Obtain the relation of phase between instantaneous current and voltage with the help of a phasor diagram for a series $LCR$ circuit.

(N/A) In a series $LCR$ circuit,the current $I$ is the same through all components at any instant.
Let the current be $I = I_m \sin(\omega t)$.
The voltage across the resistor $V_R$ is in phase with the current $I$.
The voltage across the inductor $V_L$ leads the current $I$ by $\pi/2$.
The voltage across the capacitor $V_C$ lags behind the current $I$ by $\pi/2$.
The total voltage $V$ is the phasor sum of $V_R$,$V_L$,and $V_C$. Since $V_L$ and $V_C$ are in opposite directions,their resultant is $(V_L - V_C)$.
Using the phasor diagram,the resultant voltage $V$ is given by $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.
The phase angle $\phi$ between the source voltage and current is given by $\tan \phi = \frac{V_L - V_C}{V_R} = \frac{I_m X_L - I_m X_C}{I_m R} = \frac{X_L - X_C}{R}$.
Thus,the phase relation is $\phi = \tan^{-1} \left( \frac{X_L - X_C}{R} \right)$.