An A.C circuit having supply voltage E consists of a resistor of resistance 3 Ω and an inductor of reactance 4 Ω as shown in the figure. what is the voltage across the inductor at $\mathrm{t}=\frac{\mathrm{\pi }}{\mathrm{\omega }}$ (in volt)?

Instantaneous value of current  $\mathrm{I}={\mathrm{I}}_0\sin (\mathrm{\omega t}-\mathrm{\varphi })$

Where,  ${\mathrm{I}}_0=\frac{{\mathrm{\epsilon }}_0}{\mathrm{Z}}=\frac{{\mathrm{\epsilon }}_0}{\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{I}}^2}}=2\mathrm{A}$

And phase

$\mathrm{\varphi }={\tan }^{-1}\left(\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}\right)={\tan }^{-1}\left(\frac{4}{3}\right)={53}^0$

Now, voltage across the inductors,

$\begin{array}{l}{\mathrm{V}}_{\mathrm{L}}=\mathrm{L}\frac{\mathrm{dI}}{\mathrm{dt}}=\mathrm{L}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{I}}_0\sin (\mathrm{\omega t}-\mathrm{\varphi })\right]\\ ={\mathrm{I}}_0\left(\mathrm{\omega L}\right)\cos \left(\mathrm{\omega t}-{53}^0\right)\end{array}$

At time, $\mathrm{t}=\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle \mathrm{\omega }}}$,

$\begin{array}{l}{\mathrm{V}}_{\mathrm{L}}={\mathrm{I}}_0{\mathrm{X}}_{\mathrm{L}}\cos \left(\mathrm{\pi }-{53}^{°}\right)\\ \left|{\mathrm{V}}_{\mathrm{L}}\right|=2\times 4\times \frac{3}{5}=4.8\mathrm{volt}\end{array}$