When an alternating emf, e=300 sin `100pit` volt is applied across a bulb, the peak value of current is found to be 2 A. the average power is
A. 100 W
B. 200 W
C. 300 W
D. 400 W

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1 Answers

Correct Answer - C
`overline(P)=(e_(0)I_(0))/(2)`
`=(300xx2)/(2)=300W`

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