Question

A capacitor of capacitive reactance $10\Omega$ is connected to an AC source having voltage $V=5\sin (\omega t+\frac{\pi }{6}).$ What is the value of peak current?

• A. $\frac{1}{2}A$
• B. $\frac{1}{3}A$
• C. $\frac{1}{4}A$
• D. $\frac{1}{5}A$

Answer 1

The correct option is A $\frac{1}{2}A$

For a capacitor in an a.c circuit, current leads voltage by a phase angle of $\frac{\pi }{2}$.


$i=i_0\sin (\omega t+\frac{\pi }{2}+\frac{\pi }{6})$
and peak current $i_0=\frac{V_0}{X_C}$
where $X_C=\frac{1}{\omega C}$ is the capacitive reactance.

$\therefore i_0=\frac{5}{10}=\frac{1}{2}A$