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 · 2019 · 08 Apr 2019 · Shift 1 · 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)

A purely resistive load keeps the current in phase with the voltage, so

i (t) = \frac{220}{50} sin (100 π t) = 4.4 sin (100 π t)

A with peak

I_{0} = 4.4

A. The current is half its peak when

sin (100 π t) = \frac{1}{2}

, that is

100 π t_{1} = \frac{π}{6}

, and it reaches the peak when

sin (100 π t) = 1

, that is

100 π t_{2} = \frac{π}{2}

. The interval between these instants is

Δ t = \frac{π /2 - π /6}{100 π} = \frac{π /3}{100 π} = \frac{1}{300}

\approx 3.3

ms.

JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Single correct option