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 552: 22

Answer

a) $V_1$ increases $V_2$ decreases $V_3$ drops to 0 b) $I_1$ increases $I_2$ decreases $I_3$ drops to 0 c) $\mathcal{E}$ increases d) $V_{ab}=8.49V$ e) $V_{ab}=8.61V$

Work Step by Step

Lets name resistors 1 to 3 from left to right a) Switch closed $R_{eq}=R+r+\frac{1}{\frac{2}{R}}=1.5R+r$ $I_{closed}=\frac{V}{R_{eq}}=\frac{\mathcal{E}}{1.5R+r}$ Switch open $R_{eq}=R+R+r=2R+r$ $I_{open}=\frac{V}{R_{eq}}=\frac{\mathcal{E}}{2R+r}$ When the switch is opened, the current decreases. This means the voltage across $R_1$ decreases. The previously divided current through $R_2$ and $R_3$ now go through $R_2$. This means the voltage across $R_2$ increases. Lastly, no current flows through $R_3$ so the voltage across it drops to 0. b) $I=\frac{V}{R}$, so current is proportional to V. If V increases, I increases and vice versa c) $V_{ab}=\mathcal{E}-Ir$ When the current is decreased, the right hand term $Ir$ decreases and the terminal velocity increases. d) $I_{closed}=\frac{9.0V}{(1.5)(5.50\Omega)+0.50\Omega}=1.03A$ $V_{ab}=\mathcal{E}-Ir=9.0V-(1.03A)(0.50\Omega)=8.49V$ e) $I_{open}=\frac{9.0V}{2(5.50\Omega)+0.50\Omega}=0.783A$ $V_{ab}=\mathcal{E}-Ir=9.0V-(0.783A)(0.50\Omega)=8.61V$
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