Series Resonance: A Real 1390 V Across the Capacitor
A school laboratory has a series LCR circuit with R = 100 Ω, L = 2.0 H and C = 5.0 µF, connected to a variable-frequency 220 V (rms) supply. A student sets the frequency to 50 Hz, observes a current of about 2.2 A, and finds that the voltages across the inductor and across the capacitor are each about 1390 V - far above the 220 V source voltage. He concludes that the voltmeter must be faulty. The teacher points out that the circuit is close to resonance (f₀ ≈ 50.3 Hz) and asks him to calculate the Q-factor and the bandwidth, to explain the high voltages, and to decide whether the 400 V capacitor available in the store can be used. The resistance of the inductor coil is included in R. The group must also suggest how to reduce the Q-factor to about 2 if the high voltages are to be avoided, and what price is paid in terms of current and selectivity.
1. Calculate the resonant frequency, the Q-factor and the bandwidth, and explain why the voltages across L and C can greatly exceed the source voltage.
Read the worked answer
Resonant frequency: ω₀ = 1/√(LC) = 1/√(2.0 × 5.0 × 10⁻⁶) = 1/√(1.0 × 10⁻⁵) = 316.2 rad/s, f₀ = ω₀/2π = 50.3 Hz.
At 50 Hz the circuit is almost exactly at resonance. X_L = 2πfL = 2π(50)(2.0) = 628 Ω, X_C = 1/(2πfC) = 1/[2π(50)(5.0 × 10⁻⁶)] = 637 Ω. The small difference gives a net reactance of about 9 Ω, so Z = √(100² + 9²) ≈ 100.4 Ω and I = 220/100.4 ≈ 2.19 A - matching the observed 2.2 A.
Q-factor: Q = X_L/R = 628/100 = 6.28 (equivalently ω₀L/R ≈ 6.3).
Bandwidth: Δf = f₀/Q = 50.3/6.28 ≈ 8.0 Hz (or R/2πL = 100/(2π × 2.0) = 7.96 Hz).
Why the high voltages? At resonance X_L and X_C are large and equal. The same current flows through L, C and R, so V_L = IX_L ≈ 2.2 × 628 ≈ 1380 V, V_C = IX_C ≈ 2.2 × 637 ≈ 1400 V. V_L and V_C are 180° out of phase, so their phasor sum is zero; the source only supplies the voltage across R, IR ≈ 220 V. Individually, however, the reactive voltages are Q times the supply voltage. The voltmeter is not faulty; the circuit really does develop about 1390 V across each reactance.
2. Can the 400 V capacitor be used? What rating is needed, and what other precautions follow from the result?
Read the worked answer
No. At resonance the capacitor sees approximately V_C = QV_supply = 6.28 × 220 ≈ 1380 V. A 400 V capacitor is overstressed by a factor of more than three and is likely to puncture, sometimes with a violent failure because the stored energy is ½CV² = ½(5 × 10⁻⁶)(1380)² ≈ 4.8 J.
A suitable capacitor should be rated well above the expected voltage, for example 1.5 kV or 2 kV rms, with a safety margin for mains fluctuations and for the slight rise in V_C that occurs just below resonance. The inductor must also be insulated for the same voltage, and the wiring and terminals must be rated accordingly. The supply should be isolated and the circuit enclosed, since a resonant circuit can maintain high voltages even at modest source voltage.
The general lesson is that the source voltage is not the largest voltage in a resonant circuit. Component ratings must be chosen from the Q-multiplied voltage, not from the supply.
3. How can the Q-factor be reduced to about 2, and what are the consequences for current, bandwidth and power dissipation?
Read the worked answer
At resonance Q = X_L/R_total, so to obtain Q = 2 with X_L ≈ 628 Ω: R_total = X_L/Q = 628/2 = 314 Ω. Since the coil already contributes 100 Ω, add about 214 Ω in series (a 220 Ω, high-power resistor is the nearest standard value, giving Q ≈ 1.97).
Consequences at resonance:
- Current: I = V/R_total = 220/314 = 0.70 A, down from 2.2 A.
- Reactive voltages: V_L = V_C ≈ 0.70 × 628 = 440 V, much safer but still above the 400 V capacitor rating - a 630 V or 1 kV capacitor would still be prudent.
- Bandwidth: Δf = f₀/Q = 50.3/2 = 25 Hz, so the circuit is much less selective; it responds over a broad range of frequencies instead of a sharp peak.
- Power: at resonance P = I²R_total = (0.70)²(314) ≈ 154 W, which is large for a laboratory resistor and is wasted as heat.
So lowering Q by adding resistance does reduce the voltage magnification, but it also reduces the current, broadens the resonance and dissipates substantial power. A better engineering solution is often to redesign the circuit with a smaller L/C ratio or a lower supply voltage, rather than simply adding series resistance.