Question

A $10\mu F$ capacitor is connected across a $200V$, $50Hz$ A.C. supply. The peak current through the circuit is :

• A. $0.6A$
• B. $0.6\sqrt{2}A$
• C. $\left(0.06\sqrt{2}\right)A$
• D. $0.6\pi A$

Answer 1

Answer: (B)

$0.6\sqrt{2}A$

RMS value of applied voltage = $200V$
Impedance of a capacitor is given by:
$X_C=\frac{1}{2\pi fC}$
Hence, rms current through it is:
$I=\frac{V}{X_C}$
$I=200\times 2\times \pi \times 50\times 10\times {10}^{-6}$
$I=0.2\pi A$
Peak current, $I_{max}=\sqrt{2}I$
$=0.628\sqrt{2}A$