In the shown circuit,  ${\text{R}}_{\text{1}}\text{=1}0\Omega ,\text{ L}=\frac{\sqrt{3}}{10}\text{H},{\text{R}}_{\text{2}}=\text{2}0\Omega ,\text{ C}=\frac{\sqrt{3}}{2} \text{milli}-\text{farad}$  and t is time in seconds. Then at the instant current through   ${\mathrm{R}}_1\mathrm{is}10\sqrt{2}\mathrm{A}.$ ; find the current through resistor  ${\mathrm{R}}_2$  in amperes.

For  $$\begin{gathered} {\text{R}}_{\text{1}}–{\text{ L branch R}}_{\text{1}}–\text{L} \\ {X}_L=\omega L=100\times \frac{\sqrt{3}}{10}=10\sqrt{3}\Omega ,R_1=10\Omega \\ \therefore \tan \varphi =\frac{X_L}{R_1}=\sqrt{3}or\varphi =60° \end{gathered}$$    

Hence current  lags voltage by 60º.

For  ${\text{R}}_{\text{2}}–\text{C branch}$

$X_C=\frac{1}{\omega C}=\frac{1}{100\times \frac{\sqrt{3}}{2}\times {10}^{-3}}=\frac{20}{\sqrt{3}}\Omega$

$\therefore \tan \varphi =\frac{X_C}{R_2}=\frac{1}{\sqrt{3}}or\varphi =30°$