An engineer doubles the resistance R in a series RLC circuit while keeping L and C unchanged. What happens to the resonant frequency ω₀ and the quality factor Q?
ABoth ω₀ and Q increase — more resistance creates a stronger resonance
Bω₀ stays the same (it depends only on L and C), but Q decreases — more resistance means broader, less selective resonance
Cω₀ decreases because higher resistance shifts the resonance peak to lower frequency
DQ increases because higher resistance stabilizes the energy exchange between L and C
The resonant frequency ω₀ = 1/√(LC) depends only on L and C — not on R. Doubling R leaves ω₀ unchanged. The quality factor Q = ω₀L/R = 1/(ω₀RC) is inversely proportional to R, so doubling R halves Q. A lower Q means a broader resonance peak: the circuit responds appreciably to a wider range of frequencies near ω₀ rather than being sharply selective. This is why low-resistance (high-Q) circuits are used for radio tuners where sharp frequency selectivity is essential.
Question 2 Multiple Choice
A student says: 'At resonance, the RLC circuit has zero impedance because the inductive and capacitive reactances cancel.' Why is this statement imprecise?
AIt is completely correct — at resonance, the total impedance is zero
BThe reactive parts cancel (X_L = X_C, so net reactive impedance is zero), but the resistive part R remains; total impedance Z = R, not zero. Current is maximum (I = V/R) but finite.
CAt resonance, inductive and capacitive reactances add rather than cancel
DThe statement is wrong because impedance has no reactive component in DC circuits
At resonance, X_L = X_C exactly, so the imaginary (reactive) part of the impedance cancels: Z = R + j(X_L − X_C) = R + j·0 = R. The impedance equals R — its minimum value — not zero (unless the circuit is ideal with R = 0). Current I = V₀/R is maximized but finite. Saying 'impedance is zero' confuses the cancellation of reactive components with elimination of all impedance. Only in a purely ideal, lossless LC circuit (R = 0) would Z reach zero, which is a mathematical idealization.
Question 3 True / False
Increasing the inductance L in a series RLC circuit while keeping R and C constant will lower the resonant frequency ω₀.
TTrue
FFalse
Answer: True
The resonant frequency is ω₀ = 1/√(LC). If L increases and C is constant, LC increases, so √(LC) increases, and ω₀ = 1/√(LC) decreases. Physically: a larger inductor stores more energy per unit current, slowing the energy exchange cycle between L and C. This is analogous to adding mass to a spring-mass oscillator (larger L ↔ larger m), which reduces the natural frequency ω₀ = √(k/m).
Question 4 True / False
A high-Q RLC circuit is well-suited as a wide-band amplifier because it responds strongly to a broad range of frequencies.
TTrue
FFalse
Answer: False
A high-Q circuit has a *narrow* bandwidth — it responds significantly only to frequencies very close to ω₀. This narrow selectivity makes it ideal for applications requiring discrimination between nearby frequencies (radio tuners, narrow-band filters), but exactly the opposite of a wideband amplifier. A low-Q circuit (achieved with larger R) has a broad, flat response and responds to a wider frequency range. High Q means high selectivity; low Q means wide bandwidth.
Question 5 Short Answer
Explain why the quality factor Q determines a radio tuner's ability to select one station while rejecting adjacent ones. What does a high Q mean physically in terms of the circuit's energy storage and dissipation?
Think about your answer, then reveal below.
Model answer: Q = ω₀L/R measures the ratio of energy stored per cycle to energy dissipated per cycle. A high-Q circuit (small R relative to ω₀L) stores much more energy per oscillation than it loses, so it continues to 'ring' for many cycles after excitation. In the frequency domain, this manifests as a very narrow resonance peak: only frequencies extremely close to ω₀ drive significant current, while frequencies even slightly off-resonance drive much less current. For a radio tuner, this means the circuit responds to the desired station's broadcast frequency but rejects adjacent stations broadcasting at nearby frequencies. Adjusting C changes ω₀ to match different station frequencies.
The bandwidth Δω = ω₀/Q, so higher Q directly means narrower bandwidth and sharper station selectivity. In engineering terms, Q is the figure of merit for frequency selectivity. The energy picture — high Q means energy sloshes between L and C for many cycles before R dissipates it — is why high-Q circuits 'ring' like a well-struck bell (high-Q mechanical analog) while low-Q circuits decay quickly.