A) emf is ahead of current by \[\text{ }\!\!\pi\!\!\text{ /2}\]
B) current is ahead of emf by\[\text{ }\!\!\pi\!\!\text{ /2}\]
C) current lags behind emf by\[\text{ }\!\!\pi\!\!\text{ }\]
D) current is ahead of emf by \[\text{ }\!\!\pi\!\!\text{ }\]
Correct Answer: B
Solution :
For a pure capacitance circuit current equation is \[I=\omega C{{V}_{0}}\,\,\sin \,\,(\omega t+\pi /2)\] Comparing with \[I={{I}_{0}}\,\sin (\omega t+\phi ),\]we get \[\phi =\frac{\pi }{2}\] It means that current is ahead of emf by \[\frac{\pi }{2}\] rad.
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