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  4. 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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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?

Alternating Current

Solution:

$P=\frac{1}{2} V_{o} i_{o} \cos \phi $
$\Rightarrow P=P_{\text {peak }} \cdot \cos \phi$
$\Rightarrow \frac{1}{2}\left(P_{\text {peak }}\right)=P_{\text {peak }} \cos \phi$
$\Rightarrow \cos \phi=\frac{1}{2} \Rightarrow \phi=\frac{\pi}{3}$