Answer
$E=9.8V$
Work Step by Step
$E=IR_{eq}$
$R_{eq}=(\frac{1}{r}+\frac{1}{R_1})^{-1}+R_2=11674\Omega$
$r=(1000\Omega/V)(3.0V)=3000\Omega$
$E=V_1+V_2=1.9V+IR_2$
$E=1.9V+\frac{E}{R_{eq}}R_2=1.9V+\frac{E}{11674\Omega}(9400\Omega)$
$E=9.8V$
Physics: Principles with Applications (7th Edition)
by Giancoli, Douglas C.
Published by
Pearson
ISBN 10:
0-32162-592-7
ISBN 13:
978-0-32162-592-2
Chapter 19 - DC Circuits - Problems - Page 555: 65
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