In the above circuit, C =(√ 3)/2 μ F, R_2=20 Ω, L =(√ 3)/10 H and R_1=10 Ω. Current in L - R_1 path is I_1 and in C - R_2 path it is I_2. The voltage…

In the above circuit, $\displaystyle \mathrm{C}=\frac{\sqrt{3}}{2} \mu \mathrm{~F}, \mathrm{R}_2=20 \Omega$, $\displaystyle \mathrm{L}=\frac{\sqrt{3}}{10} \mathrm{H}$ and $\displaystyle \mathrm{R}_1=10 \Omega$. Current in $\displaystyle \mathrm{L}-\mathrm{R}_1$ path is $\displaystyle \mathrm{I}_1$ and in $\displaystyle \mathrm{C}-\mathrm{R}_2$ path it is $\displaystyle \mathrm{I}_2$. The voltage of A.C source is given by, $\displaystyle \mathrm{V}=200 \sqrt{2} \sin (100 \mathrm{t})$ volts. The phase difference between $\displaystyle \mathrm{I}_1$ and $\displaystyle \mathrm{I}_2$ is :