In the circuit shown in figure :
`R = 10 Omega , L = (sqrt(3))/(10) H, R_(2) = 20 Omega` and `C = (sqrt(3))/(2) mF`. Current in `L - R_(1)` circuit is `I_(1)` in `C - R_(1)` circuit is `I_(2)` and the main current is `I`
At some instant current in `L - R_(1)` circuit is `10 A`. At the same instant current in `C - R_(2)` branch will be
A. `5A`
B. `5 sqrt(2) A`
C. `5 sqrt(6) A`
D. `5 sqrt(3) A`