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

Options

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)`

Answer

C

Explanation

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)`