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` Phase difference between `I_(1)` and `I_(2)` isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `60^(@)`

In the circuit shown in figure : `R = 10 Omega , L = (sqrt(3))/(10) H,

1.	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` Phase difference between `I_(1)` and `I_(2)` isA. `0^(@)`B. `90^(@)`C. `180^(@)`D. `60^(@)`
Answer» Correct Answer - B `Tan phi_(1) = (omega L)/(R_(1))` , `Tan phi_(2) = (1)/(omega C R_(2))` `:. Delta phi = phi_(1) ~ phi_(2)`

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