Physics: Principles with Applications (7th Edition)

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

Chapter 19 - DC Circuits - General Problems - Page 558: 94

Answer

0.40.

Work Step by Step

The power delivered to the starter is $P=I^2R_S$. $$\frac{P}{P_o}=\frac{I^2R_S}{I_o^2R_S}=(\frac{I}{I_o})^2$$ The current flowing is the battery emf divided by the equivalent resistance. The resistors are all in series, so the equivalent resistance is the sum of the resistances. $$\frac{P}{P_o}= (\frac{\epsilon/R_{eq}}{\epsilon/R_{o\;eq}})^2= (\frac{R_{o\;eq}}{R_{eq}})^2$$ $$\frac{P}{P_o}= (\frac{0.02\Omega+0.15\Omega}{0.02\Omega+0.15\Omega +0.10\Omega})^2=0.40$$
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