In an AC circuit, capacitors and inductors oppose current in a way that depends on frequency. This opposition is called reactance and is measured in ohms, like resistance, but it does not dissipate energy in the way a resistor does. Impedance combines resistance and reactance and describes the total AC opposition.
Inductive and capacitive reactance
Reactance at 60 Hz
Problem. Find the reactance of a 50 mH inductor and a 47 µF capacitor at 60 Hz.
Solution. X_L = 2π × 60 × 0.050 = 18.85 Ω. X_C = 1 ÷ (2π × 60 × 47 × 10⁻⁶) = 1 ÷ 0.01772 = 56.4 Ω.
Series impedance
For a series R-L-C circuit, the reactances subtract because they have opposite phase effects, then combine with resistance in quadrature. The magnitude of the impedance is the square root of the sum of squares.
Series impedance
Problem. A series circuit has R = 30 Ω, X_L = 50 Ω and X_C = 10 Ω. Find the impedance magnitude.
Solution. X_L − X_C = 40 Ω. Z = √(30² + 40²) = √(900 + 1600) = √2500 = 50 Ω.
- Inductive reactance rises with frequency and inductance.
- Capacitive reactance falls as frequency and capacitance rise.
- Reactance is measured in ohms but does not dissipate real power.
- Impedance magnitude combines R and the net reactance in quadrature.
- X_L = 2πfL increases with frequency.
- X_C = 1/(2πfC) decreases with frequency.
- Net reactance is X_L − X_C in a series circuit.
- Z = √(R² + (X_L − X_C)²) gives impedance magnitude.
Worked example
At 50 Hz, a 100 mH inductor has X_L = 2π × 50 × 0.100 = 31.4 Ω, while a 100 µF capacitor has X_C = 1 ÷ (2π × 50 × 100 × 10⁻⁶) = 31.8 Ω — nearly equal and opposite in a series circuit.