An alternating voltage $\mathrm{V}(\mathrm{t})=220 \sin 100 \pi \mathrm{t}$ volt is applied to a purely resistive load of $50 \Omega$. The time taken for the current to rise from half of the peak value to the peak value is:

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Alternating Current

An alternating voltage

V (t) = 220 sin 100 π t

volt is applied to a purely resistive load of

50Ω

. The time taken for the current to rise from half of the peak value to the peak value is:

Answer: (B)

For a purely resistive load, current follows the same sinusoidal variation as voltage:

i (t) = I_{0} sin (ω t)

, with

ω = 100 π rad/s

.

Current equals half the peak value when

sin (ω t_{1}) = \frac{1}{2}

, i.e.

ω t_{1} = \frac{π}{6}

.

Current reaches the peak value when

sin (ω t_{2}) = 1

, i.e.

ω t_{2} = \frac{π}{2}

.

The time taken between these two instants is

Δ t = t_{2} - t_{1} = \frac{1}{ω} (\frac{π}{2} - \frac{π}{6}) = \frac{1}{ω} \cdot \frac{π}{3} = \frac{π}{3 \times 100 π} = \frac{1}{300} s = 3.33 ms

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option