Answer
(a) $3.97\mu F$
(b) increase
(c) $143mA$
Work Step by Step
(a) We can find the required capacitance as follows:
$C=\frac{I_{rms}}{\omega \times V_{rms}}$
We plug in the known values to obtain:
$C=\frac{21\times 10^{-3}}{120\pi \times 14}$
$C=3.97\times 10^{-6}F$
$C=3.97\mu F$
(b) We know that the value of current $I_{rms}$ is directly proportional to the frequency of AC voltage. Thus, if the frequency is increased then the current will also increase.
(c) We can find the required rms current as follows:
$\omega=2\pi f$
$\omega=2\pi(410)=820\pi s^{-1}$
Now $I_{rms}=\omega CV_{rms}$
We plug in the known values to obtain:
$I_{rms}=820\pi\times 3.97\times 10^{-6}\times 14$
$I_{rms}=143\times 10^{-3}A=143mA$
