Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 27 - Circuits - Problems - Page 796: 16a

Answer

The internal resistance is $~~1000~\Omega$

Work Step by Step

$i = \frac{\mathscr{E}}{R+r}$ We can find an expression for $\mathscr{E}$: $V = \mathscr{E}-ir$ $V = \mathscr{E}-\frac{\mathscr{E}~r}{R+r}$ $V = \frac{\mathscr{E}~R}{R+r}$ $\mathscr{E} = \frac{V~(R+r)}{R}$ We can use the given values to find the internal resistance $r$: $\mathscr{E} = \frac{(0.10)~(500+r)}{500} = \frac{(0.15)(1000+r)}{1000}$ $(2)(0.10)~(500+r) = (0.15)(1000+r)$ $0.20~r-0.15~r = 150-100$ $0.05~r = 50$ $r = \frac{50}{0.05}$ $r = 1000~\Omega$ The internal resistance is $~~1000~\Omega$
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