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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 currentA. `(pi)/6`B. `(pi)/3`C. `(pi)/4`D. `(pi)/2` |
Answer» Correct Answer - B `P=1/2V_(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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