Question

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

• A. $0.6\sqrt{2}A$
• B. $0.6A$
• C. $\frac{0.6\pi }{2}A$
• D. $\frac{0.6}{\sqrt{2}}A$

Answer 1

Answer: (A)

Peak current $i_0=\sqrt{2i_{rms}}.$
In a capacitive circuit, the capacitive reactance
$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}$
Given, $f=50Hz,C=10\mu F$
$=10\times 10^{-6}F$
$\therefore$ $X_C=\frac{1}{2\times \pi \times 50\times 10\times 10^{-6}}$
$=\frac{1000}{\pi }\Omega$
and $i_{rms}=\frac{V}{X_C}=\frac{200}{1000/\pi }$
$=\frac{3.14}{5}\approx 0.6A$
Peak value of current $i_0={\sqrt{2i}}_{rms}$
$i_0=0.6\sqrt{2}$