AC circuits
The idea
When the driving voltage oscillates sinusoidally, resistors, inductors, and capacitors each respond differently. A resistor's voltage tracks the current exactly; an inductor opposes changes in current, so its voltage leads the current by 90°; a capacitor must accumulate charge before its voltage builds, so its voltage lags by 90°. Each reactive element gets a frequency-dependent effective resistance called reactance: X_L = ωL grows with frequency, X_C = 1/(ωC) shrinks with it.
For a series RLC circuit, the opposing 90° phase shifts mean reactances partially cancel: the impedance is Z = √(R² + (X_L − X_C)²), and the rms current is V_rms/Z. At the resonant frequency ω₀ = 1/√(LC) the cancellation is exact, Z drops to R alone, and the current peaks — the principle behind radio tuning. The phase angle, tanφ = (X_L − X_C)/R, tells you whether the circuit behaves inductively (current lags) or capacitively (current leads).
The misconception to retire: reactances and resistances do not simply add. A 75 Ω inductive reactance in series with a 265 Ω capacitive reactance contributes only |75 − 265| = 190 Ω to the impedance, because their voltage phasors point in opposite directions. Also, only the resistor dissipates average power — ideal inductors and capacitors borrow energy each half-cycle and return it.
Worked example
A series circuit with R = 50 Ω, L = 0.20 H, and C = 10 μF is driven by a 120 V rms, 60 Hz source. Find the impedance, the rms current, the average power dissipated, and the resonant frequency.
- Convert to angular frequency and compute reactances: ω = 2π × 60 ≈ 377 rad/s, so X_L = ωL = 377 × 0.20 ≈ 75.4 Ω and X_C = 1/(ωC) = 1/(377 × 1.0 × 10⁻⁵) ≈ 265.3 Ω.
- Combine into impedance — reactances subtract before joining R at right angles: Z = √(50² + (75.4 − 265.3)²) = √(2500 + (−189.9)²) = √(2500 + 36060) ≈ 196 Ω.
- Apply the ac Ohm's law with rms values: I_rms = V_rms/Z = 120/196 ≈ 0.61 A.
- Only the resistor dissipates on average: P = I_rms² R = 0.61² × 50 ≈ 18.7 W — far below the naive V × I = 73 W because the current runs about 75° out of phase with the voltage.
- Find resonance: f₀ = 1/(2π√(LC)) = 1/(2π√(0.20 × 1.0 × 10⁻⁵)) ≈ 112 Hz. The source at 60 Hz sits below resonance, which is why X_C dominates and the circuit behaves capacitively, with the current leading the voltage.
Answer. Z ≈ 196 Ω, I_rms ≈ 0.61 A, average power ≈ 19 W, and the circuit would resonate near 112 Hz.
Check your understanding
- Why do inductive and capacitive reactances subtract rather than add when combining into impedance?
- What happens to the current and the power factor as the driving frequency sweeps through resonance?
- Why do ideal inductors and capacitors dissipate no average power even while carrying large currents?
- How does a radio receiver use an RLC circuit to select one station's frequency out of many arriving at once?
Build the foundations first
AC circuits builds on these concepts. If any feel shaky, start there.