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\[{{i}_{0}}={{\sqrt{2i}}_{rms}}\]. 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\[{{i}_{0}}=\sqrt{2}{{i}_{rms}}\]\[{{i}_{0}}=0.6\sqrt{2}\]