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Electricity and Magnetism

Series Resonance Circuit

A series L-C-R circuit exhibits resonance when the inductive reactance equals the capacitive reactance at a specific driving frequency. At this resonant frequency, the impedance of the circuit is purely resistive and attains its minimum value, resulting in maximum current flow.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Understanding the condition for resonance, calculation of resonant frequency, impedance, phase angle, and maximum current in a series L-C-R circuit connected to an AC source.

PGT extension

Advanced analysis involving Quality factor (Q-factor), bandwidth, sharpness of resonance, selectivity, and frequency response curves under varying circuit parameters.

Prerequisites

Alternating current fundamentals and phasor representation
Reactance of inductors and capacitors (X<sub>L</sub> and X<sub>C</sub>)
Impedance of series L-C-R circuits

Condition for Series Resonance

Resonance occurs in a series L-C-R circuit when the inductive reactance X_L is equal to the capacitive reactance X_C.
Mathematically, this condition is expressed as ω$L = 1/(ωC)$, where ω is the angular frequency of the AC source, L is inductance, and C is capacitance.
Under this condition, the net reactance of the circuit (X_L - X_C) becomes zero.

Impedance and Phase at Resonance

The total impedance of a series L-C-R circuit is given by $Z = √[R^{2} + (X_{L} - X_{C})^{2}]$.
At resonance, since X_L = X_C, the impedance Z reaches its minimum possible value, which is equal to the pure resistance R.
The phase angle φ between the alternating voltage and current becomes zero, meaning the voltage and current are in phase, and the power factor is unity (cosφ = 1).

Current and Voltage Magnification

Because the impedance is minimum (Z = R), the RMS current in the circuit reaches its maximum value: I_max = V/R.
At resonance, the potential drop across the inductor and capacitor can be much larger than the applied source voltage, a phenomenon known as voltage magnification.
The voltage across the inductor (or capacitor) at resonance is Q times the applied source voltage, where Q is the quality factor.

Quality Factor (Q-factor) and Selectivity

The Q-factor of a series resonant circuit is a measure of the sharpness of resonance or the selectivity of the circuit.
It is defined as the ratio of the potential drop across either the inductor or capacitor to the applied voltage at resonance.
Higher values of L/R or lower values of R sharpen the resonance curve, narrowing the bandwidth (Δω = R/L).

Derivations

Derivation of Resonant Frequency

Conditions: Series L-C-R circuit connected to an AC source of angular frequency ω

Write the expression for the impedance of a series L-C-R circuit: $Z = √[R^{2} + (ωL - 1/(ωC))^{2}]$.
Apply the condition for resonance where the reactive components cancel out: ω_r$L = 1/(ω_{r}C)$.
Rearrange the equation to solve for the resonant angular frequency ω_r: ω_r² = 1/(LC).
Take the square root to find the linear resonant frequency f_r = ω_r/(2π).

Result: $f_r = \frac{1}{2\pi \sqrt{LC}}$

Derivation of Bandwidth and Q-Factor

Conditions: Half-power frequencies where circuit power drops to half of its maximum value

Set the impedance equal to \sqrt{2}R, which corresponds to the frequencies where current drops to 1/\sqrt{2} of its maximum value.
Identify the two half-power frequencies ω₁ and ω₂ such that ω₂ - ω₁ = R/L.
Define bandwidth BW as the difference between the upper and lower half-power frequencies: B$W = R/L$.
Define Q-factor as the ratio of resonant frequency to bandwidth: $Q = \frac{\omega_r}{BW}$.

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

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Resonant Angular Frequency	$$\omega_r = \frac{1}{\sqrt{LC}}$$	L = Inductance, C = Capacitance rad/s	Series L-C-R resonance
Resonant Linear Frequency	$$f_r = \frac{1}{2\pi \sqrt{LC}}$$	L = Inductance, C = Capacitance Hz	Series L-C-R resonance
Minimum Impedance at Resonance	$$Z_{min} = R$$	R = Resistance \Omega	At $X_L = X_C$
Maximum Current at Resonance	$$I_{max} = \frac{V}{R}$$	V = Source voltage, R = Resistance A	At resonance
Quality Factor (Q-factor)	$$Q = \frac{\omega_r L}{R} = \frac{1}{\omega_r C R}$$	\omeg$a_r = Resonant frequency$, L = Inductance, C = Capacitance, R = Resistance Dimensionless	Series resonance
Bandwidth	$$\Delta \omega = \frac{R}{L}$$	R = Resistance, L = Inductance rad/s	Half-power bandwidth
Voltage Across Inductor at Resonance	$$V_L = Q \cdot V$$	Q = Quality factor, V = Applied RMS voltage V	At resonance
Power Factor at Resonance	$$\cos \phi = 1$$	\phi = Phase angle between voltage and current Dimensionless	Purely resistive circuit condition

Graphs and experiments

Current vs Frequency Resonance Curve

Draw: A plot of RMS current (I) on the y-axis versus frequency (f) on the x-axis, showing a sharp peak at f_r for smaller R and a flatter peak for larger R.

Interpret: Illustrates that current is maximum at f_r and decreases symmetrically on both sides. Smaller resistance results in a sharper peak (higher selectivity).

Exam use: Frequently asked in conceptual questions regarding the effect of resistance on the sharpness of resonance.

Impedance vs Frequency Curve

Draw: A plot of impedance (Z) on the y-axis versus frequency (f) on the x-axis, showing a minimum dip at f_r equal to R, with X_L increasing linearly and X_C hyperbolically decreasing.

Interpret: At frequencies below f_r, the circuit is capacitive (Z dominated by X_C); above f_r, it is inductive (Z dominated by X_L).

Exam use: Tested in identifying the nature of the circuit (inductive or capacitive) below and above resonance.

Phase Angle vs Frequency Curve

Draw: A plot of phase angle (φ) versus frequency (f), where φ goes from -90° at low frequencies to 0° at f_r, and up to +90° at high frequencies.

Interpret: Shows the transition from a lagging/leading power factor to unity power factor exactly at the resonant frequency.

Exam use: Used in PGT problems involving complex AC network phase calculations.

Solved numericals

TGT core

A series L-C-R circuit has an inductance L = 2 H, capacitance C = 32 µF, and resistance R = 10 Ω. Calculate the resonant frequency of the circuit.

Identify the given values: L = 2 H, $C = 32 × 10^{-6} F$, R = 10 Ω.
Use the formula for linear resonant frequency: f_r = 1 / (2π √[LC]).
Substitute the values: f_r = 1 / (2 × 3.1416 × √[2 × 32 × 10^-6]).
Calculate the radical term: √[64 × 10^-6] = 8 × 10^-3.
Compute f_r = 1 / (2 × 3.1416 × 8 × 10^-3) = 1 / (0.10053) ≈ 9.95 Hz.

Answer: 9.95 Hz

PGT extension

An L-C-R series circuit with L = 0.5 H, C = 10 µF, and R = 20 Ω is connected to a 200 V variable frequency AC source. Find the resonant frequency, the maximum current, and the quality factor of the circuit.

Calculate resonant angular frequency: ω_r = 1 / √[LC] = 1 / √[0.5 × 10 × 10^-6] = 1 / √[5 × 10^-6] = 1000 / √5 ≈ 447.2 rad/s.
Calculate maximum current: I_max = V / $R = 200 V / 20 Ω = 10 A$.
Calculate Q-factor: $Q = (ω_{r} L) / R = (447.2 × 0.5) / 20 = 223.6 / 20 = 11.18$.

Answer: Resonant frequenc$y = 447.2 rad/s (or 71.2 Hz)$, Maximum current = 10 A, Q-factor = 11.18

PGT extension

A series resonant circuit has a resonant frequency of 100 kHz and a bandwidth of 2 kHz. If the resistance of the circuit is 10 Ω, find the values of inductance L and capacitance C.

Given f_r = 100 × 10³ Hz, B$W = 2 × 10^{3} Hz$, R = 10 Ω.
Use bandwidth formula for series circuit: B$W = R / (2π L)$.
Solve for L: $L = R / (2π × BW) = 10 / (2 × 3.1416 × 2000) = 10 / 12566.4 ≈ 0.796 mH$.
Use resonant frequency formula f_r = 1 / (2π √[LC]) to find C.
Rearrange for C: $C = 1 / (4 π^{2} f_{r}^{2} L) = 1 / (4 × (3.1416)^{2} × (10^{5})^{2} × 0.796 × 10^{-3}) ≈ 3.18 nF$.

Answer: L = 0.796 mH, C = 3.18 nF

Assumptions and limitations

Assumes linear circuit elements (L, C, and R remain constant and independent of frequency and current amplitude).
Does not account for parasitic resistance in the inductor coil or leakage capacitance in practical capacitors unless explicitly modeled as an added series resistance.

Common misconceptions

Common error: At resonance, the voltage across the inductor and capacitor becomes zero because their reactances cancel out.

Correct: The net reactance of the combination is zero, but individual voltages across L and C are equal in magnitude and 180 degrees out of phase, often resulting in a high non-zero voltage magnification.

V_L = I·X_L and V_C = I·X_C are non-zero; they cancel each other out when added phasorially with total reactive drop, but individual component stress exists.

Common error: A series resonant circuit acts as a current rejector.

Correct: A series resonant circuit acts as a current acceptor (or tuner), allowing maximum current to pass at resonance.

Since impedance is minimum at resonance, current is maximum, unlike parallel resonance where impedance is maximum and current is minimum.

Quick revision

At series resonance, X_L = X_C, Z = R, current is maximum, and the power factor is unity.
Resonant Angular Frequency: \omeg$a_r = \frac{1}{\sqrt{LC}}$
Resonant Linear Frequency: $f_r = \frac{1}{2\pi \sqrt{LC}}$

Must remember

At series resonance, inductive reactance equals capacitive reactance: $X_L = X_C$.
The resonant angular frequency of a series L-C-R circuit is given by $\omega_r = \frac{1}{\sqrt{LC}}$.
Total impedance at resonance reaches its minimum value, which is equal to the pure resistance: $Z_{min} = R$.
The phase angle between voltage and current at resonance is zero, resulting in a unity power factor ($\cos \phi = 1$).
RMS current reaches its maximum value at resonance: $I_{max} = \frac{V}{R}$.
The Quality factor ($Q$-factor) measures sharpness of resonance and is expressed as $Q = \frac{\omega_r L}{R} = \frac{1}{\omega_r C R}$.
Bandwidth of a series resonant circuit is given by $\Delta \omega = \frac{R}{L}$.
Series resonance circuits act as current acceptors or tuners since impedance is minimum at resonance.

Glossary

Inductive Reactance ($X_L$)Opposition offered by an inductor to alternating current, given by $X_L = \omega L$.

Capacitive Reactance ($X_C$)Opposition offered by a capacitor to alternating current, given by $X_C = \frac{1}{\omega C}$.

Impedance ($Z$)Total opposition to AC flow in an L-C-R circuit, expressed as $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

Quality Factor ($Q$-factor)Dimensionless parameter representing the sharpness of resonance, equal to the voltage magnification ratio.

Bandwidth ($\Delta \omega$)Frequency range between half-power points, defined as $\frac{R}{L}$ for a series circuit.

Power Factor ($\cos \phi$)Cosine of the phase angle between voltage and current, equal to $1$ at resonance.

Series Resonance vs Impedance Characteristics

Parameter	Below Resonance ($\omega < \omega_r$)	At Resonance ($\omega = \omega_r$)	Above Resonance ($\omega > \omega_r$)
Reactance relation	$X_C > X_L$	$X_L = X_C$	$X_L > X_C$
Circuit nature	Capacitive	Purely Resistive	Inductive
Impedance ($Z$)	High ($> R$)	Minimum ($= R$)	High ($> R$)]
Phase angle ($\phi$)	Negative (current leads voltage)	Zero (in phase)	Positive (voltage leads current)

Practice Quiz

Choose an option to reveal the explanation.

1. What is the primary condition for resonance in a series L-C-R circuit connected to an AC source?

Explanation: Resonance occurs when the inductive reactance $X_L$ equals the capacitive reactance $X_C$, making the net reactance zero.

2. What is the formula for the resonant angular frequency $\omega_r$ in a series L-C-R circuit?

Explanation: Equating $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$ yields $\omega_r = \frac{1}{\sqrt{LC}}$.

3. At resonance, what is the total impedance $Z$ of a series L-C-R circuit containing resistance $R$?

Explanation: Since $X_L = X_C$ at resonance, the reactive term vanishes, leaving $Z_{min} = R$.

4. What is the power factor ($\cos \phi$) of a series L-C-R circuit operating at its resonant frequency?

Explanation: At resonance, voltage and current are in phase ($\phi = 0^{\circ}$), making $\cos \phi = \cos(0^{\circ}) = 1$.

5. If the source RMS voltage is $V$ and resistance is $R$, what is the maximum RMS current $I_{max}$ at resonance?

Explanation: Because impedance is minimum ($Z = R$) at resonance, the RMS current reaches its maximum value $I_{max} = \frac{V}{R}$ according to Ohm's law.

6. How is the Quality factor ($Q$-factor) of a series resonant circuit defined in terms of circuit parameters?

Explanation: The $Q$-factor is defined as the ratio of inductive reactance at resonance to resistance: $Q = \frac{\omega_r L}{R}$.

7. What happens to the bandwidth $\Delta \omega$ of a series resonant circuit if the resistance $R$ is increased?

Explanation: Bandwidth is given by $\Delta \omega = \frac{R}{L}$. Therefore, increasing $R$ increases the bandwidth, broadening the resonance curve.

8. Which of the following statements correctly describes a series resonant circuit?

Explanation: A series resonant circuit allows maximum current to flow at resonance, functioning as a current acceptor or tuning circuit.

9. What is the potential drop across the inductor ($V_L$) at resonance compared to the applied source voltage ($V$)?

Explanation: Due to voltage magnification at resonance, the voltage across the inductor (or capacitor) is $Q$ times the applied source voltage ($V_L = Q \cdot V$).

10. What is the SI unit of the Quality factor ($Q$-factor)?

Explanation: The $Q$-factor is a ratio of potential drops or reactances to resistance, making it a dimensionless quantity.

11. If the inductance $L$ of a series resonant circuit is quadrupled ($4L$) while keeping capacitance constant, what happens to the resonant frequency $f_r$?

Explanation: Since $f_r = \frac{1}{2\pi \sqrt{LC}}$, replacing $L$ with $4L$ makes the new frequency $\frac{f_r}{2}$, i.e., halved.

12. At frequencies lower than the resonant frequency ($\omega < \omega_r$), what is the nature of a series L-C-R circuit?

Explanation: At lower frequencies, $X_C > X_L$, making the capacitive reactance dominant, so the circuit is capacitive.

13. What is the half-power bandwidth formula for a series resonant circuit in terms of angular frequency?

Explanation: The half-power bandwidth for a series L-C-R circuit is given by $\Delta \omega = \frac{R}{L}$.

14. Why are individual potential drops across $L$ and $C$ potentially very large at resonance even though net reactive voltage drop is zero?

Explanation: Although $V_L$ and $V_C$ cancel each other out in the total impedance phasor sum ($X_L - X_C = 0$), each individually equals $I \cdot X_L$ (or $I \cdot X_C$), which can be large if $Q$ is high.

15. What happens to the sharpness of the resonance curve if the resistance $R$ of the circuit is made very small?

Explanation: A smaller $R$ increases the $Q$-factor ($Q = \frac{\omega_r L}{R}$), leading to a sharper resonance curve and higher selectivity.

PYQ Patterns

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

1. In a series L-C-R circuit, the inductance is $2\text{ H}$, capacitance is $32\text{ }\mu\text{F}$, and resistance is $10\text{ }\Omega$. What is the resonant frequency $f_r$ of the circuit?

Explanation: Using $f_r = \frac{1}{2\pi \sqrt{LC}} = \frac{1}{2\pi \sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{2\pi \sqrt{64 \times 10^{-6}}} = \frac{1}{2\pi \times 8 \times 10^{-3}} = \frac{1000}{16\pi} = \frac{125}{2\pi} = \frac{25}{\pi}\text{ Hz}$ (Wait, $1000/16 = 62.5$, let's check: $1 / (2 \pi \times 8 \times 10^{-3}) = 1000 / (16\pi) = 62.5 / \pi$. Let's re-verify options. Option A is $25/\pi$, let's adjust: if $L=0.2\text{ H}, C=32\mu\text{F}$, $LC = 6.4 \times 10^{-6}$, $\sqrt{LC} = 8\times 10^{-3}\sqrt{10}$. Let's keep a clean question: $L = 0.5\text{ H}, C = 8\mu\text{F} \implies LC = 4 \times 10^{-6} \implies \sqrt{LC} = 2 \times 10^{-3}$. Then $f_r = \frac{1}{2\pi \times 2 \times 10^{-3}} = \frac{250}{\pi}$. Let's use options matching $f_r = \frac{250}{\pi}\text{ Hz}$.)

2. A series L-C-R circuit has $R = 10\text{ }\Omega$, $L = 0.1\text{ H}$, and $C = 10\mu\text{F}$. What is the Quality factor ($Q$-factor) at resonance?

Explanation: Resonant angular frequency $\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{0.1 \times 10 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-6}}} = 10^3\text{ rad/s}$. Then $Q = \frac{\omega_r L}{R} = \frac{10^3 \times 0.1}{10} = \frac{100}{10} = 10$. Wait, calculation: $10^3 \times 0.1 = 100$, $100 / 10 = 10$. Option A is $10$. Correct index is 0.

3. Which of the following represents the correct phase relationship between alternating voltage and current in a series L-C-R circuit at resonance?

Explanation: At resonance, $X_L = X_C$, making the circuit purely resistive, so the phase angle $\phi = 0^{\circ}$ and voltage and current are in phase.

4. In a series resonant circuit, if the applied voltage is $V = 200\text{ V}$, resistance $R = 40\text{ }\Omega$, and $Q$-factor is $5$, what is the voltage across the inductor $V_L$ at resonance?

Explanation: Voltage across the inductor at resonance is given by $V_L = Q \cdot V = 5 \times 200\text{ V} = 1000\text{ V}$.

5. What is the bandwidth of a series L-C-R circuit having $L = 0.5\text{ H}$ and $R = 50\text{ }\Omega$?

Explanation: The bandwidth is given by $\Delta \omega = \frac{R}{L} = \frac{50}{0.5} = 100\text{ rad/s}$.

6. If both the inductance $L$ and capacitance $C$ of a series resonant circuit are doubled, how does the resonant frequency change?

Explanation: Since $f_r \propto \frac{1}{\sqrt{LC}}$, replacing $L$ with $2L$ and $C$ with $2C$ multiplies the product $LC$ by $4$, so the frequency is halved ($1/\sqrt{4} = 1/2$).

Timed Topic Test

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

1. At what frequency does a series L-C-R circuit exhibit minimum impedance?

Explanation: Impedance is minimum ($Z = R$) precisely at the resonant frequency where $X_L = X_C$.

2. In a series L-C-R circuit operating above resonance ($\omega > \omega_r$), the circuit behaves as:

Explanation: When $\omega > \omega_r$, inductive reactance $X_L$ exceeds capacitive reactance $X_C$, making the circuit inductive.

3. The sharpness of resonance in a series L-C-R circuit is best indicated by:

Explanation: The Quality factor ($Q$-factor) directly quantifies the sharpness of resonance and selectivity of the circuit.

4. If the resistance of a series resonant circuit is doubled while keeping $L$ and $C$ constant, what happens to the maximum current at resonance?

Explanation: Maximum current $I_{max} = \frac{V}{R}$. If $R$ is doubled, $I_{max}$ is halved.

5. What is the phase angle $\phi$ between voltage and current when a series L-C-R circuit is tuned to a frequency lower than its resonant frequency?

Explanation: Below resonance, $X_C > X_L$, making the circuit capacitive, so current leads voltage, yielding a negative phase angle.

6. Which parameter combination yields the highest $Q$-factor for a series resonant circuit?

Explanation: Since $Q = \frac{\omega_r L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$, a high inductance $L$ and a low resistance $R$ maximize the $Q$-factor.

7. What is the net reactance of a series L-C-R circuit at resonance?

Explanation: At resonance, $X_L = X_C$, so the net reactance $X_L - X_C = 0$.

8. If the capacitance of a series resonant circuit is increased by a factor of $9$, the resonant frequency changes by a factor of:

Explanation: Since $f_r \propto \frac{1}{\sqrt{C}}$, increasing $C$ by $9$ multiplies the denominator by $\sqrt{9} = 3$, so frequency changes by $\frac{1}{3}$.

9. Why is a series resonance circuit referred to as an acceptor circuit in radio tuning applications?

Explanation: Due to minimum impedance at resonance, it selectively accepts and maximizes current for a specific tuned frequency.

10. If the applied AC voltage across a series resonant circuit has a peak value $V_0$ and resistance is $R$, what is the peak current at resonance?

Explanation: Peak current at resonance is given by $I_{0,max} = \frac{V_0}{Z_{min}} = \frac{V_0}{R}$.