Question

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. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{3}$

$P=\frac{1}{2}{V}_0{i}_{0}cos\varphi \Rightarrow P=P_{Peak}.cos\varphi$
$\Rightarrow \frac{1}{2}\left(P_{Peak}\right)=P_{Peak}cos\varphi \Rightarrow cos\varphi =\frac{1}{2}\Rightarrow \varphi =\frac{\pi }{3}$