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Question: A lamp consumes only 50% of peak power in an a.c. circuit. What is the phase difference between the ...

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

π6\frac{\pi}{6}

B

π3\frac{\pi}{3}

C

π4\frac{\pi}{4}

D

π2\frac{\pi}{2}

Answer

π3\frac{\pi}{3}

Explanation

Solution

P=12Voiocos⁡φ⇒P=PPeak.cos⁡φP = \frac{1}{2}V_{o}i_{o}\cos\varphi \Rightarrow P = P_{Peak}.\cos\varphi

⇒12(Ppeak)=Ppeakcos⁡φ⇒cos⁡φ=12⇒φ=π3\Rightarrow \frac{1}{2}(P_{peak}) = P_{peak}\cos\varphi \Rightarrow \cos\varphi = \frac{1}{2} \Rightarrow \varphi = \frac{\pi}{3}