Phasor-Diagram Solution for the LCR Circuit

Setting up the phasor diagram. With the current $I$ as the common reference phasor, three voltage phasors are drawn from the same origin: $V_R=IR$ along $I$; $V_L=IX_L$ perpendicular to $I$, leading it by $90^\circ$; and $V_C=IX_C$ perpendicular to $I$ but LAGGING it by $90^\circ$ -- i.e. pointing in exactly the OPPOSITE direction to $V_L$, along the same vertical line (the figure for this section shows all three explicitly).

Combining $V_L$ and $V_C$ first. Since $V_L$ and $V_C$ point in opposite directions along the same line, their resultant is simply their (signed) difference in magnitude, $V_L-V_C$ (taking the direction of the larger one as positive) -- exactly as two forces pulling in opposite directions along one line combine to a single net force.

Adding this to $V_R$. The net perpendicular phasor $(V_L-V_C)$ and the horizontal phasor $V_R$ are now at right angles to each other, exactly as in the LR/CR case, and the resultant applied-voltage phasor $V$ is again the diagonal of the rectangle they form:

The impedance and phase angle. Comparing with $V=IZ$ gives the impedance of the general series LCR circuit,

and the phase angle by which the current lags (or leads) the voltage is read off the same triangle as

$\tan\phi = \frac{X_L-X_C}{R}$ …