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
A) $\frac{\pi}{6}$
B) $\frac{\pi}{3}$
C) $\frac{\pi}{4}$
D) $\frac{\pi}{2}$
Explanation
(b) $P = \frac{1}{2}V_{0}i_{0}\cos\phi \Rightarrow P = P_{Peak}.\cos\phi$
$\Rightarrow \frac{1}{2}\left( P_{peak} \right) = P_{peak}\cos\phi \Rightarrow \cos\phi\frac{1}{2} \Rightarrow \phi = \frac{\pi}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.