Answer
Please see the work below.
Work Step by Step
(a) We know that
$I_{rms}=2\pi fCV_{rms}$
We plug in the known values to obtain:
$I_{rms}=2\pi(1.00\times 10^3Hz)(0.395\times 10^{-6}F)(20.5V)$
$I_{rms}=50.9\times 10^{-3}=50.9mA$
(b) We know that
$I^{\prime}_{rms}=I_{rms}\frac{V^{\prime}_{rms}}{V_{rms}}$
As $V^{\prime}_{rms}=2V_{rms}$
$\implies I^{\prime}_{rms}=2I_{rms}$
We plug in the known values to obtain:
$I^{\prime}_{rms}=2\times 50.9mA=101.8mA$
(c) As $I^{\prime}_{rms}=I_{rms}\frac{f^{\prime}}{f}$
We plug in the known values to obtain:
$I^{\prime}_{rms}=50.9mA\times \frac{2.0KHz}{1.0KHz}$
$I^{\prime}_{rms}=2\times 50.9mA=101.8mA$
