Power-Factor Correction in a Series LCR Circuit
In an AC laboratory a student connects a coil of resistance 20 Ω and inductance 0.10 H in series with a 50 μF capacitor across a 230 V, 50 Hz supply. He is asked to improve the power factor of this series circuit to unity. His plan is to add a second 50 μF capacitor in series with the first, arguing that 'more capacitance means more correction'. The teacher points out that the circuit is already capacitive, so adding series capacitance will make the power factor worse rather than better, and asks him to calculate before wiring. A second group suggests instead connecting an extra capacitor in parallel with the existing one, claiming that this will bring the circuit to resonance.
1. Calculate the impedance, rms current and power factor of the original circuit, and state whether the current leads or lags the supply voltage.
Read the worked answer
With $\omega=2\pi(50)=314\ \text{rad s}^{-1}$:
$X_L=\omega L=314\times0.10=31.4\ \Omega$,
$X_C=\dfrac{1}{\omega C}=\dfrac{1}{314\times50\times10^{-6}}=63.7\ \Omega$.
The net reactance is $X=X_L-X_C=31.4-63.7=-32.3\ \Omega$, negative, so the circuit is capacitive.
$Z=\sqrt{R^2+X^2}=\sqrt{20^2+32.3^2}=\sqrt{400+1043}=37.9\ \Omega$,
$I=\dfrac{V}{Z}=\dfrac{230}{37.9}=6.1\ \text{A}$.
Power factor $=\dfrac{R}{Z}=\dfrac{20}{37.9}=0.53$, and because $X_C>X_L$ the current leads the voltage. Average power is $I^2R=(6.1)^2(20)=7.4\times10^2\ \text{W}$, against an apparent power of $230\times6.1=1.4\ \text{kVA}$.
2. Evaluate the student's proposal of adding a second 50 μF capacitor in series, and then determine the correct capacitance needed for unity power factor, stating how it should be connected.
Read the worked answer
Two 50 μF capacitors in series give 25 μF, so
$X_C=\dfrac{1}{314\times25\times10^{-6}}=127\ \Omega$,
$X=31.4-127=-95.9\ \Omega$,
$Z=\sqrt{400+9197}=98\ \Omega$.
Then $I=230/98=2.3\ \text{A}$ and the power factor falls to $20/98=0.20$, still leading. The current is smaller but the circuit is far more reactive, so the student's plan makes the power factor worse, exactly as the teacher said.
Unity power factor requires $X_C=X_L=31.4\ \Omega$, hence
$C=\dfrac{1}{\omega X_C}=\dfrac{1}{314\times31.4}=1.01\times10^{-4}\ \text{F}\approx101\ \mu\text{F}$.
The existing capacitor is 50 μF, so about 51 μF must be added in parallel with it (parallel capacitances add), not in series. Replacing the 50 μF unit by a 101 μF unit would do the same job.
3. At the corrected resonance condition, calculate the current, the power dissipated, and the voltages across the inductor and the capacitor. Comment on why these voltages are a practical concern.
Read the worked answer
At resonance $Z=R=20\ \Omega$, so
$I=\dfrac{230}{20}=11.5\ \text{A}$.
The average power is $I^2R=(11.5)^2(20)=2.6\times10^3\ \text{W}$, which is the maximum the supply can deliver to this circuit.
The reactive voltages are
$V_L=IX_L=11.5\times31.4=3.6\times10^2\ \text{V}$,
$V_C=IX_C=11.5\times31.4=3.6\times10^2\ \text{V}$.
Each component carries about 360 V rms even though the supply is only 230 V, because $V_L$ and $V_C$ are equal and opposite and cancel in the loop; the quality factor is $X_L/R\approx1.6$. The practical consequence is that capacitors and coil insulation must be rated well above the supply voltage, and the large current at resonance can overheat the coil and blow a fuse. The 11.5 A current also means the wiring, not just the components, must be rated for it.