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 - Problems - Page 621: 37

Answer

$55MW$

Work Step by Step

Without transformers, the power generated will be the sum of the power lost in the wires and the power used by the mall. $P=I^2R+P_M$. Using $I=\frac{P_M}{V_M}=\frac{2.0\times10^6W}{120V}=16667A$, $P=(16667A)^22(0.100\Omega)+2.0\times10^6W=5.7556\times10^7W$ With transformers, $P_M=0.99P_w=0.99I_wV_w$ $I_w=\frac{P_M}{0.99V_w}=\frac{2.0\times10^6W}{(0.99)(1200V)}=1683.5A$ $P_{lost}=I_w^2R_w=(1683.5A)^2(0.200\Omega)=5.6684\times10^5W$ $P_w+P_{lost}=0.99P_G$ $P_G=2.6132\times10^6W$ Power saved is $5.7556\times10^7W-2.6132\times10^6W=55MW$
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