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Q.

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?

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a

π6

b

π3

c

π4

d

π2

answer is B.

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Detailed Solution

P=12Voiocosϕ⇒P=Ppeak ⋅cosϕ⇒12Ppeak =Ppeak cosϕ⇒cosϕ=12⇒ϕ=π3
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