Q.A resistor $R$, an inductor $L$ and a capacitor $C$ are connected in series to an AC source $v=v_0\sin\omega t$. Using the phasor-diagram method, derive the expression for the impedance $Z=\sqrt{R^2+(X_L-X_C)^2}$ of the circuit and the phase angle $\tan\phi=(X_L-X_C)/R$ between the current and the voltage, and state the condition under which the circuit behaves as

Setting up all three phasors. With current $I$ as reference: $V_R=IR$ in phase; $V_L=IX_L$ leading by $90^\circ$; $V_C=IX_C$ lagging by $90^\circ$ -- so $V_L$ and $V_C$ point in exactly OPPOSITE directions along the same perpendicular line.

Combining $V_L$ and $V_C$. Being opposite, they combine to a single net phasor of magnitude $(V_L-V_C)$ (taking the larger as positive), along whichever direction is larger.

Adding to $V_R$. This net phasor is perpendicular to $V_R$, so the resultant applied voltage is again a Pythagorean sum:

and the phase angle is $\tan\phi=(X_L-X_C)/R$.

Inductive versus capacitive condition. If $X_L>X_C$, $\tan\phi>0$: the circuit is net INDUCTIVE and current lags. If $X_C>X_L$, $\tan\phi<0$: the circuit is net CAPACITIVE and current leads. If $X_L=X_C$, $\phi=0$: current and voltage are exactly in phase (resonance).

Combine the opposite $V_L$/$V_C$ phasors into one net perpendicular phasor first, then add this to $V_R$ exactly as in the LR/CR cases.

• Adding $X_L$ and $X_C$ together instead of subtracting them -- their phasors point in OPPOSITE directions, not the same one.
• Not stating the inductive/capacitive condition, which depends on which reactance is larger, not merely on the sign convention chosen.