The time required for a 50 Hz alternating current to reach the peak value starting from zero
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1/200\,\mathrm{s} \)
An alternating current varies as \( i = i_0 \sin \omega t \). It starts from zero at \( \omega t = 0 \) and first reaches its peak at \( \omega t = \pi/2 \), which is one quarter of a cycle.
\( \displaystyle T = \frac{1}{f} = \frac{1}{50}\,\mathrm{s} \)
\( \displaystyle t = \frac{T}{4} = \frac{1}{4 \times 50} = \frac{1}{200}\,\mathrm{s} \)