Answer
a. 8.9$\Omega$
b. $1.2\times10^{4}C$
Work Step by Step
a. Use equation 18–2, which defines the resistance.
$$R=\frac{V}{I}=\frac{120\;V}{13.5A}=8.9\Omega$$
b. Use the definition of current, equation 18–1.
$$I=\frac{\Delta Q}{\Delta t}$$
$$\Delta Q=I\Delta t=(13.5A)(15min)( 60s/min)=1.2\times10^{4}C$$
