An ac source has a frequency of 60 Hz. The time taken for the voltage to change from peak value to zero is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{240}\,\mathrm{s} \)
An alternating voltage \( V = V_0\sin\omega t \) goes from zero to its peak, back to zero, to the negative peak and back to zero in one time period T. From a peak value to the next zero is one quarter of a period.
\( \displaystyle T = \frac{1}{f} = \frac{1}{60}\,\mathrm{s} \)
\( \displaystyle t = \frac{T}{4} = \frac{1}{240}\,\mathrm{s} \)