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In the shown AC circuit phase different ...

In the shown AC circuit phase different between current `I_(1)` and `I_(2)` is

A

`pi/2-tan^(-1) x_(L)/R`

B

`tan^(-1) (x_(L)-x_(C ))/R`

C

`pi/2+tan^(-1) x_(L)/R`

D

`tan^(-1) (x_(L)-x_(C ))/R+pi/2`

Text Solution

Verified by Experts

The correct Answer is:
C
Current leads the voltage by `pi/2` in capacitor and by `tan^(-1) X_(L)/R` in `RL` circuit phase difference in `I_(1)` and `I_(2)`
`=pi/2+tan^(-1)X_(L)/R`
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Knowledge Check

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