Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 21 - Electromagnetic Induction and Faraday's Law - General Problems - Page 623: 81

Answer

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Work Step by Step

a. There is a voltage drop across the lines because of the resistance. $$V_{out}=V_{in}-IR=42000V-(740A)(2)(0.95\Omega)=40594V\approx 41kV$$ b. Calculate the power input from P=IV. $$P_{in}=IV_{in}=(740A)(42000V)=3.1\times10^7W$$ c. The power loss is due to the current and the resistance in the wires. $$P_{loss}=I^2R=(740A)^2(2)(0.95\Omega)=1.0\times10^6W$$ d. The power output found from P=IV. $$P_{out}=IV_{out}=(740A)(40594V)=3.0\times10^7W$$ As a check, this also equals the input power minus the loss.
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