Question

If an alternating current $i=i_m\sin \omega t$ is flowing through a capacitor then voltage drop $\Delta V_C$ across capacitor $C$ will be?

• A. $-\frac{i_m}{\omega C}\sin \omega t$
• B. $-\frac{i_m}{\omega C}\cos \omega t$
• C. $-\frac{i_m}{\omega C}(\sin \omega t+\frac{\pi }{4})$
• D. $\frac{i_m}{\omega C}(\sin \omega t-\frac{\pi }{4})$

Answer 1

The correct option is A $-\frac{i_m}{\omega C}\sin \omega t$

$\Delta V_C=-X_{C}i=-\frac{1}{C\omega }{I}_{m}sin\omega t$

$=-\frac{I_m}{C\omega }sin\omega t$