In a series RLC circuit with Q = 50 driven by a 1V source at its resonant frequency, what is the approximate voltage across the capacitor?
A0V — the capacitive and inductive reactances cancel, so no voltage develops across either element
B1V — the source voltage appears entirely across the capacitor at resonance
C50V — the voltage across each reactive element is Q times the source voltage at resonance
D0.5V — the source voltage is split equally between the inductor and capacitor
At resonance, the current is maximum: I = V_s/R. The voltage across the capacitor is V_C = I · X_C = (V_s/R) · (1/ω₀C) = V_s · Q. With Q = 50 and V_s = 1V, V_C ≈ 50V. This is the voltage amplification property of resonance. The same magnitude appears across the inductor — both are Q times the source — but they are 180° out of phase and cancel in series, which is why the net impedance is only R. Students who think 'the reactances cancel so there's no voltage' confuse net voltage with individual element voltages.
Question 2 Multiple Choice
A radio designer wants to sharpen the frequency selectivity of a series RLC tuner — accepting only a very narrow band of frequencies near the station frequency and rejecting everything else. Which change accomplishes this?
AIncrease the resistance R to raise the quality factor Q
BDecrease the resistance R to raise the quality factor Q and narrow the bandwidth
CDecrease the inductance L, which shifts the resonant frequency and sharpens the peak
DIncrease the capacitance C, which reduces the bandwidth BW = R/L
Bandwidth BW = R/L = ω₀/Q. To narrow the bandwidth (higher selectivity), you need higher Q, and Q = ω₀L/R. For fixed L and resonant frequency, reducing R increases Q and decreases bandwidth. Increasing R does the opposite — it broadens the response and reduces selectivity. This is why low-loss components (high-Q coils with low resistance) are valued in tuned circuits: they give sharp, selective frequency responses.
Question 3 True / False
At the resonant frequency of a series RLC circuit, the voltage across the inductor and the voltage across the capacitor are equal in magnitude but 180° out of phase.
TTrue
FFalse
Answer: True
At resonance ω₀ = 1/√(LC), so X_L = ω₀L and X_C = 1/(ω₀C) are equal in magnitude (X_L = X_C). Both are driven by the same current I, so V_L = I·X_L and V_C = I·X_C are also equal in magnitude. However, V_L leads the current by 90° while V_C lags the current by 90° — making them 180° apart in phase. They cancel in series (their phasor sum is zero), so the total reactive voltage is zero and all source voltage appears across R. This is why impedance at resonance equals R.
Question 4 True / False
At resonance in a series RLC circuit, the voltages across the inductor and capacitor are both zero because the reactive elements are effectively 'short-circuited' by their cancellation.
TTrue
FFalse
Answer: False
The reactive elements do not become short circuits at resonance — they still carry the full current I = V_s/R and each develops a voltage of magnitude Q·V_s across them individually. What cancels is the *net* series voltage across the L-C combination (V_L + V_C = 0 as phasors), because they are equal and opposite. Each element individually has a large voltage — up to Q times the source — which is precisely the voltage amplification that makes resonance useful. Thinking 'cancellation means zero voltage' confuses series voltage addition with individual element voltages.
Question 5 Short Answer
Explain why the voltage across the capacitor in a series RLC circuit can be much larger than the source voltage at resonance, and what circuit parameter determines by how much.
Think about your answer, then reveal below.
Model answer: At resonance, the series impedance is minimized to Z = R, so current is at its maximum: I = V_s/R. The voltage across the capacitor is V_C = I · X_C = (V_s/R) · (1/ω₀C). This equals V_s · (1/(ω₀RC)) = V_s · Q, where Q = 1/(ω₀RC) = ω₀L/R is the quality factor. So V_C = Q · V_s. For a high-Q circuit (small R relative to the reactive impedances), Q can be large, and the capacitor voltage greatly exceeds the source. The same factor applies to the inductor. They don't violate energy conservation because they are 180° out of phase and cancel in series — energy sloshes between L and C with only R absorbing power from the source.
This Q-fold voltage amplification is why resonance is useful in practice: radio receivers use high-Q resonant circuits to amplify the tiny voltages of desired signals while rejecting interference. It also explains why high-voltage components must be specified carefully in resonant circuits — the capacitor and inductor must withstand Q times the source voltage, which can exceed component ratings even when the source is modest. The quality factor Q = ω₀L/R = 1/(ω₀CR) is the single most important characteristic of a resonant circuit, encoding both its selectivity (narrow vs. broad bandwidth) and its voltage amplification.