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In as shown in the figure AC circuit, ph...

In as shown in the figure AC circuit, phase difference between currents `I_(1) and I_(2)` is

A

`tan^(-1)((X_(C))/(R))-(pi)/(2)`

B

`tan^(-1)((X_(C))/(R))`

C

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

D

`(pi)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Let `V_(s) = V_(s) sin omegat`
`I_(1) = I_(01) sin (omegat - pi//2)`
`I_(2) = I_(02) sin (omegat + theta)`
`tan theta = ((x_(c ))/(R ))` : So phase diffreence `= theta + (pi)/(2)`
`tan^(-1) ((X_(c ))/(R )) + (pi)/(2)`
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