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

Options

A. `5A`
B. `5 sqrt(2) A`
C. `5 sqrt(6) A`
D. `5 sqrt(3) A`

Answer

D

Explanation

`i_(1) = (200 sqrt(2))/(sqrt(R_(1)^(2) + (omega L)^(2))) sin (100 t - (pi)/(3))`
and `i_(2) = (200 sqrt(2))/(sqrt(R_(2)^(2) + ((1)/(omega C))^(2))) sin (100 t + (pi)/(6))`