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Electricity and Magnetism
Series LCR Circuit in Alternating Current
A series LCR circuit consists of a resistor, an inductor, and a capacitor connected in series across an alternating voltage source. It exhibits electrical resonance at a specific frequency where the inductive and capacitive reactances cancel each other out.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Basic understanding of impedance, phase angle, resonant frequency, and maximum current condition in an AC series circuit containing R, L, and C components.
PGT extension
Detailed analytical treatment of sharpness of resonance, Q-factor, half-power frequencies, band-width, and phasor diagram analysis with complex impedance.
Prerequisites
Alternating current fundamentals (RMS and peak values)
Reactance of pure inductor and pure capacitor
Phasor algebra and vector addition
Impedance and Phasor Representation
The total opposition offered by a series LCR circuit to the flow of alternating current is called impedance (Z).
Impedance is represented as a complex number: $Z = R + j(X_L - X_C)$, where $X_L = \omega L$ and $X_C = 1/(\omega C)$.
The magnitude of impedance is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Phase Angle and Power Factor
The phase angle (phi) between the alternating voltage and current is given by ($\tanphi = (X_L - X_C) / R)$.
When X_L > X_C, the circuit is inductive, current lags voltage, and (phi) is positive.
When X_C > X_L, the circuit is capacitive, current leads voltage, and (phi) is negative.
Electrical Resonance
Resonance occurs 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.
The resonant angular frequency is (omeg$a_0 = 1 / \sqrt{LC})$.
Quality Factor and Sharpness of Resonance
The Quality factor (Q-factor) measures the sharpness of resonance and voltage magnification across L or C at resonance.
Q-factor is defined as the ratio of resonant frequency to bandwidth: $Q = \omega_0 / (Deltaomega) = (\omega_0 L) / R = 1 / (\omega_0 C R)$.
A higher Q-factor indicates a sharper resonance curve and a more selective circuit.
Derivations
Impedance and Phase Angle of Series LCR Circuit
Conditions: AC source $E = E_0 \sin(\omega t) connected across series combination of R$, L, and C.
Let the alternating current in the circuit be represented by $i = i_0 \sin(\omega t)$.
The potential difference across resistor R is $V_R = i_0 R \sin(\omega t) (in phase with current)$.
The potential difference across inductor L is $V_L = i_0 X_L \sin(\omega t + \pi/2) (leading current$ by 90 degrees).
The potential difference across capacitor C is $V_C = i_0 X_C \sin(\omega t - \pi/2) (lagging current$ by 90 degrees).
Using phasor addition, the total voltage $E_0 = \sqrt{V_{R0}^2 + (V_{L0} - V_{C0})^2} = \sqrt{(i_0 R)^2 + (i_0 X_L - i_0 X_C)^2}$.
Expressing impedance as $Z = E_0 / i_0$, we obtain $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Result: $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \text{}$ and } $\tanphi = \frac{\omega L - \frac{1}{\omega C}}{R}$
Resonant Frequency and Bandwidth
Conditions: Condition for resonance $X_L = X_C$ and half-power frequencies where power drops to half of maximum.
Draw: A bell-shaped curve plotting RMS current (I_rms) on the y-axis against frequency (f) on the x-axis, showing peaks at f_0 for different values of resistance R.
Interpret: Smaller resistance values yield sharper peaks with higher maximum current, whereas larger resistance values flatten the resonance curve.
Exam use: Frequently tested in conceptual questions regarding the effect of damping (resistance) on resonance sharpness.
Phasor Diagram of Series LCR Circuit
Draw: A vector diagram with current vector along the reference axis, V_R in phase, V_L leading by 90 degrees, and V_C lagging by 90 degrees.
Interpret: Demonstrates vector addition of voltages in an AC circuit to obtain total applied voltage vector.
Exam use: Used to derive impedance formula and determine phase relationships.
Impedance vs Frequency Graph
Draw: A hyperbolic-parabolic curve showing individual reactances X_L (linear) and X_C (inverse), and total impedance Z having a minimum value equal to R at resonance frequency f_0.
Interpret: Visually displays how total impedance drops to its minimum resistive component at the resonant frequency.
Exam use: Asked in matching and graphical identification questions in PGT physics exams.
Solved numericals
TGT core
A 100 Omega resistor, a 0.5 H inductor, and a 10 mu F capacitor are connected in series across a 200 V, 50 Hz AC source. Calculate the impedance of the circuit.
Calculate inductive reactance: $X_L = 2pi f L = 2 \times 3.1416 \times 50 \times 0.5 = 157.08 \omega$.
Find the resonant frequency and the Q-factor of a series LCR circuit containing an inductor of 2 H, a capacitor of 32 mu F, and a resistance of 10 Omega.
Calculate Q-factor: $Q = \omega_0 L / R = (125 \times 2) / 10 = 25$.
Answer: Resonant frequency = 19.89 Hz, Q-factor = 25
PGT extension
A series LCR circuit has R = 20 Omega, L = 0.4 H, and C = 100 mu F. Determine the bandwidth of the resonance curve.
Recall the formula for bandwidth in a series LCR circuit: Deltaomeg$a = R / L$.
Substitute the given values: R = 20 Omega and L = 0.4 H.
Deltaomeg$a = 20 / 0.4 = 50 rad/s$.
Answer: 50 rad/s
Assumptions and limitations
Assumes ideal lumped circuit elements where resistance, inductance, and capacitance are localized and independent of frequency (skin effect and dielectric losses are neglected).
Common misconceptions
Common error: At resonance, the voltage across the inductor and capacitor individually is zero.
Correct: At resonance, the net reactive voltage (V_L - V_C) is zero, but individual voltages V_L and V_C can be very high and equal in magnitude.
Individual reactances cancel each other out in series due to opposite phase angles, but finite current still produces potential drops across L and C.
Common error: The resonant frequency of a series LCR circuit depends on the resistance R.
Correct: The resonant frequency depends only on inductance L and capacitance C (omeg$a_0 = 1/\sqrt{LC})$.
Resistance only affects the damping and the sharpness of the resonance curve (bandwidth), not the frequency at which $X_L = X_C$.
Quick revision
At series resonance, $X_L = X_C$, impedance is minimum (Z = R), current is maximum, and power factor is unity.
Inductive Reactance: $X_L = \omega L = 2pi f L$
Capacitive Reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2pi f C}$
Must remember
The total opposition offered by a series LCR circuit to alternating current is called impedance, given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Electrical resonance occurs when inductive reactance equals capacitive reactance ($X_L = X_C$), making impedance minimum ($Z = R$) and current maximum.
The resonant angular frequency of a series LCR circuit is $\omega_0 = \frac{1}{\sqrt{LC}}$ and is independent of circuit resistance $R$.
The phase angle $\phi$ between voltage and current is given by $\tan\phi = \frac{X_L - X_C}{R}$. When $X_L > X_C$, current lags voltage; when $X_C > X_L$, current leads voltage.
The Quality factor ($Q$-factor) measures the sharpness of resonance and is defined as $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$.
Glossary
Impedance (Z)The total effective resistance offered by an AC circuit containing resistance, inductance, and capacitance, measured in ohms ($Omega$).
Inductive Reactance ($X_L$)The opposition offered by an inductor to alternating current, given by $X_L = \omega L = 2\pi f L$.
Capacitive Reactance ($X_C$)The opposition offered by a capacitor to alternating current, given by $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
Resonant Frequency ($f_0$)The frequency at which inductive reactance equals capacitive reactance, resulting in maximum current amplitude.
Quality Factor (Q-factor)A dimensionless parameter that characterizes the resonance sharpness and voltage magnification of a resonant circuit.
Inductive vs Capacitive Behavior in Series LCR Circuit
Condition
Reactance Relation
Phase Angle ($\phi$)
Power Factor ($\cos\phi$)
Inductive Circuit
$X_L > X_C$
Positive ($\phi > 0$), current lags voltage
Lagging power factor
Resonant Circuit
$X_L = X_C$
Zero ($\phi = 0$), voltage and current in phase
Unity ($\cos\phi = 1$)
Capacitive Circuit
$X_C > X_L$
Negative ($\phi < 0$), current leads voltage
Leading power factor
Practice Quiz
Choose an option to reveal the explanation.
1. What is the SI unit of electrical impedance ($Z$) in a series LCR circuit?
Explanation: Impedance is the total opposition to alternating current expressed in ohms ($\omega$), analogous to resistance in DC circuits.
2. The formula for the impedance of a series LCR circuit connected to an AC source of angular frequency $\omega$ is:
Explanation: Impedance is the phasor sum of resistance $R$ and net reactance $(X_L - X_C)$, giving $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
3. At electrical resonance in a series LCR circuit, the net reactance of the circuit is:
Explanation: At resonance, inductive reactance equals capacitive reactance ($X_L = X_C$), so the net reactance $X_L - X_C = 0$.
4. What is the resonant frequency formula for a series LCR circuit in terms of inductance $L$ and capacitance $C$?
Explanation: Equating $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$ yields angular frequency $\omega_0 = \frac{1}{\sqrt{LC}}$, giving frequency $f_0 = \frac{1}{2\pi\sqrt{LC}}$.
5. When $X_L > X_C$ in a series LCR circuit, what is the nature of the phase angle $\phi$ and current behavior?
Explanation: When $X_L > X_C$, the circuit is inductive, $\tan\phi = \frac{X_L - X_C}{R}$ is positive, meaning current lags the alternating voltage.
6. How does changing the resistance $R$ affect the resonant frequency $f_0$ of a series LCR circuit?
Explanation: The resonant frequency depends strictly on $L$ and $C$ ($\omega_0 = \frac{1}{\sqrt{LC}}$). Resistance only alters damping and curve sharpness (bandwidth).
7. The Quality factor ($Q$-factor) of a series LCR circuit is expressed as:
Explanation: The $Q$-factor is defined as the ratio of inductive reactance at resonance to resistance, $Q = \frac{\omega_0 L}{R}$.
8. What happens to the current in a series LCR circuit at resonance when connected to a constant AC voltage source?
Explanation: At resonance, impedance is minimum ($Z = R$), which causes the alternating current to reach its maximum possible value ($I_{max} = \frac{V}{R}$).
9. What is the expression for the average power consumed in a series LCR circuit?
Explanation: Average power in an AC circuit is given by $P_{avg} = V_{rms} I_{rms} \cos\phi$, where $\cos\phi$ is the power factor.
10. What is the bandwidth $\Delta \omega$ of a series LCR circuit in terms of circuit parameters $R$ and $L$?
Explanation: The bandwidth for a series resonant circuit is given by $\Delta \omega = \frac{R}{L}$, indicating that higher resistance widens the resonance curve.
11. At resonance in a series LCR circuit, the individual voltages across the inductor ($V_L$) and capacitor ($V_C$) are:
Explanation: At resonance $X_L = X_C$, so $V_L = I X_L$ equals $V_C = I X_C$ in magnitude, but they are in phase opposition ($180^{\circ}$ apart), cancelling each other's net voltage drop.
12. What is the power factor of a series LCR circuit at resonance?
Explanation: At resonance, the phase angle $\phi = 0$, so the power factor is $\cos(0^{\circ}) = 1$ (unity power factor).
13. If the capacitance $C$ of a series LCR circuit is quadrupled (4C), what happens to the resonant frequency $f_0$?
Explanation: Since $f_0 \propto \frac{1}{\sqrt{C}}$, replacing $C$ with $4C$ scales the frequency by $\frac{1}{\sqrt{4}} = \frac{1}{2}$ (halved).
14. A higher Quality ($Q$) factor in a series LCR circuit signifies:
Explanation: A high $Q$-factor indicates a sharp resonance curve with narrow bandwidth, making the circuit highly selective to the resonant frequency.
15. What is the unit of inductive reactance ($X_L$)?
Explanation: Reactance is a measure of opposition to AC and is dimensionally equivalent to resistance, measured in ohms ($\omega$).
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A series LCR circuit has resistance $R = 10\ \Omega$, inductance $L = 2\ \text{H}$, and capacitance $C = 32\ \mu\text{F}$. What is the resonant angular frequency $\omega_0$?
2. In a series LCR circuit, the potential difference across resistance $R$ is $80\ \text{V}$, across inductor $L$ is $60\ \text{V}$, and across capacitor $C$ is $120\ \text{V}$. What is the total applied RMS voltage?
3. A circuit contains an inductor of inductance $L$ and a capacitor of capacitance $C$ in series with a resistor $R$. If the angular frequency of the source is $\omega$, the condition for maximum current is:
Explanation: Maximum current occurs at resonance, which requires $X_L = X_C$, meaning $\omega L = \frac{1}{\omega C}$.
4. What is the $Q$-factor of a series LCR circuit with $L = 0.5\ \text{H}$, $C = 8\ \mu\text{F}$, and $R = 10\ \Omega$ at resonance?
5. The phase difference between alternating current and voltage in a series LCR circuit at resonance is:
Explanation: At resonance, $X_L = X_C$, making the impedance purely resistive ($Z = R$). Thus, voltage and current are completely in phase with a phase difference of zero.
6. If the RMS current in a series LCR circuit at resonance is $2\text{ A}$ and the resistance is $50\ \Omega$, what is the average power dissipated in the circuit?
Explanation: At resonance, power factor $\cos\phi = 1$. Average power $P_{avg} = I_{rms}^2 R = (2)^2 \times 50 = 4 \times 50 = 200\ \text{W}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following components determines the bandwidth of a series LCR circuit resonance curve?
Explanation: Bandwidth is given by $\Delta \omega = \frac{R}{L}$, which depends directly on both resistance $R$ and inductance $L$.
2. In a series LCR circuit, if the frequency of the AC source is lower than the resonant frequency ($f < f_0$), the circuit behaves as:
Explanation: When frequency is lower than resonance, capacitive reactance $X_C = \frac{1}{2\pi f C}$ increases and exceeds inductive reactance $X_L$, making the circuit capacitive.
3. What is the phase angle $\phi$ when $X_L = 30\ \Omega$, $X_C = 10\ \Omega$, and $R = 20\ \Omega$ in a series LCR circuit?
4. What is the voltage magnification factor across an inductor or capacitor at resonance in a series LCR circuit?
Explanation: The voltage magnification at resonance across $L$ or $C$ relative to the source voltage is exactly equal to the $Q$-factor of the circuit ($V_L = Q \cdot V$).
5. Which statement is correct regarding the power factor of a series LCR circuit?
Explanation: Power factor is $\cos\phi$, where $\phi$ is the phase angle between voltage and current. Since $-90^{\circ} \le \phi \le 90^{\circ}$, $\cos\phi$ always lies between 0 and 1.
6. If both $L$ and $C$ of a series LCR circuit are doubled, the resonant frequency $f_0$ becomes:
Explanation: Since $f_0 = \frac{1}{2\pi\sqrt{LC}}$, replacing $L$ with $2L$ and $C$ with $2C$ changes product $LC$ to $4LC$, multiplying the frequency by $\frac{1}{\sqrt{4}} = \frac{1}{2}$.
7. An alternating voltage $V = V_0 \sin(\omega t)$ is applied across a series LCR circuit. At resonance, the current phasor and voltage phasor are:
Explanation: At resonance, there is no phase difference ($\phi = 0$), so current and voltage phasors point in the exact same direction.
8. What is the impedance of a series LCR circuit at resonance if $R = 25\ \Omega$, $L = 0.1\ \text{H}$, and $C = 40\ \mu\text{F}$?
Explanation: At resonance, net reactance is zero ($X_L = X_C$), so the impedance equals the resistance $R$, which is $25\ \Omega$.
9. What happens to the sharpness of resonance when the resistance $R$ of a series LCR circuit is decreased?
Explanation: Decreasing resistance $R$ reduces bandwidth ($\Delta\omega = R/L$) and increases the $Q$-factor, making the resonance curve sharper and more selective.
10. A series LCR circuit has $X_L = 50\ \Omega$, $X_C = 50\ \Omega$, and $R = 10\ \Omega$. If the RMS voltage is $200\ \text{V}$, what is the RMS current?
Explanation: Since $X_L = X_C$, the circuit is at resonance. Impedance $Z = R = 10\ \Omega$. RMS current $I_{rms} = \frac{V_{rms}}{R} = \frac{200}{10} = 20\text{ A}$.