Question

What is the peak current in a capacitor connected to an AC source?

Options

• \[i_m=\omega Cv_m=v_mX_C\]

• \[i_m=\omega Cv_m=\frac{v_m}{X_C}\]

• \[i_m=\frac{Cv_m}{\omega}=v_mX_C\]

• \[i_m=\frac{\omega v_m}{C}=\frac{v_m}{X_C}\]

Answer 1

The amplitude of \[i=\omega Cv_m\cos\omega t\] is \[\omega Cv_m\]. Using capacitive reactance, the same peak current is \[\frac{v_m}{X_C}\].