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: 24

Answer

a) The voltage through $R_1$ will increase The voltage through $R_2$ will increase from 0 The voltage through $R_3$ will decrease The voltage through $R_4$ will decrease b) The current through $R_1$ will increase The current through $R_2$ will increase from 0 The current through $R_3$ will decrease The current through $R_4$ will decrease c) The power output will increase d) Our predictions are correct.

Work Step by Step

a) When we close the circuit, we are adding a resistor that is in parallel, meaning the total resistance will decrease. Therefore, the current through the circuit will increase. The voltage through $R_1$ will increase The voltage through $R_2$ will increase from 0 The voltage through $R_3$ will decrease The voltage through $R_4$ will decrease, because $R_2$, $R_3$, and $R_4$ are connected in parallel and the current needs to divide into 3 paths. b) From Ohm's Law $V=IR$, we know current and voltage are proportional. If voltage increases, current increases. The current through $R_1$ will increase The current through $R_2$ will increase from 0 The current through $R_3$ will decrease The current through $R_4$ will decrease c) Power can be calculated by the following equations: $P=I^2R=\frac{V^2}{R}$ The current, voltage, and power output are proportional and the resistance is inversely proportional to power output. The total resistance decreases, so power output will increase. d) Switch open $R_{eq}=R_1+\frac{1}{\frac{1}{R_3}+\frac{1}{R_4}}=232.5\Omega$ $I_{1}=\frac{V}{R_{eq}}=0.0946A$ $V_3=V_4$, so $I_3=I_4$ $V_3=22.0V-(0.0946A)(155\Omega)=7.3V$ $I_3=\frac{V_3}{R_3}=\frac{7.3V}{155\Omega}=0.0473A$ $I_{1}=0.0946A$ $I_{2}=0A$ $I_3=I_4=0.0473A$ Switch closed $R_{eq}=R_1+\frac{1}{\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}=207\Omega$ $I_{1}=\frac{V}{R_{eq}}=0.106A$ $V_2=V_3=V_4$, so $I_2=I_3=I_4$ $V_3=22.0V-(0.106A)(155\Omega)=5.5V$ $I_3=\frac{V_3}{R_3}=\frac{5.5V}{155\Omega}=0.0355A$ $I_{1}=0.106A$ $I_2=I_3=I_4=0.0355A$ Our predictions are correct.
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