In the circuit shown in figure: `R_(1)=10Omega, L+(sqrt3)/(10)H, R_(2)=20Omega and C-(sqrt3)/(2)mF`. Current in `L-R_(1) "circuit is" I_(1) "in" C-R_(2) "circuit is" I_(2)` and the main current is I.

At some instant current in `L-R_(1)` circuit is 10A. At the same instant current in `C-R_(2)` circuit will be

Explanation

For `R_(1)`-L branch
`X_(L)=omegaL=100xx(sqrt(3))/10=10 sqrt(3)Omega,R_(1)=10 Omega`
`:. tan phi=(X_(L))/(R_(1))=sqrt(3)` or `phi=60^(@)`
Hence current `I_(1)` lags voltage by `60^(@)` For `R_(2)-C` branch
`X_(C)=1/(omegaC)=1/(100xx(sqrt(3))/2xx10^(-3))=20/(sqrt(3)) Omega`
`:. tan phi=(X_(C))/(R_(2))=1/(sqrt(3)) ` or `phi=30^(@)`
Hence current `I_(2)` leads voltage by `30^(@)`
`:. ` The phase difference between `I_(1)` and `I_(2)` branch is `90^(@)`
The maximum current through `R_(1)-L` branch is
`=(V_(0))/(sqrt(R_(1)^(2)+omega^(2)L^(2)))=(200sqrt(2))/(sqrt(10^(2)+(10sqrt(3))^(2)))=10sqrt(2) amp`.
Hence when current through `R_(1)-L` branch is `10 sqrt(2)` amp. the current through `R_(2)-C` branch will be zero.