Question

A capacitor of capacitance $12\mu \text{F}$ is joined to an $\text{AC}$ source of frequency $200\text{Hz}$. The $rms$ current in the circuit is $2\text{A}$. The average energy stored in the capacitor is

• A. $0.01\text{J}$
• B. $0.02\text{J}$
• C. $0.2\text{J}$
• D. $0.1\text{J}$

Answer 1

The correct option is D $0.1\text{J}$

The impedance $Z$ of a purely capacitive circuit is equal to the capacitive reactance $X_C$.

$\therefore Z=X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$

$=\frac{1}{(2\pi \times 200)(12\times {10}^{-6})}\approx 66.3\Omega$

The $rms$ voltage across the capacitor is

$V_{rms}=i_{rms}Z=2.0\times 66.3=132.6\text{V}$

The average energy stored in the capacitor is

$E=\frac{1}{2}C{V}_{rms}^2$

$E=\frac{1}{2}\times (12\times {10}^{-6})\times (132.6{)}^2$

$\Rightarrow E\approx 0.1J$

Hence, $\left(D\right)$ is the correct answer.