Answer
$0.10K\Omega$
Work Step by Step
We know that
$Z=\frac{V_{rms}}{I_{rms}}$
We plug in the known values to obtain:
$Z=\frac{95}{0.72}=130\Omega$
Now we can find the required resistance as follows:
$R=\sqrt{Z^2-(\frac{1}{\omega C})^2}$
We plug in the known values to obtain:
$R=\sqrt{(130)^2+[\frac{1}{2\pi(150)(13\times 10^{-6})}]^2}$
$R=0.10K\Omega$
