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Electricity and Magnetism

Series and Parallel L-C-R Circuits in Alternating Current

An L-C-R circuit consists of a resistor, an inductor, and a capacitor connected in series or parallel across an alternating voltage source. It exhibits resonance phenomena governed by the interplay of resistance, inductive reactance, and capacitive reactance.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Series LCR circuit impedance, resonance frequency, phase angle, and basic power factor concepts.

PGT extension

Parallel LCR resonance, quality factor (Q-factor), sharpness of resonance, bandwidth, and vector/phasor analysis of complex AC networks.

Prerequisites

Alternating current fundamentals and phasor representation
Reactance of pure resistors, inductors, and capacitors
Complex numbers and vector addition for harmonic oscillations

Impedance and Phasor Diagram of Series L-C-R Circuit

The total opposition to AC flow in a series L-C-R circuit is called impedance (Z), combining resistance R and net reactance $X = X_{L} - X_{C}$.
Voltage leads current by a phase angle φ when X_L > X_C (inductive circuit) and lags when X_L < X_C (capacitive circuit).
The impedance is expressed vectorially as $Z = R + j(X_{L} - X_{C}) with magnitude Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.

Series Resonance and Resonance Frequency

Resonance occurs in a series L-C-R circuit when the inductive reactance equals the capacitive reactance (X_L = X_C).
At resonance, the net reactance is zero, impedance is minimum (Z = R), and current is maximum for a given voltage source.
The angular resonance frequency is given by omeg$a_0 = 1/\sqrt{LC}$, making the current strictly in phase with the applied voltage (φ = 0).

Quality Factor (Q-factor) and Selectivity

The Q-factor of a series resonant circuit measures the sharpness of resonance and voltage magnification across L or C relative to R.
It is defined as the ratio of resonant inductive reactance (or capacitive reactance) to the circuit resistance: $Q = \omega_0 L / R = 1 / (\omega_0 C R)$.
A high Q-factor signifies a narrow bandwidth and high selectivity, crucial in radio tuning circuits.

Parallel L-C-R (Rejector) Circuit

In a parallel L-C-R circuit, an inductor (with resistance r) and a capacitor are connected in parallel across an AC source.
At parallel resonance, the circuit offers maximum impedance to the source and draws minimum (or nearly zero) current.
The parallel resonant frequency is slightly different from the series case, given by omeg$a_0 = \sqrt{\frac{1}{LC} - \frac{r^2}{L^2}}$.

Derivations

Impedance and Phase Angle of Series L-C-R Circuit

Conditions: AC voltage $V = V_0 \sin(\omega t) applied across series combination of R$, L, and C

Write the potential drops across R, L, and C as $V_R = I R (in phase with I)$, $V_L = I X_L (leading I$ by 90°), and $V_C = I X_C (lagging I$ by 90°).
Apply Kirchhoff's voltage law using phasor addition: $V_0^2 = V_{R0}^2 + (V_{L0} - V_{C0})^2$.
Substitute $V_0 = I_0 Z$, $V_{R0} = I_0 R$, $V_{L0} = I_0 X_L$, and $V_{C0} = I_0 X_C into the equation$ and divide by I_0^2.
Obtain the impedance magnitude and phase angle using \tan(phi) = \frac{X_L - X_C}{R}.

Result: $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \text{}$ and } ph$i = \tan^{-1}\left(\frac{\omega L - \frac{1}{\omega C}}{R}\right)$

Bandwidth and Quality Factor of Series L-C-R Circuit

Conditions: Series resonant circuit driven by a constant voltage source of variable frequency

Define bandwidth (Delta omega) as the frequency interval between the two half-power points where current drops to 1/sqrt{2} of its maximum value.
Set the total impedance at half-power frequencies to Z = sqrt{2} R, meaning |omega L - \frac{1}{omega C}| = R.
Solve the resulting quadratic equations for the lower (omega_1) and upper (omega_2) cutoff frequencies: omega_2 - omeg$a_1 = \frac{R}{L}$.
Express the quality factor as $Q = \frac{\omega_0}{\delta \omega} = \frac{\omega_0 L}{R}$.

Result: $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R} \text{}$ and } Delta omega = \frac{R}{L}

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Inductive Reactance	$$X_L = \omega L = 2 \pi f L$$	f = frequency, L = inductance Ohm (Omega)	Valid for sinusoidal AC
Capacitive Reactance	$$X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$$	f = frequency, C = capacitance Ohm (Omega)	Valid for sinusoidal AC
Series Impedance	$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$	R = resistance, $X_L = inductive reactance$, $X_C = capacitive reactance$ Ohm (Omega)	Series LCR combination
Series Resonance Frequency	$$f_0 = \frac{1}{2 \pi \sqrt{LC}}$$	L = inductance, C = capacitance Hertz (Hz)	$X_L = X_C$
Power Factor	$$\cos(\phi) = \frac{R}{Z}$$	R = resistance, Z = impedance Dimensionless	Steady-state AC circuit
Quality Factor	$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$$	omeg$a_0 = resonant angular frequency$, L = inductance, C = capacitance, R = resistance Dimensionless	Series resonant circuit
Average Power Dissipation	$$P_{avg} = V_{rms} I_{rms} \cos(\phi)$$	$V_{rms} = root mean square voltage$, $I_{rms} = root mean square current$, cos(phi) = power factor Watt (W)	Sinusoidal AC source
Parallel Resonance Frequency	$$f_0 = \frac{1}{2 \pi}\sqrt{\frac{1}{LC} - \frac{r^2}{L^2}}$$	r = internal resistance of inductor, L = inductance, C = capacitance Hertz (Hz)	Parallel L-C-R circuit with lossy inductor

Graphs and experiments

Current vs Frequency Resonance Curve

Draw: A bell-shaped curve plotting root-mean-square current (I_{rms}) on the y-axis against AC frequency (f) on the x-axis for different values of resistance R.

Interpret: Shows a sharp peak at $f = f_0$. Lower resistance curves yield narrower, taller peaks (high Q-factor), while higher resistance curves flatten out.

Exam use: Frequently used in conceptual MCQs asking how resistance affects sharpness of resonance and peak current.

Phasor Diagram of Series L-C-R Circuit

Draw: A vector diagram showing current phasor along the reference axis, V_R in phase, V_L pointing 90° ahead, V_C pointing 90° behind, and resultant voltage V vectorially.

Interpret: Visualizes Kirchhoff's voltage law in alternating current and clearly defines the phase angle φ between voltage and current.

Exam use: Tested in numericals determining phase lag/lead and instantaneous voltage equations.

Impedance vs Frequency Curve

Draw: A hyperbolic-looking V-shaped curve plotting impedance Z on the y-axis versus frequency f on the x-axis, achieving a minimum value equal to R at f_0.

Interpret: Demonstrates that total impedance drops to its lowest value at resonance because reactive components cancel out completely.

Exam use: Tested in questions concerning minimum impedance and maximum current conditions in radio receivers.

Solved numericals

TGT core

A series L-C-R circuit contains a resistance of 10 Omega, an inductance of 0.2 H, and a capacitance of 50 muF connected to a 200 V, 50 Hz AC supply. Calculate the impedance of the circuit.

Calculate inductive reactance: $X_L = 2 \pi f L = 2 \times \pi \times 50 \times 0.2 = 62.8 \omega$.
Calculate capacitive reactance: $X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \times \pi \times 50 \times 50 \times 10^{-6}} = 63.66 \omega$.
Compute net reactance: $X = X_L - X_C = 62.8 - 63.66 = -0.86 \omega$.
Calculate total impedance: $Z = \sqrt{R^2 + X^2} = \sqrt{10^2 + (-0.86)^2} \approx \sqrt{100 + 0.74} = 10.04 \omega$.

Answer: 10.04 Omega

PGT extension

A series L-C-R circuit with L = 2 H, C = 32 muF, and R = 10 Omega is driven by a variable frequency source of peak voltage 100 V. Find the resonant frequency, the maximum current at resonance, and the quality factor of the circuit.

Calculate resonant angular frequency: omeg$a_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{2 \times 32 \times 10^{-6}}} = \frac{1}{\sqrt{64 \times 10^{-6}}} = \frac{1}{8 \times 10^{-3}} = 125 \text{ rad/s}$.
Convert to linear frequency: $f_0 = \frac{\omega_0}{2 \pi} = \frac{125}{2 \pi} \approx 19.89 \text{ Hz}$.
Calculate maximum current at resonance: I_{0,max} = \frac{V_0}{R} = \frac{100}{10} = 10 \text{ A}.
Calculate Q-factor: $Q = \frac{\omega_0 L}{R} = \frac{125 \times 2}{10} = 25$.

Answer: $f_0 = 19.89 Hz$, $I_{max} = 10 A$, Q = 25

PGT extension

In a parallel L-C-R circuit, L = 5 H, C = 2 muF, and the internal resistance of the inductor is r = 20 Omega. Determine the parallel resonance frequency and the dynamic impedance of the circuit at resonance.

Calculate parallel resonance angular frequency: omeg$a_0 = \sqrt{\frac{1}{LC} - \frac{r^2}{L^2}} = \sqrt{\frac{1}{5 \times 2 \times 10^{-6}} - \frac{20^2}{5^2}}$.
Evaluate terms under root: \frac{1}{10^{-5}} = 10^5 = 100,000 \text{ and } \frac{400}{25} = 16.
Find omega_0 = sqrt{100,000 - 16} approx sqrt{100,000} approx 316.22 \text{ rad/s}.
Calculate dynamic impedance at resonance: Z_d = \frac{L}{C r} = \frac{5}{(2 \times 10^{-6}) \times 20} = \frac{5}{4 \times 10^{-5}} = 125,000 Omega = 125 \text{ k}Omega.

Answer: $f_0 = 50.33 Hz$, $Z_d = 125 kOmega$

Assumptions and limitations

Components are assumed to be ideal (pure R, pure L, pure C) except where internal resistance of inductors is explicitly factored into PGT derivations.
The source voltage is assumed to be perfectly sinusoidal with single fixed frequency or smoothly variable frequency.
Stray capacitances and mutual inductances connecting circuit wiring are neglected in standard theoretical analysis.

Common misconceptions

Common error: At series resonance, the voltage across the inductor and capacitor individually is zero.

Correct: Individual voltages V_L and V_C are non-zero and can be very large, but they are equal in magnitude and 180° out of phase, thus canceling each other out in the total circuit sum.

Cancellation happens across the reactive pair (V_L - $V_C = 0)$, but each component still sustains a large potential drop equal to Q times the supply voltage.

Common error: The impedance of a series L-C-R circuit is simply the arithmetic sum of R, X_L, and X_C.

Correct: Impedance is the vector (phasor) sum $Z = \sqrt{R^2 + (X_L - X_C)^2}$, because reactive voltages maintain a 90° phase difference relative to resistive voltage.

Algebraic addition ignores the phase quadrature between resistive and reactive voltage drops.

Common error: Parallel resonance occurs at exactly the same frequency as series resonance (omeg$a_0 = 1/\sqrt{LC})$.

Correct: Parallel resonance frequency is affected by the ohmic resistance of the inductor, shifting the resonance condition slightly.

The exact parallel resonance formula includes a damping term dependent on the inductor's internal resistance r.

Quick revision

Series resonance condition: $X_L = X_C$, yielding minimum impedance Z = R and maximum current.
Resonant frequency formula: $f_0 = \frac{1}{2 \pi \sqrt{LC}}$.
Power factor equals cos(phi) = R/Z, becoming unity at resonance.
Q-factor represents voltage magnification and selectivity, given by $Q = \frac{\omega_0 L}{R}$.
Parallel LCR circuit acts as a rejector circuit offering maximum impedance at resonance.

Must remember

In a series L-C-R circuit, impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
At series resonance, $X_L = X_C$, impedance is minimum ($Z = R$), and current is maximum.
The angular resonance frequency for a series circuit is $\omega_0 = \frac{1}{\sqrt{LC}}$.
Quality factor $Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 C R}$ measures voltage magnification and resonance sharpness.
A parallel L-C-R circuit acts as a rejector circuit, offering maximum impedance at resonance.

Glossary

Impedance (Z)The total opposition offered by an AC circuit to the flow of alternating current, measured in Ohms ($Omega$).

Inductive Reactance ($X_L$)Opposition offered by an inductor to AC, given by $X_L = \omega L = 2 \pi f L$.

Capacitive Reactance ($X_C$)Opposition offered by a capacitor to AC, given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.

Quality Factor (Q-factor)A dimensionless parameter that characterizes the resonance sharpness and voltage magnification of a resonant circuit.

Power Factor ($\cos \phi$)The ratio of resistance to impedance in an AC circuit, representing the cosine of the phase angle between voltage and current.

Series vs Parallel L-C-R Resonance

Feature	Series Resonance	Parallel Resonance
Impedance at Resonance	Minimum ($Z = R$)	Maximum
Current at Resonance	Maximum	Minimum
Circuit Role	Acceptor circuit	Rejector circuit
Resonance Frequency	$f_0 = \frac{1}{2\pi\sqrt{LC}}$	$f_0 = \frac{1}{2\pi}\sqrt{\frac{1}{LC} - \frac{r^2}{L^2}}$

Practice Quiz

Choose an option to reveal the explanation.

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

Explanation: At resonance, inductive reactance equals capacitive reactance ($X_L = X_C$), making their net difference (reactance) zero.

2. The SI unit of impedance ($Z$) is:

Explanation: Impedance is a measure of total opposition to current flow, and its SI unit is the Ohm ($\Omega$).

3. In an inductive series L-C-R circuit where $X_L > X_C$, the phase angle $\phi$ indicates that:

Explanation: When $X_L > X_C$, the circuit is inductive, meaning the voltage leads the current by a phase angle $\phi$.

4. What is the formula for angular resonance frequency in a series L-C-R circuit?

Explanation: Equating $X_L = X_C$ gives $\omega L = \frac{1}{\omega C}$, leading to $\omega_0 = \frac{1}{\sqrt{LC}}$.

5. What does a high Q-factor signify in a resonant circuit?

Explanation: A high Q-factor means sharp resonance, narrow bandwidth, and high selectivity, which is ideal for radio tuning.

6. The power factor of a purely resistive circuit is:

Explanation: In a purely resistive circuit, voltage and current are in phase ($\phi = 0^{\circ}$), so $\cos(0^{\circ}) = 1$.

7. What happens to the current in a series L-C-R circuit when it is tuned to resonance?

Explanation: At series resonance, impedance is minimum ($Z = R$), which causes the current to reach its maximum possible value.

8. The average power dissipation in an AC circuit is given by:

Explanation: Average power in an AC circuit depends on the rms values of voltage and current and the power factor $\cos(\phi)$.

9. A parallel L-C-R circuit is commonly referred to as a:

Explanation: A parallel L-C-R circuit offers maximum impedance at resonance and blocks (rejects) current at the resonant frequency, making it a rejector circuit.

10. If the frequency of an AC source increases, the capacitive reactance ($X_C$) will:

Explanation: Capacitive reactance is given by $X_C = \frac{1}{2\pi f C}$, showing that it is inversely proportional to frequency $f$.

11. Which term accounts for the damping in the parallel resonance frequency formula containing a lossy inductor?

Explanation: The parallel resonance frequency is given by $\omega_0 = \sqrt{\frac{1}{LC} - \frac{r^2}{L^2}}$, where $\frac{r^2}{L^2}$ accounts for the internal resistance of the inductor.

12. In a series L-C-R circuit, the vector relation for impedance is written as:

Explanation: Impedance combines resistance and net reactance in quadrature, yielding $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

13. What is the unit of the Q-factor?

Explanation: The Q-factor is a ratio of reactances to resistance ($\frac{\omega_0 L}{R}$), making it dimensionless.

14. At series resonance, the individual voltages across the inductor ($V_L$) and capacitor ($V_C$) are:

Explanation: At resonance $V_L$ and $V_C$ are equal in magnitude but $180^{\circ}$ out of phase, thus canceling each other's terminal voltage contribution.

15. What is the expression for inductive reactance?

Explanation: Inductive reactance is directly proportional to angular frequency and inductance, expressed as $X_L = \omega L = 2\pi f L$.

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 $R = 10\ \Omega$, $L = 2\ \text{H}$, and $C = 2\ \mu\text{F}$ connected across a $200\ \text{V}$ variable frequency AC source. At resonance, the current in the circuit is:

Explanation: At resonance, impedance $Z = R = 10\ \Omega$. Therefore, current $I = \frac{V}{R} = \frac{200}{10} = 20\ \text{A}$.

2. In a series L-C-R circuit, the voltage across the resistance, inductor, and capacitor are $30\ \text{V}$, $60\ \text{V}$, and $20\ \text{V}$ respectively. The total source voltage is:

Explanation: Total source voltage $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{30^2 + (60 - 20)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\ \text{V}$.

3. The resonant frequency of a series L-C-R circuit is $f_0$. If both $L$ and $C$ are doubled, the new resonant frequency will be:

Explanation: Resonant frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$. If $L \to 2L$ and $C \to 2C$, product $LC$ becomes $4LC$, making the new frequency $\frac{f_0}{2}$.

4. What is the quality factor (Q-factor) of a coil with inductance $L = 0.5\ \text{H}$, resistance $R = 10\ \Omega$ at a resonant frequency of $100\ \text{rad/s}$?

Explanation: Using $Q = \frac{\omega_0 L}{R} = \frac{100 \times 0.5}{10} = \frac{50}{10} = 5$.

5. For a parallel L-C-R circuit to exhibit resonance, which condition must be satisfied regarding susceptances?

Explanation: In parallel AC circuits, resonance occurs when the imaginary parts (susceptances) cancel out, meaning inductive susceptance equals capacitive susceptance.

6. If the power factor of an AC circuit is $0.5$ and the RMS voltage is $200\ \text{V}$ with an RMS current of $2\ \text{A}$, the average power consumed is:

Explanation: Average power $P_{avg} = V_{rms} I_{rms} \cos(\phi) = 200 \times 2 \times 0.5 = 200\ \text{W}$.

Timed Topic Test

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

1. In a series L-C-R circuit, if the angular frequency $\omega$ is greater than the resonance angular frequency $\omega_0$, the circuit is:

Explanation: When $\omega > \omega_0$, $X_L > X_C$, making the inductive reactance dominant and the circuit inductive.

2. The bandwidth of a series resonant circuit is given by:

Explanation: The bandwidth of a series resonant circuit is $\Delta \omega = \frac{\omega_0}{Q} = \frac{R}{L}$.

3. Which of the following components causes a phase lag of current behind voltage by $90^{\circ}$ in an ideal AC circuit?

Explanation: In an ideal inductor, current lags voltage by $90^{\circ}$ due to self-induced back EMF.

4. At parallel resonance, the impedance of a parallel L-C-R circuit is:

Explanation: A parallel L-C-R circuit offers maximum impedance at resonance, restricting current flow from the source.

5. If capacitance $C$ in a series resonant circuit is decreased, the resonance frequency will:

Explanation: Since $f_0 = \frac{1}{2\pi\sqrt{LC}}$, decreasing $C$ reduces the denominator, causing the resonance frequency to increase.

6. The phase angle $\phi$ in a series L-C-R circuit is evaluated using which trigonometric ratio?

Explanation: From the impedance phasor triangle, $\tan(\phi) = \frac{X_L - X_C}{R}$.

7. Why is a parallel L-C-R circuit termed a 'rejector' circuit?

Explanation: Because it offers maximum impedance at resonance, it draws minimal current at the resonant frequency, effectively rejecting that frequency from the line current.

8. What is the peak value of voltage if the RMS voltage is $220\ \text{V}$?

Explanation: Peak voltage $V_0 = \sqrt{2} V_{rms} = 1.414 \times 220 \approx 311\ \text{V}$.

9. In a series L-C-R circuit operating at resonance, the power factor is equal to:

Explanation: At resonance, $X_L = X_C$, so impedance $Z = R$, leading to $\cos(\phi) = \frac{R}{Z} = 1$.

10. How does internal resistance ($r$) of an inductor affect parallel resonance?

Explanation: The presence of internal resistance $r$ introduces a damping factor $\frac{r^2}{L^2}$ which modifies the parallel resonant frequency formula.