Q. A $ 10\,\mu F $ capacitor is connected across a $200 \,V, 50\, Hz$ AC supply. The peak current through the circuit is:
Jharkhand CECEJharkhand CECE 2004Alternating Current
Solution:
Peak current $i_{0}=\sqrt{2 i_{r m s}} .$
In a capacitive circuit, the capacitive reactance
$X_{C}=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$
Given, $f=50 \,Hz , C=10 \,\mu F$
$ =10 \times 10^{-6} F $
$\therefore $ $X_{C}=\frac{1}{2 \times \pi \times 50 \times 10 \times 10^{-6}}$
$=\frac{1000}{\pi} \Omega$
and $i_{r m s}=\frac{V}{X_{C}}=\frac{200}{1000 / \pi}$
$=\frac{3.14}{5} \approx 0.6 A$
Peak value of current $i_{0}=\sqrt{2 i}_{r m s}$
$ i_{0}=0.6 \sqrt{2}$
