A 100 pF capacitor is connected to a 230 V. 50Hz AC source. The rms value of conduction current will be

Question

Please log in or register to answer this question.

Devanshi · Answer 1 · 2020-06-01T04:52:02+0000

RMS value of conduction current, \(I=\frac{V}{X_C}\)
where, \(X_C\)=capacitative reactance.
\(X_C=\frac{1}{\omega C}\)

\(I=V\omega C=230V\times300rad/s\times10^{-10}F\\I=6.9\times10^{-6}A\)