Question

A capacitor of capacitive reactance $5\Omega$ is connected with $\text{AC}$ source having voltage $V=3\sin (\omega t+\frac{\pi }{6})$. Then

• A. $I_{rms}=\frac{3}{\sqrt{2}}\text{A}$
• B. $I_{rms}=\frac{3}{5\sqrt{2}}\text{A}$
• C. Instantaneous current in the circuit is $\frac{3}{5}\sin (\omega t+\frac{2\pi }{3})$
• D. Instantaneous current in the circuit is $\frac{3}{5\sqrt{2}}\sin (\omega t+\frac{2\pi }{3})$

Answer 1

Answer: (B), (C)

Instantaneous current in the circuit is $\frac{3}{5}\sin (\omega t+\frac{2\pi }{3})$

Given:

$X_C=5\Omega ;V=3\sin (\omega t+\frac{\pi }{6})$

On comparing the given voltage with
$V=V_0\sin (\omega t+\varphi )$

We have, $V_0=3\text{V}$

$\Rightarrow V_{rms}=\frac{V_0}{\sqrt{2}}=\frac{3}{\sqrt{2}}\text{V}$

Therefore, the $rms$ current in the circuit is

$I_{rms}=\frac{V_{rms}}{X_C}=\frac{3}{5\sqrt{2}}\text{A}$

We know that in a capacitor, current leads voltage by $\frac{\pi }{2}$.

Therefore, the instantaneous current is given by

$I=\frac{V_0}{X_C}\sin (\omega t+\frac{\pi }{6}+\frac{\pi }{2})$

$=\frac{3}{5}\sin (\omega t+\frac{\pi }{6}+\frac{\pi }{2})=\frac{3}{5}\sin (\omega t+\frac{2\pi }{3})$

Hence, option $\left(B\right)$ and $\left(C\right)$ are the correct answers.