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: 83

Answer

$R=100\Omega$.

Work Step by Step

Assume the switches are both open. The circuit is a simple series circuit. Use Ohm’s Law to find the current. $$I=\frac{6.0V}{50\Omega +20 \Omega + 10\Omega}=0.075 A$$ Now assume the switches are both closed. The $20 \Omega$ resistor is in parallel with the unknown R. Use Kirchhoff’s loop rule on the outer loop of the circuit, using the current through the $20 \Omega$ resistor that we just found. $$6.0V-I(50\Omega)-(0.075)(20 \Omega)=0$$ $$I=0.090A$$ This is the current in the parallel combination of the $20 \Omega$ resistor and the unknown R. Since 0.075 A flows in the $20 \Omega$ resistor, 0.015 A must flow through R. The voltage drop across R is the same as the drop across the $20 \Omega$ resistor because they are in parallel. $$V_{20}=(0.075A)(20\Omega)=IR=(0.015A)(R)$$ $$R=100\Omega$$
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