Power-factor correction

Objective

Add a parallel capacitor to an inductive load to raise the power factor and cut the supply current.

Procedure

1. Run: the supply current is lower than in the previous experiment.
2. Remove the capacitor (set C to 0.1 µF) and compare.

Theory

The load admittance is G − jB; a capacitor C = XL/(ω(R²+XL²)) = 15.9 µF cancels the susceptance so the source sees only the conductance: the supply current falls to PF × its old value while the load power is unchanged.

Questions to answer

1. What happens if the capacitor is too large?

Components used

Ceramic Capacitor 0.1uF (104), Function generator (sine), Inductor 100uH, Oscilloscope (2-channel), Resistor 220 ohm 1/4W

About the model

Ideal wires; component models as listed in the catalogue (no parasitics, no temperature drift). Built without a breadboard so the schematic stays readable.

More Basic Electrical Lab experiments

• Ohm's law
• Equivalent resistance (series + parallel)
• Lamp circuits in series and in parallel
• Kirchhoff's current law
• Kirchhoff's voltage law