A series RLC circuit has R = 10 Ω, L = 100 mH, C = 100 μF, and is driven by a 1V sinusoidal source at resonance. The quality factor Q ≈ 10. What is the approximate voltage across the capacitor at resonance?
A1 V — the capacitor voltage equals the source voltage at resonance
B0 V — the capacitor and inductor voltages cancel, so the capacitor contributes nothing
C10 V — the capacitor voltage is Q times the source voltage
D0.707 V — the half-power condition applies at resonance
In a series RLC circuit at resonance, the voltages across the inductor and capacitor are each Q times the source voltage — but they are equal in magnitude and opposite in phase, so they cancel in the total circuit voltage. With Q = 10 and V_source = 1V, V_C = V_L = QV_source = 10V. This voltage magnification is the subtlest and most counterintuitive result in resonance analysis. Option A reflects the common misconception that 'the voltages cancel, so they must each be 1V.' They cancel in *sum* but individually they are Q times larger. This is why high-Q resonant circuits in power systems can be dangerous.
Question 2 Multiple Choice
Below the resonant frequency, a series RLC circuit behaves like which type of load?
AInductive — current lags the source voltage
BCapacitive — current leads the source voltage
CPurely resistive — the reactive elements cancel exactly at all sub-resonant frequencies
DOpen circuit — no current flows below resonance
Below resonance (ω < ω₀), the capacitive reactance 1/(ωC) is large and dominates over the inductive reactance ωL, so the net reactance is capacitive (negative imaginary). In a capacitive circuit, current leads voltage. Above resonance (ω > ω₀), the inductive reactance dominates, the net reactance is inductive, and current lags voltage. At resonance, the reactances cancel exactly and the circuit is purely resistive, with current in phase with voltage and at maximum magnitude.
Question 3 True / False
In a series RLC circuit at resonance, the voltages across the inductor and capacitor individually exceed the source voltage by a factor equal to the quality factor Q.
TTrue
FFalse
Answer: True
This is the voltage magnification property of resonant circuits. At resonance, the current is I = V/R (maximum). The voltage across the inductor is V_L = IωL = (V/R)ω₀L = V × (ω₀L/R) = QV. Similarly, V_C = I/(ωC) = QV. Both V_L and V_C equal QV, but they are 180° out of phase with each other, so they cancel in the KVL loop. A high-Q series circuit is effectively an AC voltage amplifier for the reactive elements, even though the source 'sees' only the resistance R.
Question 4 True / False
In a parallel RLC circuit at resonance, the impedance is at a minimum and the circuit draws maximum current from the source.
TTrue
FFalse
Answer: False
This describes series resonance, not parallel resonance. The parallel circuit is the dual of the series circuit: at resonance, the parallel impedance is at a MAXIMUM (the admittance is minimum), and the circuit draws MINIMUM current from the source. Internally, large circulating currents flow between L and C — they exchange energy back and forth — but these cancel in the external circuit, so the source 'sees' only a high resistance load. This is why parallel resonant circuits are used as tank circuits in oscillators and as high-impedance notch elements in filter design.
Question 5 Short Answer
Explain the physical mechanism of resonance in an RLC circuit in terms of energy exchange between the inductor and capacitor.
Think about your answer, then reveal below.
Model answer: At resonance, the inductor and capacitor exchange energy back and forth at exactly the resonant frequency ω₀ = 1/√(LC). When the capacitor is fully charged, it begins to discharge through the inductor, building up a magnetic field. When the capacitor is discharged, the inductor's collapsing magnetic field drives current that recharges the capacitor with opposite polarity. This oscillation continues at the natural frequency set by L and C. At resonance, the energy stored in the electric field of the capacitor and the magnetic field of the inductor are equal on average. The only energy actually consumed comes from the resistance, which dissipates the circulating energy as heat.
The quality factor Q measures how much energy is stored relative to how much is lost per cycle: Q = 2π × (peak energy stored)/(energy dissipated per cycle). High Q means the circuit stores much more energy than it loses per oscillation — the energy sloshes back and forth many times before being significantly dissipated. This is why high-Q circuits have narrow bandwidth and sharp frequency selectivity: they 'remember' the resonant frequency strongly and reject off-resonance excitation.