The time required for a 50 Hz alternating current to reach the peak value starting from zero

1/200 s

The correct answer is Option (2) → 1/200 s

Given:

Frequency of AC, $f = 50 \ \text{Hz}$

Time period: $T = \frac{1}{f} = \frac{1}{50} = 0.02 \ \text{s}$

AC current: $i(t) = I_\text{max} \sin(\omega t)$, where $\omega = 2 \pi f$

Time to reach peak from zero (quarter period of sine wave): $t = \frac{T}{4} = \frac{0.02}{4} = 0.005 \ \text{s}$

Time required: $t = \frac{1}{200} \ \text{s}$