Series LCR Resonance and Voltage Magnification
A student assembles a series LCR circuit with R = 50 Ω, L = 2.0 H, and C = 10 μF connected to a 220 V, 50 Hz AC source. Expecting the current to be V/R = 4.4 A, the student is surprised when the milliammeter in the circuit reads only about 0.70 A. The student then varies the frequency of the source while keeping the voltage fixed. At one frequency the current rises sharply to a maximum, and the voltage across the capacitor measured by a high-impedance voltmeter is much larger than 220 V. The student worries that this violates Kirchhoff's voltage law. The teacher asks the student to calculate the reactances at 50 Hz, to find the resonant frequency and the current at resonance, and to explain how the capacitor voltage can exceed the source voltage without violating KVL.
1. Calculate X_L, X_C, the impedance Z, the current, and the phase angle at 50 Hz. Is the current lagging or leading the source voltage? Why is the student's 4.4 A estimate wrong?
Read the worked answer
ω = 2π(50) = 314 rad/s.
X_L = ωL = (314)(2.0) = 628 Ω.
X_C = 1/(ωC) = 1/[(314)(10×10⁻⁶)] = 318 Ω.
Net reactance = X_L − X_C = 310 Ω.
Z = √(R² + (X_L − X_C)²) = √(50² + 310²) ≈ 314 Ω.
I = V/Z = 220/314 ≈ 0.70 A.
tan φ = (X_L − X_C)/R = 310/50 = 6.2, so φ ≈ 81°. The current lags the voltage because the circuit is net inductive.
The student's 4.4 A would require Z = R = 50 Ω, which happens only at resonance where X_L = X_C. At 50 Hz they do not cancel.
2. Find the resonant frequency and the current at resonance. Explain why the current is maximum even though L and C are still present.
Read the worked answer
f₀ = 1/(2π√(LC)) = 1/[2π√(2.0×10×10⁻⁶)] = 35.6 Hz.
At resonance X_L = X_C, so Z = R = 50 Ω and I = 220/50 = 4.4 A.
L and C still store energy, but their voltages are equal in magnitude and opposite in phase at every instant. The reactive drops cancel, leaving only the resistor to limit the current. That is why the current is maximum at resonance.
3. At resonance, calculate the rms voltage across the capacitor and the quality factor Q. Explain why V_C can be larger than the source emf and why KVL is not violated.
Read the worked answer
At resonance, ω₀ = 2π(35.6) = 224 rad/s. X_C = 1/(ω₀C) = 1/[(224)(10×10⁻⁶)] ≈ 447 Ω. V_C = I X_C = (4.4)(447) ≈ 1.97×10³ V.
Q-factor = ω₀L/R = (224)(2.0)/50 ≈ 8.9; V_C ≈ QV = 8.9×220 ≈ 1.96×10³ V, consistent.
KVL is not violated because the voltages must be added as phasors. At resonance V_L and V_C are 180° out of phase, so V_L − V_C = 0. The source voltage equals V_R = 220 V. Each reactive voltage can be much larger than the source, but their phasor sum cancels; this is series resonance voltage magnification.