In the shown AC circuit phase difference between currents I₁ and I₂ is

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$\frac{π}{2} - \tan^{- 1} \frac{X_{L}}{R}$

$\tan^{- 1} (\frac{X_{L} - X_{C}}{R})$

$\frac{π}{2} + \tan^{- 1} \frac{X_{L}}{R}$

$\tan^{- 1} (\frac{X_{L} + X_{C}}{R}) + \frac{π}{2}$

answer is C.

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Detailed Solution

In the upper branch containing C, current $I_{1}$ leads supply voltage by $\frac{π}{2}$ .

In the lower branch containing L and R, current $I_{2}$ lags supply voltage by $ϕ = \tan^{- 1} \frac{X_{L}}{R}$ .

Hence the total phase difference between the $I_{1}$ and $I_{2}$ is $\frac{π}{2} + \tan^{- 1} \frac{X_{L}}{R}$ .

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