## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 18 - Problems - Page 700: 44

#### Answer

(a) The equivalent resistance is $23~\Omega$ (b) $I = 8~A$

#### Work Step by Step

(a) We can find the equivalent resistance of the $12~\Omega$ and the $24~\Omega$ resistors: $\frac{1}{R'} = \frac{1}{12~\Omega}+ \frac{1}{24~\Omega}$ $\frac{1}{R'} = \frac{1}{8~\Omega}$ $R' = 8~\Omega$ We can find the equivalent resistance with the $15~\Omega$ resistor: $R_{eq} = 8~\Omega+ 15~\Omega$ $R_{eq} = 23~\Omega$ The equivalent resistance is $23~\Omega$ (b) We can find the total current in the circuit: $I = \frac{V}{R_{eq}}$ $I = \frac{276~V}{23~\Omega}$ $I = 12~A$ We can find the potential difference across the $15~\Omega$ resistor: $\Delta V = (12~A)(15~\Omega)$ $\Delta V = 180~V$ Thus the potential difference across the $12~\Omega$ resistor is $276~V-180~V$ which is $96~V$. We can find the current in the $12~\Omega$ resistor: $I = \frac{\Delta V}{R}$ $I = \frac{96~V}{12~\Omega}$ $I = 8~A$

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