How does the sign of the phase angle phi,by w | vedclass

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(D) The phase angle $\phi$ by which the supply voltage leads the current in an $LCR$ series circuit is given by the relation: $	an \phi = \frac{X_L -

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It changes from negative to positive.

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(D) The phase angle $\phi$ by which the supply voltage leads the current in an $LCR$ series circuit is given by the relation: $	an \phi = \frac{X_L - X_C}{R} = \frac{2 \pi 
u L - \frac{1}{2 \pi 
u C}}{R}$.At very low frequencies $(
u 	o 0)$,the capacitive reactance $X_C = \frac{1}{2 \pi 
u C}$ is very large,making $X_L - X_C$ negative. Thus,$	an \phi$ is negative,meaning the voltage lags behind the current (or the phase angle is negative).As the frequency $
u$ increases and reaches the resonant frequency $
u_r = \frac{1}{2 \pi \sqrt{LC}}$,the inductive reactance $X_L$ equals the capacitive reactance $X_C$,so $X_L - X_C = 0$ and $\phi = 0$.At very high frequencies $(
u 	o \infty)$,the inductive reactance $X_L = 2 \pi 
u L$ becomes very large,making $X_L - X_C$ positive. Thus,$	an \phi$ is positive,meaning the voltage leads the current (or the phase angle is positive).Therefore,as the frequency increases from very low to very high values,the sign of the phase angle $\phi$ changes from negative to positive.

How does the sign of the phase angle $\phi$,by which the supply voltage leads the current in an $LCR$ series circuit,change as the supply frequency is gradually increased from very low to very high values?

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