Question

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

• A. $66.3\text{V}$
• B. $132.6\text{V}$
• C. $94\text{V}$
• D. $47\text{V}$

Answer 1

The correct option is B $132.6\text{V}$

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

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

$\Rightarrow Z=\frac{1}{2\pi fC}$

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

$\Rightarrow Z=66.3\Omega$

The $rms$ voltage across the capacitor be,

$V_{rms}=i_{rms}Z=2.0\times 66.3\approx 133\text{V}$

Hence, $\left(B\right)$ is the correct option.