Q
JEE MAIN 2019
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 $\mathrm{I}_1$...