An  alternating voltage $\mathrm{V}\left(\mathrm{t}\right)=220$  sin 100πt volts is applied to a purely resistive load of 50Ω   The time taken for the current to rise from half of the peak value in millisecond is $\mathrm{x}\times 1.1,$  then  $x=?$

As   $\mathrm{V}\left(\mathrm{t}\right)=220\sin 100\pi t$ $\to \mathrm{V}\left(\mathrm{t}\right)=220\sin \mathrm{\pi t}$
So  $I\left(\mathrm{t}\right)=\left(220/50\right)\sin 100\pi t$  $\to \mathrm{I}\left(\mathrm{t}\right)=\frac{220}{50}\sin 100\mathrm{\pi t}$
l.e., $\mathrm{I}={\mathrm{I}}_{\mathrm{m}}=$ sin 100πt $\to \mathrm{I}={\mathrm{I}}_{\mathrm{n}}\sin 100\mathrm{\pi t}$
For  $\mathrm{l}={\mathrm{l}}_{\mathrm{m}}$
${\mathrm{t}}_1=\left(\mathrm{\pi }/2\right)\times \left(1/100\mathrm{\pi }\right)=\left(1/200\right)$ sec, $\to {\mathrm{t}}_1=\frac{\mathrm{\pi }}{2}\times \frac{1}{100\mathrm{\pi }}=\frac{1}{200}\sec$
And for  $\mathrm{l}={\mathrm{l}}_{\mathrm{m}}/2$
$\Rightarrow \left({\mathrm{l}}_{\mathrm{m}}/2\right)={\mathrm{l}}_{\mathrm{m}}\sin \left(100\pi t_2\right)$$\to {\mathrm{I}}_{\mathrm{m}/2}={\mathrm{I}}_{\mathrm{m}}\sin \left(100{\mathrm{\pi t}}_2\right)$
$\to {\mathrm{I}}_{\mathrm{m}/2}={\mathrm{I}}_{\mathrm{m}}\sin \left(100{\mathrm{\pi t}}_2\right)\to \mathrm{\pi }/6=100{\mathrm{\pi t}}_2$
$\begin{array}{l}\Rightarrow {\mathrm{t}}_2=(1/600)\mathrm{S}\\ \therefore {\mathrm{t}}_{\mathrm{req}}=(1/200)-(1/600)=(2/600)=(1/300)\mathrm{S}=3.3\mathrm{ms}\end{array}$