Q. An AC source is rated at 220 V, 50 Hz. The time taken for voltage to change from its peak value to zero is
A) 50 s
B) 0.02 s
C) 5 s
D) $5 \times 10^{- 3}s$
Explanation
(d) Frequency $n =$50Hz
Time period $T = \frac{1}{50}$ s
Time taken for voltage to change from its peak value to zero
$= \frac{T}{4} = \frac{1}{4 \times 50} = \frac{1}{200} = 5 \times 10^{- 3}s\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.