JEE Main 2019 — Alternating Current Question with Solution

In an AC resistive circuit, current and voltage are in phase.

So, $$I=\frac{V}{R}\Rightarrow I=\frac{220}{50}\sin (100\pi t)$$ ..... (i)

$$\therefore$$ Time period of one complete cycle of current is

$$T=\frac{2\pi }{\omega }=\frac{2\pi }{100\pi }=\frac{1}{50}s$$

So, current reaches its maximum value at

$$t_1=\frac{T}{4}=\frac{1}{200}s$$

When current is half of its maximum value, then from Eq.(i), we have

$$I=\frac{I_{\max }}{2}=I_{\max }\sin (100\pi t_2)$$

$$\Rightarrow \sin (100\pi t_2)=\frac{1}{2}\Rightarrow 100\pi t_2=\frac{5\pi }{6}$$

So, instantaneous time at which current is half of maximum value is $$t_2=\frac{1}{120}s$$

Hence, time duration in which current reaches half of its maximum value after reaching maximum value is

$$\Delta t=t_2-t_1=\frac{1}{120}-\frac{1}{200}=\frac{1}{300}s=3.3$$ ms