Physics - EMI/AC Question with Solution | TestHub

$I_{rms}=6.9\text{μA};V_{rms}=230\mathrm{V};\omega =600rad{\text{s}}^{-1}$

Current in capacitor $I_{rms}=\frac{V_{rms}}{X_C}$, where, $X_C$ is capacitive reactance given by $X_C=\frac{1}{\omega \mathrm{C}}$

Now, current, $I_{rms}=\left(V_{rms}\right)\times \left(\omega \mathrm{C}\right)$

So, capacitance is $C=\frac{I_{rms}}{V_{rms}\omega }=\frac{6.9\times 10^{-6}}{230\times 600}=50\text{pF}$