LCR Circuit, Resonance and Power in AC Circuits
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LCR Circuit, Resonance and Power in AC Circuits

notes with clean derivations, phasor diagrams, power concepts, resonance formulas, Q-factor and solved CBSE numerical problems.

Section 1: Series LCR Circuit

Series Combination of R, L and C

In a series LCR circuit, a resistor R, an inductor L and a capacitor C are connected one after another with an AC supply.

V = V0 sinωt
  • Resistance: opposition offered by resistor, measured in Ω.
  • Inductive reactance: XL = ωL.
  • Capacitive reactance: XC = 1/ωC.
  • Impedance: total opposition to AC, measured in Ω.
R L C AC: V = V₀ sinωt I

Impedance, Current and Phase Angle

Since all three elements are in series, the same current flows through R, L and C. The voltage across R is in phase with current, voltage across L leads current by 90°, and voltage across C lags current by 90°.

Net reactance = XL − XC
Z² = R² + (XL − XC)²
Z = √[R² + (XL − XC)²]
I = V/Z
I = I0 sin(ωt − φ)
tanφ = (XL − XC)/R
cosφ = R/Z
Physical meaning of phase angle: φ tells how much the current is shifted from the applied voltage. If XL > XC, the circuit is inductive and current lags. If XC > XL, the circuit is capacitive and current leads. If XL = XC, current and voltage are in phase.

Section 2: Phasor Diagrams

To analyze a series LCR circuit, draw the phasor diagram by taking current I along the positive X-axis. Then VR is along the positive X-axis, VL is along the positive Y-axis and VC is along the negative Y-axis. The applied voltage V is the vector resultant of VR and the net reactive voltage.

Case 1: XL > XC

Vᵣ, I Vₗ V꜀ V φ

Here VL > VC, so the resultant voltage lies above the X-axis. The circuit behaves inductively and current lags voltage.

tanφ = (XL − XC)/R Pavg = VrmsIrms cosφ

Case 2: XC > XL

Vᵣ, I Vₗ V꜀ V φ

Here VC > VL, so the resultant voltage lies below the X-axis. The circuit behaves capacitively and current leads voltage.

tanφ = (XC − XL)/R Pavg = VrmsIrms cosφ

Case 3: XL = XC

V = Vᵣ, I Vₗ V꜀ vertical components cancel

The vertical components cancel. Resultant voltage equals VR. The circuit is at resonance, current and voltage are in phase.

φ = 0, cosφ = 1 Pavg = VrmsIrms

Section 3: Power in AC Circuits

Average Power Derivation

V = V0 sinωt I = I0 sin(ωt − φ)
Pavg = (1/2π) ∫02π VI d(ωt)
Pavg = (1/2π) ∫02π V0I0 sinωt sin(ωt − φ) d(ωt)
Using 2 sinA sinB = cos(A − B) − cos(A + B)
sinωt sin(ωt − φ) = 1/2 [cosφ − cos(2ωt − φ)]
Pavg = V0I0/(4π) ∫02π [cosφ − cos(2ωt − φ)] d(ωt)
Average value of cos(2ωt − φ) over one complete cycle = 0
Pavg = V0I0 cosφ/2
Since Vrms = V0/√2 and Irms = I0/√2
Pavg = VrmsIrms cosφ

Active Power

P = VrmsIrms cosφ

Unit: Watt. It is the useful power actually consumed by the circuit.

Reactive Power

Q = VrmsIrms sinφ

Unit: VAR. It oscillates between the source and reactive elements.

Apparent Power

S = VrmsIrms

Unit: VA. It is the product of rms voltage and rms current.

Power Triangle

The power triangle gives the relation between active power, reactive power and apparent power.

S² = P² + Q²
Practical significance: A high power factor means more useful power is obtained for the same current. Industries improve power factor to reduce current losses, voltage drop and electricity penalties.
P = VI cosφ Q = VI sinφ S = VI φ

Section 4: Series Resonance

Definition and Condition

Resonance in a series LCR circuit is the condition at which inductive reactance becomes equal to capacitive reactance. The net reactance becomes zero and the circuit behaves like a pure resistance.

At resonance: XL = XC
ωL = 1/ωC
ω² = 1/LC
ω0 = 1/√LC
f0 = 1/(2π√LC)
  • Net reactance = 0
  • Impedance Z = R, minimum
  • Current Imax = V/R, maximum
  • φ = 0 and power factor = 1
f₀ I f maximum current

Section 5: Quality Factor

The quality factor measures the sharpness of resonance in a series LCR circuit. A larger Q means a sharper resonance curve and better selectivity.

Q = ω0L/R Q = 1/(ω0CR) Q = (1/R)√(L/C)
At resonance, voltage across L or C may become much larger than the source voltage. This is why Q-factor is important in radio tuning and filter circuits.

Solved CBSE Numerical Problems on LCR Circuit and Resonance

A 2 μF capacitor, 100 Ω resistor and 8 H inductor are connected in series with an AC source. Find the frequency for maximum current and the maximum current if peak emf is 200 V.

Given Data

C = 2 μF = 2 × 10−6 F, R = 100 Ω, L = 8 H, V0 = 200 V.

Formula Used
f0 = 1/(2π√LC), I0max = V0/R
Step-by-step Solution

Maximum current flows at resonance. √LC = √(8 × 2 × 10−6) = √(16 × 10−6) = 4 × 10−3.

f0 = 1/[2π × 4 × 10−3] = 39.8 Hz. This frequency is called the resonant frequency. At resonance, Z = R, so I0max = 200/100 = 2 A.

Final Answer: f0 = 39.8 Hz, resonant frequency; maximum peak current = 2 A.

Common Mistake: Using rms voltage with peak current formula.

Exam Tip: At maximum current in a series LCR circuit, immediately use XL = XC and Z = R.

A 40 Ω resistor, 3 mH inductor and 2 μF capacitor are connected in series to a 110 V, 5000 Hz AC source. Calculate the current.

Given Data

R = 40 Ω, L = 3 mH = 3 × 10−3 H, C = 2 μF, V = 110 V, f = 5000 Hz.

Formula Used
XL = 2πfL, XC = 1/(2πfC), Z = √[R² + (XL − XC)²]
Step-by-step Solution

XL = 2π × 5000 × 0.003 = 94.25 Ω. XC = 1/(2π × 5000 × 2 × 10−6) = 15.92 Ω.

Net reactance = 94.25 − 15.92 = 78.33 Ω. Z = √(40² + 78.33²) = 87.95 Ω. I = 110/87.95 = 1.25 A.

Final Answer: I = 1.25 A.

Common Mistake: Forgetting to convert mH and μF into SI units.

Exam Tip: First identify whether the circuit is inductive or capacitive from XL − XC.

A capacitor, 100 Ω resistor and inductor L = 4/π² H are connected across 200 V, 50 Hz. Find capacitance and current when current is in phase with voltage.

Given Data

R = 100 Ω, L = 4/π² H, V = 200 V, f = 50 Hz.

Formula Used
At resonance: C = 1/(ω²L), I = V/R
Step-by-step Solution

Current is in phase with voltage, so the circuit is at resonance. ω = 2πf = 100π rad s−1.

C = 1/[(100π)² × (4/π²)] = 1/40000 F = 25 × 10−6 F = 25 μF. I = 200/100 = 2 A.

Final Answer: C = 25 μF, I = 2 A.

Common Mistake: Treating “in phase” as zero current instead of resonance.

Exam Tip: In phase in series LCR means φ = 0, XL = XC.

A 50 μF capacitor, 0.05 H inductor and 48 Ω resistor are connected to ε = 310 sin314t. Find reactance, nature and phase angle.

Given Data

C = 50 μF, L = 0.05 H, R = 48 Ω, ω = 314 rad s−1.

Formula Used
XL = ωL, XC = 1/ωC, tanφ = |XL − XC|/R
Step-by-step Solution

XL = 314 × 0.05 = 15.7 Ω. XC = 1/(314 × 50 × 10−6) = 63.7 Ω.

Net reactance = XL − XC = −48 Ω, so the circuit is capacitive. tanφ = 48/48 = 1, hence φ = 45°.

Final Answer: Reactance = 48 Ω capacitive; phase angle = 45°, current leads voltage.

Common Mistake: Missing the negative sign of XL − XC.

Exam Tip: If XC is larger, write “current leads voltage.”

An LCR series circuit has L = 100 mH, C = 100 μF, R = 120 Ω and ε = 30 sin100t V. Find impedance, peak current and resonant frequency.

Given Data

L = 0.1 H, C = 100 μF = 10−4 F, R = 120 Ω, V0 = 30 V, ω = 100 rad s−1.

Formula Used
Z = √[R² + (ωL − 1/ωC)²], I0 = V0/Z, f0 = 1/(2π√LC)
Step-by-step Solution

XL = 100 × 0.1 = 10 Ω. XC = 1/(100 × 10−4) = 100 Ω.

Z = √[120² + (10 − 100)²] = √(14400 + 8100) = 150 Ω. I0 = 30/150 = 0.2 A.

f0 = 1/[2π√(0.1 × 10−4)] = 50.3 Hz ≈ 50 Hz.

Final Answer: Z = 150 Ω, peak current = 0.2 A, resonant frequency ≈ 50 Hz.

Common Mistake: Confusing source angular frequency 100 rad s−1 with resonant frequency.

Exam Tip: Use the given ω for impedance, but use L and C only for resonance frequency.

A 12 Ω resistance and inductance 0.05/π H are in series across 130 V, 50 Hz. Calculate current and voltages across R and L.

Given Data

R = 12 Ω, L = 0.05/π H, V = 130 V, f = 50 Hz.

Formula Used
XL = 2πfL, Z = √(R² + XL²), VR = IR, VL = IXL
Step-by-step Solution

XL = 2π × 50 × (0.05/π) = 5 Ω. Z = √(12² + 5²) = 13 Ω.

I = 130/13 = 10 A. VR = 10 × 12 = 120 V. VL = 10 × 5 = 50 V.

Final Answer: I = 10 A, VR = 120 V, VL = 50 V.

Common Mistake: Adding 120 V and 50 V algebraically to get source voltage.

Exam Tip: In AC, element voltages are phasors; use vector addition.

A capacitor, 5 Ω resistor and 50 mH inductor are in series with 100 V, 50 Hz. Voltage is in phase with current. Find C and impedance.

Given Data

R = 5 Ω, L = 50 mH = 0.05 H, f = 50 Hz.

Formula Used
C = 1/ω²L, Z = R at resonance
Step-by-step Solution

Since voltage is in phase with current, the circuit is at resonance. ω = 2πf = 100π rad s−1.

C = 1/[(100π)² × 0.05] = 2.02 × 10−4 F. At resonance, Z = R = 5 Ω.

Final Answer: C = 2.02 × 10−4 F, Z = 5 Ω.

Common Mistake: Using 100 V in the capacitance formula.

Exam Tip: Resonance capacitance depends on f and L, not on applied voltage.

In an AC circuit with constant supply voltage and variable frequency, for what frequency will voltage across resistance R be maximum?

Given Data

The supply voltage is constant and frequency is variable. The circuit is a series LCR circuit as shown in the referenced figure.

Formula Used
VR = IR; VR is maximum when I is maximum, i.e. at f0 = 1/(2π√LC)
Step-by-step Solution

For constant supply voltage, VR becomes maximum when current I becomes maximum. In a series LCR circuit, current is maximum at resonance, where XL = XC.

Substituting the L and C values from the figure in f0 = 1/(2π√LC) gives f0 = 500 Hz.

Final Answer: Frequency = 500 Hz.

Common Mistake: Maximizing VR independently of current.

Exam Tip: In a series circuit, maximum VR means maximum current, hence resonance.

A 50 Hz AC source is connected to a 50 mH inductor and a bulb. Find capacitance to be connected in series for maximum brightness.

Given Data

f = 50 Hz, L = 50 mH = 0.05 H.

Formula Used
For maximum brightness: XL = XC, C = 1/ω²L
Step-by-step Solution

The bulb glows brightest when current is maximum, which occurs at resonance. ω = 2πf = 100π rad s−1.

C = 1/[(100π)² × 0.05] = 2.02 × 10−4 F.

Final Answer: C = 2.02 × 10−4 F.

Common Mistake: Taking maximum brightness as maximum resistance.

Exam Tip: Bulb brightness depends on current; maximum current occurs at resonance.

A 200 km telegraph wire has capacitance 0.014 μF per km and carries AC of 50 kHz. Find series inductance for minimum impedance. Take π = √10.

Given Data

Total C = 200 × 0.014 μF = 2.8 μF = 2.8 × 10−6 F, f = 50 kHz = 5 × 104 Hz, π² = 10.

Formula Used
For minimum impedance: L = 1/ω²C, ω = 2πf
Step-by-step Solution

ω = 2π × 5 × 104 = 105π rad s−1. Therefore ω² = 1010π² = 1011.

L = 1/(1011 × 2.8 × 10−6) = 3.57 × 10−6 H = 0.00357 mH.

Final Answer: For the stated 50 kHz, L = 0.00357 mH. Note: the printed answer 0.357 mH would correspond to 5 kHz, not 50 kHz.

Common Mistake: Missing that capacitance is given per km and must be multiplied by 200 km.

Exam Tip: If a supplied answer differs by 100 times, recheck whether kHz has been copied correctly.

A series LCR circuit has L = 2.0 H, C = 32 μF, R = 10 Ω. Find resonant angular frequency and Q value.

Given Data

L = 2.0 H, C = 32 μF = 32 × 10−6 F, R = 10 Ω.

Formula Used
ω0 = 1/√LC, Q = ω0L/R
Step-by-step Solution

LC = 2 × 32 × 10−6 = 64 × 10−6. √LC = 8 × 10−3.

ω0 = 1/(8 × 10−3) = 125 rad s−1. Q = (125 × 2)/10 = 25.

Final Answer: For the stated values, ω0 = 125 rad s−1, Q = 25. Note: the printed answer 62.5 rad s−1, 12.5 is not obtained from L = 2.0 H and C = 32 μF.

Common Mistake: Taking √(64 × 10−6) as 16 × 10−3 instead of 8 × 10−3.

Exam Tip: Q can be checked quickly by Q = (1/R)√(L/C).

A series LCR circuit is connected to 200 V, 50 Hz. Voltages across R, C and L are 200 V, 250 V and 250 V. Explain the voltage paradox and find current if R = 40 Ω.

Given Data

V = 200 V, VR = 200 V, VC = 250 V, VL = 250 V, R = 40 Ω.

Formula Used
V = √[VR² + (VL − VC)²], I = VR/R
Step-by-step Solution

The voltages cannot be added algebraically because VR, VL and VC are phasors. Since VL = VC, their opposite phasors cancel.

Thus V = VR = 200 V. Current through the resistor is I = VR/R = 200/40 = 5 A.

Final Answer: The paradox is resolved by phasor addition; I = 5 A.

Common Mistake: Adding 200 + 250 + 250 = 700 V as scalar voltages.

Exam Tip: VL and VC are 180° opposite in the phasor diagram.

An inductor 200 mH, capacitor 400 μF and resistor 10 Ω are in series with a 50 V variable-frequency source. Find angular frequency for maximum power, effective current and Q-factor.

Given Data

L = 200 mH = 0.2 H, C = 400 μF = 4 × 10−4 F, R = 10 Ω, V = 50 V.

Formula Used
ω0 = 1/√LC, I = V/R at resonance, Q = ω0L/R
Step-by-step Solution

Maximum power dissipation occurs at resonance. LC = 0.2 × 4 × 10−4 = 8 × 10−5.

ω0 = 1/√(8 × 10−5) = 111.8 rad s−1. Effective current I = 50/10 = 5 A.

Q = (111.8 × 0.2)/10 = 2.236 = √5.

Final Answer: ω0 = 111.8 rad s−1, I = 5 A, Q = √5.

Common Mistake: Reporting only current and forgetting angular frequency.

Exam Tip: Maximum power in series LCR always means resonance and cosφ = 1.

A sinusoidal voltage of peak value 10 V is applied to series LCR with R = 10 Ω, C = 1 μF and L = 1 H. Find peak voltage across inductor at resonance and Q-factor.

Given Data

V0 = 10 V, R = 10 Ω, C = 1 μF = 10−6 F, L = 1 H.

Formula Used
At resonance: I0 = V0/R, XL = ω0L, ω0 = 1/√LC, Q = XL/R
Step-by-step Solution

ω0 = 1/√(1 × 10−6) = 1000 rad s−1. Therefore XL = 1000 × 1 = 1000 Ω.

I0 = 10/10 = 1 A. Peak voltage across inductor = I0XL = 1 × 1000 = 1000 V.

Q = XL/R = 1000/10 = 100.

Final Answer: Peak VL = 10³ V, Q = 100.

Common Mistake: Assuming voltage across each element must be less than source voltage.

Exam Tip: At resonance, VL and VC may be Q times the source voltage.

PYQs - AC Circuits, Resonance and Power in AC Circuits
Alternating Current PYQs Hub • CBSE • NEET • IIT JEE

PYQs - AC Circuits, Resonance and Power in AC Circuits

Exam-focused Physics question bank with revealable solutions, formulas, mock tests, revision sheets and premium coaching-style practice for school boards, NEET, JEE, IB, IGCSE and A-Level Physics.

Important Exam Notes

Most Frequently Asked

  • RMS value and average value of sinusoidal AC.
  • Impedance and phase angle in RLC circuits.
  • Resonance condition, resonant frequency and Q-factor.
  • Power factor, wattless current and transformer efficiency.

Common Mistakes

  • Mixing peak, rms and average values.
  • Adding VR, VL, VC algebraically instead of as phasors.
  • Forgetting μF, mH and kHz conversions.
  • Using f instead of angular frequency ω = 2πf.

High Weightage Concepts

  • Series resonance and bandwidth.
  • Quality factor and selectivity.
  • Power triangle: P, Q and S.
  • Step-up and step-down transformer relations.

Formula Popup Cards

Quick Revision Cards

RMS and Average

Vrms = V0/√2 and Vavg = 2V0/π for half-cycle average of sine wave.

Reactance Rule

Inductor opposes high frequency: XL = ωL. Capacitor opposes low frequency: XC = 1/ωC.

Power Rule

Real power is consumed only by resistance: P = VrmsIrmscosφ.

Mock Test Section

Mini Test 1

  1. Find Vrms for V0 = 200 V.
  2. Calculate XL for L = 0.2 H, f = 50 Hz.
  3. State the resonance condition.
  4. Define wattless current.
  5. Find power factor if R = 30 Ω and Z = 50 Ω.

Answer Key: 141.4 V; 62.8 Ω; XL = XC; current component that consumes no average power; 0.6.

Mini Test 2

  1. Derive Z for series RLC circuit.
  2. Find f0 for L = 1 H, C = 1 μF.
  3. Draw a capacitive phasor relation.
  4. State transformer voltage ratio.
  5. Explain why current is maximum at resonance.

Answer Key: Z = √[R² + (XL - XC)²]; 159 Hz; current leads voltage; Vs/Vp = Ns/Np; because net reactance is zero and Z = R.

Full Chapter Test

  1. 10 MCQs from RMS, average and reactance.
  2. 5 numerical problems from RL, RC and RLC circuits.
  3. 3 assertion-reason questions from resonance and power factor.
  4. 2 long-answer derivations: average power and transformer relation.

Answer Key Pattern: Use formulas from the final revision sheet. Numerical answers must include SI units and phase nature.

Final Revision Section

Resonance Formula Sheet

XL = XC ω0 = 1/√LC f0 = 1/(2π√LC)

AC Circuit Formula Sheet

Z = √[R² + (XL - XC)²] tanφ = (XL - XC)/R

Transformer Formula Sheet

Vs/Vp = Ns/Np VpIp = VsIs

Power Factor Summary

Power factor is cosφ. It is 1 at resonance, lagging in inductive circuits and leading in capacitive circuits.

Phasor Diagram Summary

Take current along X-axis. VR lies along current, VL is upward and VC is downward.

Last Minute Notes

Check SI units, use rms values for average power, and always mention whether current leads or lags.

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