question_answer

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

A) \[0.6\sqrt{2}A\]

B) \[0.6\,A\]

C)  \[\frac{0.6\pi }{2}A\]

D)  \[\frac{0.6}{\sqrt{2}}A\]

Correct Answer: A

Solution :

 Key Idea: Peak current \[{{i}_{0}}=\sqrt{2{{i}_{rms}}}.\] In a capacitive circuit, the capacitive reactance \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\pi fC}\] 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 \[{{i}_{rms}}=\frac{V}{{{X}_{C}}}=\frac{200}{1000/\pi }=\frac{3.14}{5}\approx 0.6\,A\] Peak value of current \[{{i}_{0}}={{\sqrt{2i}}_{rms}}\] \[{{i}_{0}}=0.6\sqrt{2}\]