Why Does Adding Resistance Lower Q in a Series RLC but Raise It in a Parallel One?

Studyloaf Team•October 10, 2026•Last verified October 10, 2026•7 min read

Because the resistor sees a different quantity in each circuit. In a series RLC the resistor carries the full resonant current, so more resistance means more power lost per cycle and Q = ω0L/R falls; in a parallel RLC it sits across the resonant voltage, so more resistance draws less current, wastes less power, and Q = R/(ω0L) rises. With L = 1 mH and C = 10 nF, a 10 Ω series resistor and a 10 kΩ parallel resistor both give Q = 31.6, and doubling each sends Q to 15.8 and 63.2 respectively.

What you'll learn

  • What Q measures, in terms of energy and of bandwidth
  • The series and parallel formulas, and the one picture that explains why they move in opposite directions
  • A worked example with one L and C used both ways, including the bandwidths
  • How a small series loss can be swapped for an equivalent large parallel resistor
  • The mistakes that come from mixing the two formulas up

What exactly is Q?

The quality factor compares how much energy a resonant circuit stores with how fast it loses it: Q = ω0 × (peak energy stored) ÷ (average power dissipated), evaluated at resonance. A high Q means energy sloshes between the inductor and the capacitor for many cycles before the resistor burns it off. For these second-order circuits the same number also sets the sharpness of the resonance: Q = f0 ÷ BW, where BW is the half-power (−3 dB) bandwidth. Both circuits resonate at ω0 = 1/√(LC), so with the same L and C they share a resonant frequency. Only the loss mechanism differs.

Conventions used here

The series circuit is driven by a voltage source and its response is the current; the parallel circuit is driven by a current source and its response is the voltage across the tank (equivalently its impedance). Bandwidth is the width between the two frequencies where the response falls to 1/√2 of its peak, which is half the peak power. Components are ideal: the only loss is the named R.

Why do the two formulas move in opposite directions?

At resonance in the series circuit, the inductor's and capacitor's reactances cancel, the whole source voltage appears across R, and the same current I flows through all three elements. The energy stored is set by that current, ½LI², while the power lost is ½I²R. Doubling R doubles the loss for the same stored energy, so Q = ω0L/R halves. In the parallel circuit the three elements share one voltage V. The stored energy is ½CV², while the power lost is ½V²/R. Doubling R halves the loss, so Q = R/(ω0L) doubles. A useful way to remember it: a series resistor of zero and a parallel resistor of infinity are both lossless, so both give infinite Q.

Worked example: one tank, two ways

A series RLC circuit and a parallel RLC circuit with the same L and CTop: a voltage source drives R, L and C in series, with R equal to 10 ohms, L 1 millihenry and C 10 nanofarads. Bottom: a current source drives R, L and C in parallel, with R equal to 10 kilohms and the same L and C. Both resonate at 50.33 kilohertz and both have Q of 31.6.Series: R = 10 Ω, Q = 31.6RLCParallel: R = 10 kΩ, Q = 31.6RLCboth: L = 1 mH, C = 10 nF, f0 = 50.33 kHz
The same L = 1 mH and C = 10 nF, with a 10 Ω resistor in series (top) or a 10 kΩ resistor in parallel (bottom). Both resonate at 50.33 kHz with Q = 31.6.
  • Resonant frequency: ω0 = 1/√(1 mH × 10 nF) = 316,228 rad/s, so f0 = 50.33 kHz.
  • The characteristic impedance √(L/C) = √(10⁵) = 316.2 Ω equals ω0L and 1/(ω0C) at resonance.
  • Series, R = 10 Ω: Q = 316.2 ÷ 10 = 31.6. Bandwidth = f0 ÷ Q = R/(2πL) = 1.59 kHz.
  • Parallel, R = 10 kΩ: Q = 10,000 ÷ 316.2 = 31.6. Bandwidth = 1/(2πRC) = 1.59 kHz.
  • Double each resistor. Series with 20 Ω: Q = 15.8 and the bandwidth widens to 3.18 kHz. Parallel with 20 kΩ: Q = 63.2 and the bandwidth narrows to 0.80 kHz.
Series RLC: doubling R halves QComputed current magnitude of the series RLC, normalised to its peak, against frequency from 44 to 56.6 kilohertz. With R equal to 10 ohms (solid) the half-power bandwidth is 1.59 kilohertz and Q is 31.6. With R equal to 20 ohms (dashed) the peak is twice as wide: bandwidth 3.18 kilohertz, Q 15.8.444750535600.250.5.7071frequency (kHz)|I| / Imaxsolid: 10 ΩQ = 31.6dashed: 20 ΩQ = 15.8
Series RLC current against frequency, computed exactly and normalised to the peak. Doubling R from 10 Ω to 20 Ω doubles the bandwidth and halves Q.
Parallel RLC: doubling R doubles QComputed impedance magnitude of the parallel RLC, normalised to its peak, against frequency from 44 to 56.6 kilohertz. With R equal to 10 kilohms (solid) the half-power bandwidth is 1.59 kilohertz and Q is 31.6. With R equal to 20 kilohms (dashed) the peak is twice as narrow: bandwidth 0.80 kilohertz, Q 63.2.444750535600.250.5.7071frequency (kHz)|Z| / Rsolid: 10 kΩQ = 31.6dashed: 20 kΩQ = 63.2
Parallel RLC impedance against frequency, computed exactly and normalised to R. Doubling R from 10 kΩ to 20 kΩ halves the bandwidth and doubles Q.

Measuring the −3 dB width on these computed curves gives 1.59, 3.18, 1.59 and 0.80 kHz, matching the formulas to within the sampling step. For a series RLC the bandwidth R/(2πL) is exact, not an approximation, and likewise 1/(2πRC) for the parallel circuit.

Why do 10 Ω in series and 10 kΩ in parallel give the same Q?

Near resonance a small series resistance Rs can be replaced by a large parallel resistance Rp = Rs(1 + Q²) without changing the circuit's behaviour at that frequency. With Rs = 10 Ω and Q = 31.6, that is 10 × (1 + 1000) = 10,010 Ω, almost exactly the 10 kΩ of the parallel example. So the two circuits above are really the same lossy tank described two ways. This conversion is how a real inductor's winding resistance, which is in series, gets folded into a parallel tank model.

Common mistakes

  • Using Q = ω0L/R for a parallel circuit. It gives 0.0316 instead of 31.6 for the 10 kΩ example.
  • Expecting a bigger parallel resistor to lower Q. It removes loss, so Q goes up.
  • Computing bandwidth as f0 × Q instead of f0 ÷ Q.
  • Forgetting that ω0 = 1/√(LC) does not depend on R in either circuit, so the resonant frequency stays put while Q changes.
  • Quoting bandwidth in rad/s and comparing it with f0 in hertz. R/L is in rad/s; divide by 2π for hertz.

Check your understanding

  • Q1. In the series circuit, R is reduced to 5 Ω. What are Q and the bandwidth? Answer: Q = 316.2 ÷ 5 = 63.2, and BW = 5/(2π × 1 mH) = 796 Hz.
  • Q2. In the parallel circuit, R is reduced to 5 kΩ. What is Q? Answer: Q = 5000 ÷ 316.2 = 15.8.
  • Q3. What resistor gives Q = 10 with this L and C, in series and in parallel? Answer: series R = 316.2 ÷ 10 = 31.6 Ω; parallel R = 10 × 316.2 = 3.16 kΩ.

Related

Official source

Last verified

The GATE 2027 EC syllabus lists sinusoidal steady state analysis and the time and frequency domain analysis of RL, RC and RLC circuits under Section 2, Networks, Signals and Systems (checked against the official PDF from IIT Madras on 10 October 2026). The Q values were checked from the formulas, from the energy definition and from the −3 dB width of the exact computed response.

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