In the given AC circuit,the phase difference | vedclass

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(B) In the capacitor branch,the current $I_1$ leads the voltage $V$ by $\frac{\pi}{2}$.Thus,the phase of $I_1$ relative to voltage is $\phi_1 = +\frac

Vedclass · Accepted Answer

$\frac{\pi}{2} + 	an^{-1}\left(\frac{X_L}{R}ight)$

Vedclass · Answer

(B) In the capacitor branch,the current $I_1$ leads the voltage $V$ by $\frac{\pi}{2}$.Thus,the phase of $I_1$ relative to voltage is $\phi_1 = +\frac{\pi}{2}$.In the $RL$ branch,the current $I_2$ lags the voltage $V$ by an angle $\phi_2$,where $	an \phi_2 = \frac{X_L}{R}$.Therefore,the phase of $I_2$ relative to voltage is $\phi_2 = -	an^{-1}\left(\frac{X_L}{R}ight)$.The phase difference $\Delta \phi$ between $I_1$ and $I_2$ is given by $\Delta \phi = \phi_1 - \phi_2$.$\Delta \phi = \frac{\pi}{2} - \left[-	an^{-1}\left(\frac{X_L}{R}ight)ight] = \frac{\pi}{2} + 	an^{-1}\left(\frac{X_L}{R}ight)$.

In the given $AC$ circuit,the phase difference between $I_1$ and $I_2$ is:

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