A 100 nF capacitor is to form part of a filter connected across a 240 V 50 Hz mains supply. What current will flow in the capacitor?
First we must find the reactance of the capacitor:
X_{C}={\frac{1}{2\pi\times50\times100\times10^{-9}}}=31.8\times10^{3}=31.8\mathrm{~k}\OmegaThe r.m.s. current flowing in the capacitor will thus be:
I_{\mathrm{c}}={\frac{V_{\mathrm{c}}}{X_{\mathrm{c}}}}={\frac{240}{31.8\times10^{3}}}=7.5\times10^{-3}=7.5\operatorname*{mA}