Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 762: 129

Answer

(a) $329\Omega $ (b) $794\Omega $

Work Step by Step

(a) We know that $\epsilon=IR_{eq}$ We plug in the known values to obtain: $12.0V=(0.075)(R_{eq})$ $\implies R_{eq}=160\Omega $ Now we can find the value of $ R_3$ as $\frac{1}{160}=\frac{1}{225}+\frac{1}{(225+R_3)}$ $\frac{1}{160}=\frac{450+R_3}{50625+225R_3}$ This simplifies to: $ R_3=329\Omega $ (b) Since $ R_2$ and $ R_3$ are in series, they share the same voltage. We know that $ I_2=\frac{2.65}{225}$ $ I_2=0.0117A $ and $ I_2R_3=12.0-2.65=9.35$ $\implies R_3=\frac{9.35}{0.0117}=794\Omega $
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