A) e.m.f. is ahead of current by p / 2
B) Current is ahead of e.m.f. by p / 2
C) Current lags behind e.m.f. by p
D) Current is ahead of e.m.f. by p
Correct Answer: B
Solution :
For purely capacitive circuit \[e={{e}_{0}}\sin \omega t\] \[i={{i}_{o}}\sin \left( \omega t+\frac{\pi }{2} \right)\] i.e. current is ahead of emf by \[\frac{\pi }{2}\]
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