Answer
(a) $1.16K\Omega$
(b) $0.137\mu F$
(c) $0.86mA$
(d) $4.3mA$
Work Step by Step
(a) We know that
$I_{rms}=\omega CV_{rms}$
This can be rearranged as:
$\frac{1}{\omega C}=\frac{V_{rms}}{I_{rms}}$
$\implies X_C=\frac{0.5}{0.43\times 10^{-3}}$
$\implies X_C=1.16\times 10^3\Omega=1.16K\Omega$
(b) We know that
$C=\frac{1}{\omega X_C}$
We plug in the known values to obtain:
$C=\frac{1}{2000\times \pi \times 1.162\times 10^3}$
$\implies C=0.137\mu F$
(c) The rms current can be determined as
$\omega=2\pi f$
$\omega=2\pi \times 2000=4000\pi s^{-1}$
Now $I_{rms}=\omega C\times V_{rms}$
$I_{rms}=4000\pi \times 0.137\times 10^{-6}\times 0.5$
$\implies I_{rms}=0.860mA$
(d) We know that
$\omega=2\pi f$
$\omega=2\pi (10000)=20000\pi s^{-1}$
Now $I_{rms}=20000\pi\times 0.137\times 10^{-6}\times 5$
$\implies I_{rms}=4.3mA$