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In the circuit shown in figure : R = 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)` is

A

`0^(@)`

B

`90^(@)`

C

`180^(@)`

D

`60^(@)`

Text Solution

Verified by Experts

The correct Answer is:
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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