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 - Problems - Page 553: 26

Answer

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Work Step by Step

Apply Kirchhoff’s loop rule to the circuit Begin at the upper left corner of the diagram, and proceed clockwise. We assume that the 18V battery will “win” against the 12V battery, and that the current flows clockwise. $$-I(2.0\Omega)+18V-I(4.8\Omega)-12V-I(1.0\Omega)=0$$ $$I=\frac{6V}{7.8\Omega}=0.769A$$ Now find the terminal voltage for each battery. Sum the potential differences across the internal resistance and the emf, going from the negative terminal toward the positive (i.e., in this case, left to right). Larger battery: $$V_{terminal}= -I(2.0\Omega)+18V \approx 16V$$ For the smaller battery, because the current is being driven through it (i.e., it is being recharged) there is a voltage rise as we cross its internal resistance from left to right: $$V_{terminal}=I(1.0\Omega)+12V \approx 13V$$
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