 0 3 0.5 0.5 2 11 22 3600 3600 = 441 10 J 441 kJ t w Pdt vi d d t b.) 1 hr 10¢ Cost = 441kJ 1.23¢ 3600s kWhr 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 3 3 9 a b 1 4 4 b c 2 2 4 a d 5 7 35 b d 6 2 12 c d 8 5 40 So Total power received = 9 4 4 35 12 40 24 0 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 = 9 4 4 35 12 40 0 We conclude that the reference direction of the element connected between nodes and b that has been reversed.
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HW1solutions - ECE35 Win09 Homework 1 Solutions P1.5-4 a.)...

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