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 557: 81

Answer

11 volts.

Work Step by Step

First find the current in the entire circuit, from the given resistance and the power dissipated in it. $$P=I^2R$$ $$I=\sqrt{\frac{P_{33}}{R_{33}}}=\sqrt{\frac{0.80W}{ 33\Omega}}=0.1557A$$ The circuit consists of a $68\Omega$ resistor in parallel with a $85\Omega$ resistor, and the combination is in series with the $33\Omega$ resistor. $$R_{eq}=(\frac{1}{68\Omega }+\frac{1}{85\Omega })^{-1}+33\Omega=70.78\Omega$$ From Ohm’s law, V=IR, the battery voltage V can be found by multiplying the circuit current by the circuit’s equivalent resistance. $$V=IR_{eq}=(0.1557A)( 70.78\Omega)=11V$$
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