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Electricity and Magnetism

Quality Factor (Q-factor) in Alternating Current Circuits

The quality factor (Q-factor) is a dimensionless parameter that characterizes the resonance sharpness and energy storage capacity of an underdamped resonant electrical circuit. It represents the ratio of maximum energy stored per cycle to the energy dissipated per radian.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Definition of Q-factor for series L-C-R circuits, dependence on circuit parameters (L, C, R), resonance frequency, and bandwidth relationships.

PGT extension

Derivation of Q-factor using potential drop magnification across reactive components at resonance, parallel LCR circuits, and energy-based definitions.

Prerequisites

Series L-C-R circuit resonance conditions
Impedance and phase angle in AC circuits
Power dissipation and RMS values in AC

Physical Definition and Significance

Q-factor is defined as 2π times the maximum energy stored divided by the energy dissipated per cycle.
It quantifies the sharpness of the resonance curve: a higher Q-factor implies a narrower resonance peak and lower damping.
In a radio receiver or tuner, a high Q-factor ensures high selectivity for a specific frequency.

Q-factor in a Series L-C-R Circuit

At resonance, inductive reactance equals capacitive reactance (ω_r$L = 1/(\omega_{r}C))$.
The Q-factor equals the ratio of resonant angular frequency to bandwidth (ω_r / Δω).
It can also be expressed as the ratio of voltage across either inductor or capacitor to the applied source voltage at resonance.

Bandwidth and Selectivity

Bandwidth (Δω) is defined as the frequency interval between the two half-power points where power drops to half its maximum value.
The relation between bandwidth, resonant frequency, and Q-factor is given by Δ$\omega = \omega_{r} / Q$.
Damping resistance R broadens the bandwidth and lowers the Q-factor.

Q-factor in a Parallel L-C-R Circuit

In a parallel resonant circuit, the Q-factor is given by the ratio of the parallel resistance to the inductive reactance at resonance.
It determines the magnification of circulating current in the parallel branch compared to the total source current.
High parallel resistance yields a high Q-factor and sharper rejection of off-resonance frequencies.

Derivations

Derivation of Q-factor for Series L-C-R Circuit using Voltage Magnification

Conditions: Series L-C-R circuit operating at angular resonance frequency ω_r

At resonance, inductive reactance X_L = ω_rL and capacitive reactance X_C = 1/(ω_rC) are equal, and circuit impedance is minimum: Z = R.
The maximum voltage across the inductor at resonance is V_L = I_max X_L = (V_rms / R) ω_rL.
The Q-factor is defined as the voltage magnification: $Q = V_{L} / V_{rms} = (\omega_{r}L) / R$.

Result: $Q = \frac{\omega_{r}L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$

Derivation of Q-factor from Bandwidth and Resonance Frequency

Conditions: Series L-C-R circuit near resonance

The half-power frequencies occur where the total impedance $Z = R\sqrt{}2$, making the current drop to I_max / &sqrt;2.
The difference between the two half-power frequencies defines the bandwidth: Δ$\omega = \omega_{2} - \omega_{1}$.
Solving the resonant circuit equations yields the half-power bandwidth as Δ$\omega = R / L$.
Substituting ω_r = 1 / &sqrt;LC and Δ$\omega = R / L gives the ratio Q = \omega_{r} / \Delta\omega$.

Result: $Q = \frac{\omega_{r}}{\Delta\omega} = \frac{\omega_{r}L}{R}$

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Q-factor (Fundamental Energy Definition)	$$Q = 2\pi \frac{E_{stored}}{E_{dissipated \, per \, cycle}}$$	$E_stored = maximum energy stored$, $E_dissipated = energy lost per cycle$ Dimensionless	Underdamped harmonic oscillation
Q-factor for Series L-C-R Circuit (Reactance form)	$$Q = \frac{\omega_{r}L}{R} = \frac{1}{\omega_{r}CR}$$	ω_r = resonant frequency, L = inductance, C = capacitance, R = resistance Dimensionless	Series resonance
Q-factor in terms of Circuit Parameters	$$Q = \frac{1}{R}\sqrt{\frac{L}{C}}$$	L = inductance, C = capacitance, R = resistance Dimensionless	Series L-C-R circuit
Q-factor and Bandwidth Relation	$$Q = \frac{\omega_{r}}{\Delta\omega} = \frac{f_{r}}{\Delta f}$$	ω_r = resonant frequency, Δ$\omega = bandwidth between half-power points$ Dimensionless	Narrowband approximation (Q > 5)
Voltage Magnification at Resonance	$$V_{L} = V_{C} = Q \cdot V_{source}$$	V_L, V_C = voltages across L and C, V_source = applied source voltage Volts (V)	Series resonance
Bandwidth Formula	$$\Delta\omega = \frac{R}{L}$$	R = resistance, L = inductance rad s^-1	Series L-C-R circuit
Q-factor for Parallel L-C-R Circuit	$$Q = R_{p}\sqrt{\frac{C}{L}} = \frac{R_{p}}{\omega_{r}L}$$	R_p = parallel equivalent resistance, L = inductance, C = capacitance Dimensionless	Parallel resonance circuit
Half-power frequencies	$$\omega_{1,2} = \mp \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^{2} + \frac{1}{LC}}$$	ω₁ = lower cutoff frequency, ω₂ = upper cutoff frequency rad s^-1	Series L-C-R circuit

Graphs and experiments

Frequency Response Curve (Current vs Frequency) for Different Q-factors

Draw: A plot of rms current (I_rms) on the y-axis versus frequency (f) on the x-axis, showing multiple resonance peaks for different values of resistance R (high R gives a broad, flat peak; low R gives a sharp, high peak).

Interpret: Lower resistance R leads to a higher Q-factor, resulting in a sharper resonance peak and narrower bandwidth (Δf).

Exam use: Directly tested in conceptual MCQs asking how altering R, L, or C affects the sharpness of tuning.

Impedance vs Frequency Resonance Curve

Draw: A curve plotting total circuit impedance Z on the y-axis against frequency f on the x-axis, showing a distinct minimum at f_r equal to R.

Interpret: At resonance, impedance is purely resistive and attains its minimum value, causing current to reach its maximum.

Exam use: Used to evaluate understanding of impedance minima and phase angle changes in series resonance.

Phase Angle vs Frequency Curve

Draw: A plot of phase angle φ between voltage and current against frequency f, transitioning from negative values (capacitive) below f_r to zero at f_r, and positive values (inductive) above f_r.

Interpret: The slope of the phase curve at resonance is directly related to the Q-factor of the circuit.

Exam use: Tested in matching-type or assertion-reason questions concerning circuit behavior off-resonance.

Solved numericals

TGT core

A series L-C-R circuit has an inductance L = 2 H, capacitance $C = 32 \muF$, and resistance $R = 10 \Omega$. Calculate the Q-factor of the circuit at resonance.

Identify the given parameters: L = 2 H, $C = 32 \times 10^{-6} F$, $R = 10 \Omega$.
Use the formula for Q-factor: $Q = (1 / R) \times \sqrt{L / C}$.
Substitute the values: $Q = (1 / 10) \times \sqrt{}(2 / (32 \times 10^{-6}))$.
Simplify the term inside the square root: 2 / (32 × 10^-6) = 1 / (16 × 10^-6) = 62500.
Take the square root: &sqrt;62500 = 250.
Calculate Q: $Q = (1 / 10) \times 250 = 25$.

Answer: 25

PGT extension

A series resonant circuit has a resonant frequency of 100 kHz and a bandwidth of 2 kHz. If the inductance is 0.5 mH, find the resistance R and the capacitance C of the circuit.

Identify the given values: f_r = 100 kH$z = 10^{5} Hz$, Δ$f = 2 kHz = 2 \times 10^{3} Hz$, $L = 0.5 mH = 0.5 \times 10^{-3} H$.
Calculate the Q-factor using $Q = f_{r} / \Deltaf = 10^{5} / (2 \times 10^{3}) = 50$.
Use the bandwidth formula Δ$\omega = R / L$, where Δ$\omega = 2\pi \Deltaf$. Thus, $R = (2\pi \Deltaf) L$.
Calculate R: $R = 2 \times 3.1416 \times (2 \times 10^{3}) \times (0.5 \times 10^{-3}) = 6.283 \Omega$.
Use the Q-factor formula $Q = \omega_{r}L / R$ to find ω_r = 2πf_r = 2 × 3.1416 × 10⁵ = 6.283 × 10⁵ rad/s.
Alternatively, use resonant frequency formula ω_r = 1 / &sqrt;LC to find C: $C = 1 / (\omega_{r}^{2} L)$.
Calculate C: $C = 1 / ((6.283 \times 10^{5})^{2} \times 0.5 \times 10^{-3}) \approx 5.07 \times 10^{-9} F = 5.07 nF$.

Answer: $R = 6.28 \Omega$, C = 5.07 nF

PGT extension

In a series L-C-R circuit with L = 0.1 H, $C = 10 \muF$, and $R = 20 \Omega$, an alternating source of 20 V (rms) is applied at the resonant frequency. Find the voltage across the capacitor at resonance.

Calculate the resonant angular frequency: ω_r = 1 / &sqrt;L$C = 1 / \sqrt{0.1 \times 10 \times 10^{-6}} = 1 / \sqrt{}10^{-6} = 10^{3} rad/s$.
Calculate the Q-factor of the circuit: $Q = \omega_{r}L / R = (10^{3} \times 0.1) / 20 = 100 / 20 = 5$.
At resonance, the voltage across the capacitor is magnified by the Q-factor: V_C = Q × V_source.
Substitute the values: V_C = 5 × 20 V = 100 V.

Answer: 100 V

Assumptions and limitations

The formulas assume a linear, time-invariant lumped-parameter R-L-C circuit.
The approximation $Q = \omega_{r} / \Delta\omega$ is highly accurate for Q > 5 (narrowband condition).
Stray capacitances and inductances of connecting wires are neglected.

Common misconceptions

Common error: A higher Q-factor means lower voltage across reactive components at resonance.

Correct: A higher Q-factor means higher voltage magnification across the inductor and capacitor at resonance.

Q represents the amplification factor V_L / V_source = ω_rL / R; thus, larger Q gives larger reactive voltages.

Common error: Q-factor has units of frequency or ohms.

Correct: Q-factor is a completely dimensionless quantity.

It is a ratio of energies or reactances to resistance, making all units cancel out.

Common error: Increasing resistance R increases the Q-factor of a series LCR circuit.

Correct: Increasing resistance R decreases the Q-factor and broadens the resonance curve.

R is in the denominator of the Q formula ($Q = \omega_{r}L / R)$, so damping increases energy loss and reduces Q.

Quick revision

Q-factor is the ratio of resonant frequency to bandwidth: $Q = \omega_{r} / \Delta\omega$.
In series L-C-R, $Q = \omega_{r}L / R = (1 / R)\sqrt{L / C}$.
Q-factor measures voltage magnification across L or C at resonance: V_C = Q V.
Higher Q-factor implies sharper tuning and lower energy dissipation per radian.

Must remember

Q-factor is defined as $2\pi$ times the maximum energy stored divided by the energy dissipated per cycle.
In a series $L-C-R$ circuit, the Q-factor is given by $Q = \frac{\omega_{r}L}{R} = \frac{1}{\omega_{r}CR} = \frac{1}{R}\sqrt{\frac{L}{C}}$.
Q-factor represents voltage magnification at resonance: $V_{L} = V_{C} = Q \cdot V_{source}$.
Bandwidth is related to resonant frequency and Q-factor by $\Delta\omega = \frac{\omega_{r}}{Q} = \frac{R}{L}$.
Q-factor is a dimensionless quantity.
In a parallel $L-C-R$ circuit, the Q-factor is given by $Q = R_{p}\sqrt{\frac{C}{L}} = \frac{R_{p}}{\omega_{r}L}$.

Glossary

Q-factorA dimensionless parameter that measures the sharpness of resonance and energy storage relative to energy dissipation in a resonant circuit.

Bandwidth ($\Delta\omega$)The frequency interval between the two half-power points where power drops to half its maximum value.

Voltage MagnificationThe amplification of voltage across the inductor or capacitor relative to the applied source voltage at resonance, equal to $Q$.

SelectivityThe ability of a circuit to respond to a specific resonant frequency while rejecting all other frequencies, directly proportional to the Q-factor.

Damping Resistance ($R$)The resistance in an $L-C-R$ circuit that causes energy dissipation, broadening bandwidth and lowering the Q-factor.

Series vs Parallel L-C-R Circuit Q-Factor

Feature	Series L-C-R Circuit	Parallel L-C-R Circuit
Q-Factor Expression	$Q = \frac{\omega_{r}L}{R} = \frac{1}{R}\sqrt{\frac{C}{L}}$ (Wait, correct is $\frac{1}{R}\sqrt{\frac{L}{C}}$)	$Q = R_{p}\sqrt{\frac{C}{L}} = \frac{R_{p}}{\omega_{r}L}$
Resistance Role	Series resistance $R$ is in the denominator; smaller $R$ gives higher $Q$.	Parallel resistance $R_{p}$ is in the numerator; larger $R_{p}$ gives higher $Q$.
Magnification	Magnifies voltage across reactive components ($V_L = V_C = Q V_{source}$).	Magnifies circulating current in the parallel branch compared to source current.

Practice Quiz

Choose an option to reveal the explanation.

1. What is the fundamental definition of the Q-factor in alternating current circuits?

Explanation: Q-factor is fundamentally defined as $2\pi$ times the ratio of the maximum energy stored to the energy dissipated per cycle.

2. Which of the following units is appropriate for the Q-factor?

Explanation: Q-factor is a ratio of energies or reactances to resistance, making it completely dimensionless.

3. In a series $L-C-R$ circuit, what is the expression for the Q-factor in terms of circuit parameters $L$, $C$, and $R$?

Explanation: Using $\omega_r = 1/\sqrt{LC}$ in $Q = \frac{\omega_r L}{R}$, we get $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$.

4. How does an increase in resistance $R$ affect the Q-factor of a series $L-C-R$ circuit?

Explanation: Since $Q = \frac{\omega_r L}{R}$, increasing $R$ increases energy dissipation, which decreases the Q-factor and broadens the resonance curve.

5. What is the relationship between the resonant frequency $\omega_r$, bandwidth $\Delta\omega$, and Q-factor?

Explanation: The Q-factor is defined as the ratio of the resonant frequency to the bandwidth: $Q = \frac{\omega_r}{\Delta\omega}$.

6. At resonance in a series $L-C-R$ circuit, the voltage across the inductor $V_L$ is related to the source voltage $V_{source}$ by:

Explanation: Voltage magnification at resonance dictates that $V_L = V_C = Q \cdot V_{source}$.

7. What is the expression for the bandwidth ($\Delta\omega$) of a series $L-C-R$ circuit?

Explanation: Bandwidth in a series $L-C-R$ circuit is given by $\Delta\omega = \frac{\omega_r}{Q} = \frac{\omega_r}{({\omega_r L / R})} = \frac{R}{L}$.

8. What happens to the selectivity of a radio tuner circuit when its Q-factor is high?

Explanation: A higher Q-factor implies a narrower resonance peak, which provides higher selectivity for tuning into a specific frequency.

9. In a parallel $L-C-R$ resonance circuit, the Q-factor is given by:

Explanation: For a parallel $L-C-R$ circuit, the Q-factor is expressed as $Q = R_p \sqrt{\frac{C}{L}} = \frac{R_p}{\omega_r L}$.

10. If the inductance of a series $L-C-R$ circuit is quadrupled while keeping resistance and capacitance constant, how does the Q-factor change?

Explanation: Since $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$, $Q$ is proportional to $\sqrt{L}$. Multiplying $L$ by 4 multiplies $Q$ by $\sqrt{4} = 2$.

11. Which of the following correctly describes the half-power frequencies $\omega_1$ and $\omega_2$ relative to resonance frequency $\omega_r$ in a symmetrical approximation?

Explanation: The half-power frequencies are symmetrically located around the resonant frequency separated by bandwidth $\Delta\omega$, so $\omega_{1,2} \approx \omega_r \mp \frac{\Delta\omega}{2}$.

12. What does a low Q-factor signify regarding the damping of an oscillating circuit?

Explanation: A low Q-factor indicates high energy dissipation per cycle, which corresponds to heavy damping and a wide resonance curve.

13. In a parallel resonant circuit, a high parallel resistance ($R_p$) results in:

Explanation: Since $Q = R_p \sqrt{\frac{C}{L}}$, a high parallel resistance $R_p$ yields a high Q-factor and sharper rejection of unwanted frequencies.

14. Which circuit parameter determines the magnification of circulating current in a parallel $L-C-R$ circuit at resonance?

Explanation: The Q-factor in a parallel circuit determines the magnification of circulating current in the parallel branch compared to the total source current.

15. If $R$ is decreased to zero in an ideal series $L-C-R$ circuit, what happens to the Q-factor theoretically?

Explanation: Since $Q = \frac{\omega_r L}{R}$, as $R \to 0$, the Q-factor approaches infinity, representing an undamped resonant circuit.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. A series $L-C-R$ circuit has resistance $R = 10\;\Omega$, inductance $L = 0.1\;\text{H}$, and capacitance $C = 10\;\mu\text{F}$. What is the Q-factor of the circuit at resonance?

Explanation: Using $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{10}\sqrt{\frac{0.1}{10 \times 10^{-6}}} = \frac{1}{10}\sqrt{10000} = \frac{100}{10} = 10$ (Wait: $\sqrt{10000} = 100$; $100 / 10 = 10$. Let's recalculate: $\sqrt{0.1 / 10^{-5}} = \sqrt{10000} = 100$. $Q = 100 / 10 = 10$). Wait, option A is 10. Let's check correct_index for 10. Option index 0 is 10.

2. In a series resonant circuit, if the resonant frequency is $f_r = 1000\;\text{Hz}$ and the bandwidth $\Delta f = 50\;\text{Hz}$, the Q-factor of the circuit is:

Explanation: $Q = \frac{f_r}{\Delta f} = \frac{1000}{50} = 20$.

3. The potential drop across an inductor of inductance $2\text{ H}$ in a series $L-C-R$ circuit at resonance is $200\;\text{V}$. If the applied source voltage is $20\;\text{V}$, what is the Q-factor of the circuit?

Explanation: Voltage magnification at resonance is given by $V_L = Q \cdot V_{source}$. Thus, $Q = \frac{V_L}{V_{source}} = \frac{200}{20} = 10$.

4. If the bandwidth of a series $L-C-R$ circuit is $40\;\text{rad s}^{-1}$ and its inductance is $0.5\;\text{H}$, what is the resistance $R$ of the circuit?

Explanation: Bandwidth is given by $\Delta\omega = \frac{R}{L}$. Therefore, $R = \Delta\omega \cdot L = 40 \times 0.5 = 20\;\Omega$.

5. Which of the following statements about Q-factor and energy is correct?

Explanation: By definition, $Q = 2\pi \times \frac{E_{stored}}{E_{dissipated \, per \, cycle}}$.

6. In a parallel $L-C-R$ circuit, if $R_p = 1000\;\Omega$, $L = 10\;\text{mH}$, and $C = 100\;\text{pF}$, what is the expression for the Q-factor?

Explanation: For a parallel $L-C-R$ circuit, the Q-factor is given by $Q = \frac{R_p}{\omega_r L} = R_p \sqrt{\frac{C}{L}}$. Thus option A is correct.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. A series $L-C-R$ circuit has $L = 0.2\;\text{H}$, $C = 5\;\mu\text{F}$, and $R = 50\;\Omega$. What is the resonant angular frequency $\omega_r$?

Explanation: $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.2 \times 5 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-6}}} = 1000\;\text{rad s}^{-1}$.

2. Using the circuit parameters from the previous question ($L = 0.2\;\text{H}$, $C = 5\;\mu\text{F}$, $R = 50\;\Omega$), what is the Q-factor at resonance?

Explanation: $Q = \frac{1}{R}\sqrt{\frac{L}{C}} = \frac{1}{50}\sqrt{\frac{0.2}{5 \times 10^{-6}}} = \frac{1}{50}\sqrt{40000} = \frac{200}{50} = 4$.

3. What is the bandwidth of the circuit described in the preceding questions ($L = 0.2\;\text{H}$, $R = 50\;\Omega$)?

Explanation: Bandwidth $\Delta\omega = \frac{R}{L} = \frac{50}{0.2} = 250\;\text{rad s}^{-1}$.

4. If a source voltage of $10\;\text{V}$ (rms) is applied across the series $L-C-R$ circuit with $Q = 4$, what is the voltage across the capacitor at resonance?

Explanation: $V_C = Q \cdot V_{source} = 4 \times 10\;\text{V} = 40\;\text{V}$.

5. Which condition must hold for the narrowband approximation formula $Q = \frac{\omega_r}{\Delta\omega}$ to be highly accurate?

Explanation: The narrowband approximation assumes $Q > 5$, ensuring that the resonance peak is sharp and symmetrical.

6. If both inductance $L$ and capacitance $C$ of a series $L-C-R$ circuit are doubled while resistance $R$ remains unchanged, how does the Q-factor change?

Explanation: $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$. If $L$ and $C$ are both doubled, the ratio $\frac{L}{C}$ remains constant, so $Q$ is unchanged.

7. In terms of half-power frequencies $\omega_1$ and $\omega_2$, the Q-factor of a series resonance circuit can also be expressed as:

Explanation: Since bandwidth $\Delta\omega = \omega_2 - \omega_1$, $Q = \frac{\omega_r}{\Delta\omega} = \frac{\omega_r}{\omega_2 - \omega_1}$.

8. What is the primary consequence of having a very low Q-factor in an audio filter or tuning circuit?

Explanation: A low Q-factor means a wide bandwidth and poor selectivity, allowing unwanted neighboring frequencies to pass through.

9. Why does a parallel resonant circuit (tank circuit) exhibit maximum impedance at resonance?

Explanation: At parallel resonance, the reactive branch currents cancel out in phase, making the total circuit impedance purely resistive and maximum (or near-maximum), minimizing source current.

10. If a circuit designer wants to increase the Q-factor of a series $L-C-R$ tuner without changing the resonant frequency, what adjustment should be made?

Explanation: To increase $Q = \frac{\omega_r L}{R}$, one must decrease $R$ (reduce damping) while keeping $\omega_r = 1/\sqrt{LC}$ constant by adjusting $L$ and $C$ inversely.