QJEEM19Ph4ece: JEE Main 2019 Physics - Class 12 JEEM19Ph4ece Solution: An alternating voltage v(t) = 220 sin 100 t volt is applied to…

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

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

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

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

So, current reaches its maximum value at

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

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

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

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

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

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

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