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: 87

Answer

$9.0\Omega$.

Work Step by Step

Find the current in the circuit from the bulb’s power rating. $$P=IV$$ $$I=\frac{P}{V}=\frac{2.0W}{3.0V}=0.667A$$ Note that the bulb is to have a voltage of 3.0 V across it, so from Kirchhoff’s loop law, the resistor R has 6.0 V across it. Find the resistance of R from Ohm’s Law. $$R=\frac{V_R}{I_R}=\frac{6.0V}{0.667A}=9.0\Omega$$
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