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A lamp consumes only 50% of peak power i...

A lamp consumes only `50%` of peak power in an `a.c.` circuit. What is the phase difference between the applied voltage and the circuit current

A

`(pi)/(6)`

B

`(pi)/(3)`

C

`(pi)/(4)`

D

`(pi)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`P = (1)/(2) xx V_(0) i_(0) cos phi implies P = P_("peak") cos phi`
`implies (1)/(2) (P_("peak")) = P_("peak") cos phi implies cos phi (1)/(2) implies phi = (pi)/(3)`
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