Q.A series LCR circuit with $R=100\ \Omega$, $L=0.5\ \text{H}$ and $C=10\ \mu\text{F}$ is connected to a $200\ \text{V}$ (rms), $50\ \text{Hz}$ AC source. Calculate

Using a phasor (rotating vector) diagram, the voltages across R, L and C add vectorially (since they are out of phase with the common current) to give $Z=\sqrt{R^2+(X_L-X_C)^2}$.

In a series L-C-R circuit connected to an AC source, the same current $I = I_0\sin\omega t$ flows through all three elements, so current is taken as the reference phasor.

• Voltage across $R$: $V_R = IR$, in phase with $I$ (phasor along the current direction).
• Voltage across $L$: $V_L = IX_L$, leads $I$ by $90°$ (phasor drawn $90°$ ahead).
• Voltage across $C$: $V_C = IX_C$, lags $I$ by $90°$ (phasor drawn $90°$ behind).

Since $V_L$ and $V_C$ are along the same line but opposite directions (both perpendicular to $I$, one up one down), their resultant is $(V_L - V_C)$, perpendicular to $V_R$.

By the phasor (vector) diagram, the resultant applied voltage is the vector sum: …