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