LCR Circuits & Resonance
Vector bookkeeping, resonance and tuning.
Lesson at a glanceShow the key points
- Phasors: R along x, 1/ωC up, ωL down; Z = √(R² + X²), tan φ = X/R (39.11).
- LCR: Z = √(R² + (1/ωC − ωL)²) (39.12), tan φ (39.13).
- Resonance ω₀ = 1/√(LC) (39.14): Z = R, maximum current; small R gives a sharp peak.
- LC circuit: energy oscillates between C and L.
With several elements in series, a neat vector trick gives the impedance and phase at once. It reveals resonance: the frequency at which a circuit responds most strongly, the principle behind every radio tuner.
The vector (phasor) method
For a series circuit with ℰ = ℰ₀ sin ωt:
- Draw R as a vector along the x-axis.
- Draw the capacitive reactance 1/ωC along +y (π/2 ahead) and the inductive reactance ωL along −y (π/2 behind).
- Add them. The resultant’s length is the impedance Z; its angle with the x-axis is the phase factor φ.
On switching on, the current behaves in a complicated way, but soon settles into this steady state at the source’s frequency.
CR and LR circuits
CR: Z = √(R² + 1/ω²C²), tan φ = 1/ωCR, current leads. LR: Z = √(R² + ω²L²), tan φ = ωL/R, current lags.
The LCR circuit
The two reactances point in opposite directions, so the net reactance is X = 1/ωC − ωL:
- XC > XL: φ > 0, current leads (capacitive).
- XC < XL: φ < 0, current lags (inductive).
- XC = XL: net reactance zero, current in phase, Z = R.
L = 0.10 H, C = 25 µF and R = 30 Ω are in series with ℰ = (20 V) sin(400 t). Find the impedance, peak current, phase and resonant frequency.
A coil of resistance 40 Ω and inductance 0.40 H is connected to 230 V, 50 Hz. Find the rms current. Then a 60 µF capacitor is added in series: find the new rms current.
Resonance
Vary ω and the peak current i₀ = ℰ₀/Z is greatest when 1/ωC = ωL:
A small R makes the resonance sharp: the current is large only very close to ν₀. Such a circuit used near ν₀ is a resonant circuit.
LC oscillations
With R = 0 the current at resonance would be infinite for any emf: a finite current can flow with no source at all. Connect a charged capacitor to a pure inductor: it discharges through the inductor, whose emf then recharges it the other way, and so on, at ν = 1/2π√(LC). Energy swaps between the capacitor’s electric field and the inductor’s magnetic field.
- 1Phasors: R along x, 1/ωC up, ωL down; Z = √(R² + X²), tan φ = X/R (39.11).
- 2LCR: X = 1/ωC − ωL (39.12–39.13); leads if capacitive, lags if inductive.
- 3Resonance at ω₀ = 1/√(LC) (39.14): Z = R, i₀ maximal; sharper for small R; used in tuning.
- 4LC circuit: energy oscillates between C and L at 1/2π√(LC).
Check yourself
Keep exploring
Hand-picked videos, simulations and readings that explain this idea from a different angle.