Q.Suppose the frequency of the source in the previous example can be varied.
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Resonance in AC Circuits
A series circuit containing a resistor R, an inductor L and a capacitor C driven by an AC source exhibits resonance — a sharp condition at which the circuit responds most strongly.
The Competing Reactances
In a series RLC circuit the inductor and capacitor oppose the current in opposite senses. Their reactances are
XL=ωL,XC=ωC1
where ω=2πf is the angular frequency. As frequency rises, XL grows while XC shrinks. The total impedance is
Z=R2+(XL−XC)2
The Resonance Condition
At one special frequency the two reactances become exactly equal and cancel:
XL=XC⇒ω0L=ω0C1⇒ω0=LC1
The corresponding resonant frequency is
f0=2πLC1
At this frequency the impedance falls to its minimum, Z=R (purely resistive), so the current reaches its maximum value
Imax=RVrms
Because the reactances cancel, the source voltage and current are exactly in phase — the power factor is 1 at resonance.
Physical Picture
At resonance energy sloshes back and forth entirely between the inductor's magnetic field and the capacitor's electric field, cycle after cycle. The source only has to make up the small amount of energy lost as heat in R. This is the electrical analogue of a swing pushed at its natural frequency: a small periodic drive builds a large oscillation.
Sharpness and the Q-factor
How sharply the current peaks around f0 is measured by the quality factor:
Q=Rω0L=R1CL
A large Q (small R) gives a tall, narrow resonance curve — the circuit is highly selective, responding to a very narrow band of frequencies. A small Q gives a broad, flat peak.
Why It Matters …
Why this formula?
Resonance in AC Circuits: Why the Key Formulas Hold
Resonance in an AC circuit occurs when the inductive reactance (XL) and capacitive reactance (XC) exactly cancel each other out. Let's build the understanding step-by-step.
1. The Core Condition for Resonance
Consider a series RLC circuit (resistor R, inductor L, capacitor C) driven by an AC voltage source V=V0sin(ωt).
The total impedance Z of the series combination is:
Z=R+j(XL−XC)
where:
- XL=ωL (inductive reactance)
- XC=ωC1 (capacitive reactance)
- j=−1
Why resonance happens:
The circuit "wants" to let maximum current flow. The opposition to current comes from both resistance and reactance. But reactance can be negative (capacitive) or positive (inductive). When they are equal in magnitude but opposite in sign, they cancel:
XL=XC
This is the fundamental condition — not a formula to memorize, but a logical consequence of impedance minimization.
2. Deriving the Resonant Frequency
From XL=XC:
ωL=ωC1
Multiply both sides by ω:
ω2LC=1
Thus:
ω0=LC1
Since ω=2πf, the resonant frequency in hertz is:
f0=2πLC1
Why this makes sense:
- A larger L or C means the circuit takes longer to "oscillate" — lower frequency.
- A smaller L or C means faster oscillations — higher frequency.
- The product LC controls the natural time scale of the circuit.
3. What Happens at Resonance — Key Consequences
(a) Impedance is Minimum (Purely Resistive)
At resonance, XL−XC=0, so:
Z=R+j(0)=R
Why: The reactive parts cancel, leaving only the resistance. The circuit behaves like a pure resistor.
(b) Current is Maximum
From Ohm's law for AC:
I=ZV
At resonance, Z=R (minimum possible), so current is maximum:
Imax=RV
Why: The opposition to current is smallest when reactance cancels.
(c) Voltage Across L and C Can Be Very Large
The voltage across the inductor:
VL=I⋅XL=RV⋅ω0L
The voltage across the capacitor:
VC=I⋅XC=RV⋅ω0C1
Since XL=XC at resonance, VL=VC in magnitude, but they are 180° out of phase — they cancel each other in the loop.
Why this is important:
If R is small, VL and VC can be many times larger than the source voltage V. This is called voltage magnification — a key concept for tuned circuits and filters.
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Using the previous example's data: R=3 Ω, L=25.48 mH, C=796 μF, Vrms=200 V.
- Resonant frequency:
f0=2πLC1=2π(25.48×10−3)(796×10−6)1≈35.4 Hz
- At resonance the reactances cancel, so Z=R: Z=R=3 Ω,I=ZV=3200≈66.7 A …
With R=3 Ω, L=25.48 mH, C=796 μF and Vrms=200 V, resonance occurs at f0=2πLC1≈35.4 Hz, where Z=R=3 Ω, I=V/R≈66.7 A and P=V2/R≈13.3 kW.
At resonance the inductive and capacitive reactances of the series LCR circuit become equal and cancel, leaving a purely resistive impedance. This is where the impedance is smallest and the current is largest.
(a) Resonant frequency
Resonance requires XL=XC, i.e. ω0L=1/ω0C, giving ω0=1/LC and
f0=2πLC1
Substituting L=25.48×10−3 H and C=796×10−6 F:
LC=(25.48×10−3)(796×10−6)=2.03×10−5 s2,
LC=4.50×10−3 s,
f0=2π×4.50×10−31≈35.4 Hz.
(b) Impedance, current and power at resonance
Since XL=XC, the net reactance is zero and
Z=R2+(XL−XC)2=R=3 Ω.
The rms current is then maximum:
I=ZV=3200≈66.7 A. …
Method: Resonance Condition in Series RLC Circuit
This method applies when an AC source is connected to a series combination of a resistor (R), an inductor (L), and a capacitor (C). Resonance occurs when the inductive reactance equals the capacitive reactance.
Step 1 — Write the resonance condition
At resonance:
XL=XC
Where:
- XL=2πfL (inductive reactance)
- XC=2πfC1 (capacitive reactance)
Step 2 — Solve for resonant frequency f0
Set XL=XC:
2πf0L=2πf0C1
Multiply both sides:
(2πf0)2LC=1
Thus:
f0=2πLC1
Answer (a): The resonant frequency is f0=2πLC1
Step 3 — Impedance at resonance
At resonance, XL=XC, so they cancel each other. The total impedance is purely resistive:
Z=R
Impedance at resonance: Z=R
Step 4 — Current at resonance
Using Ohm’s law for AC circuits:
I=ZV=RV
Where V is the RMS voltage of the source.
Current at resonance: I=RV
Step 5 — Power dissipated at resonance …
Here are the common mistakes students make on resonance in AC circuits, along with how to avoid each — tailored for Indian exam accuracy (JEE, NEET, CBSE).
1. Using the Wrong Formula for Resonant Frequency
Mistake:
Students often confuse the formula for resonance in a series LCR circuit with that of a parallel circuit, or they forget the square root.
Correct formula (series LCR):
f0=2πLC1
How to avoid:
- Memorise: Resonance occurs when XL=XC.
- Derive quickly:
ωL=ωC1⇒ω2=LC1⇒f=2πLC1
- Never write f0=2π1CL — that’s wrong.
2. Forgetting That Impedance is Minimum (Not Maximum) at Resonance
Mistake:
Thinking Z is maximum at resonance (confusing with parallel resonance or voltage across L/C).
Correct:
At resonance, XL=XC, so:
Z=R2+(XL−XC)2=R
Impedance is minimum and purely resistive.
How to avoid:
- Remember: Resonance = minimum opposition to current.
- In a series circuit, current is maximum → impedance must be minimum.
3. Calculating Current Without Using the Correct Impedance
Mistake:
Using I=V/(XL−XC) or forgetting that Z=R at resonance.
Correct:
At resonance:
I0=RV
How to avoid:
- Always first find Z at resonance.
- If f=f0, then Z=R — no reactance left.
4. Power Dissipation Formula Error
Mistake:
Using P=VrmsIrmscosϕ but forgetting that at resonance cosϕ=1.
Correct:
At resonance, ϕ=0, so:
P=VrmsIrms=Irms2R
How to avoid:
- At resonance, circuit is purely resistive → power factor = 1.
- So P=Vrms2/R also works.
5. Mixing Up RMS and Peak Values
Mistake:
Using peak voltage V0 in formulas meant for RMS values, or vice versa.
Correct approach:
- If source voltage is given as V=V0sin(ωt), then Vrms=V0/2.
- Use RMS values for power and current calculations unless asked for peak.
How to avoid:
- Check the problem statement: “220 V” usually means RMS.
- Write explicitly:
Irms=RVrms
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Showing the 12 most recent of 14 on this concept.
- GUJCET 2026Set x1 markMCQQ.A charged 30 μF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? (A) 11 rad/s (B) 1100 rad/s (C) 110 rad/s (D) 11000 rad/s
›Reveal solutionSolution
ω=1/LC≈1100 rad/s.
LC=(27×10−3)(30×10−6)=8.1×10−7,LC=9.0×10−4. …
- GUJCET 2025Set 031 markMCQQ.In which of the following AC circuit, we get the value of power factor 1 at resonance condition? (A) LCR series circuit (B) CR series circuit (C) Only inductor (L) circuit (D) LR series circuit
›Reveal solutionSolution
[!TLDR]
An LCR series circuit has power factor 1 at resonance, because the reactances cancel and the impedance is purely resistive.
Concept
Power factor cosϕ=ZR equals 1 only when the net reactance is zero. Resonance (XL=XC) is defined only for a circuit containing both L and C.
Solution
- In a series LCR circuit, Z=R2+(XL−XC)2.
- At resonance XL=XC, so Z=R (purely resistive). …
- GUJCET 2024Set 131 markMCQQ.In LCR series a.c. circuit at resonance the value of power factor will be ________. (A) ∞ (B) 1 (C) −1 (D) 0
›Reveal solutionSolution
At resonance the reactances cancel; impedance =R, so power factor =1. …
- GUJCET 2023Set 091 markMCQQ.A pure inductor of 25.48 mH and a pure resistor of 8 Ω are connected in series with an A.C. source of frequency 50 Hz. The phase difference between current (I) and voltage (V) in this circuit is ______. (A) 45° (B) 30° (C) 60° (D) 90°
›Reveal solutionSolution
In a series RL circuit the phase angle is ϕ=tan−1(XL/R).
Concept: Inductive reactance
XL=2πfL=2π(50)(25.48×10−3)≈8 Ω.
Since XL=R=8 Ω, …
- GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQQ.Which of the following combination should be selected for better tuning of an LCR a.c. circuit used for communication?(a) R = 15 ohm, L = 3.5 H, C = 30 microF(b) R = 25 ohm, L = 2.5 H, C = 45 microF(c) R = 20 ohm, L = 1.5 H, C = 35 microF(d) R = 25 ohm, L = 1.5 H, C = 45 microF
›Reveal solutionSolution
Sharp tuning means high Q = (1/R) sqrt(L/C); computing Q for each set, option (a) (low R, high L/C) gives the largest value.
Good (sharp) tuning requires a high quality factor: Q = (1/R) sqrt(L/C).
Approximate Q for each:
- (1/15) sqrt(3.5/30e-6) = (1/15)(342) approximately 23.
- (1/25) sqrt(2.5/45e-6) = (1/25)(236) approximately 9.4.
- (1/20) sqrt(1.5/35e-6) = (1/20)(207) approximately 10.4. …
- GUJCET 2022Set 171 markMCQQ.A charged 10 μF capacitor is connected to a 16 mH inductor. What is the angular frequency of free oscillations of the circuit? (A) 250 rad s−1 (B) 25 rad s−1 (C) 1111 rad s−1 (D) 2500 rad s−1
›Reveal solutionSolution
Free (undamped) LC oscillations have angular frequency ω=1/LC.
Concept. A charged capacitor discharging through an inductor exchanges energy between the electric and magnetic fields, oscillating at the natural frequency ω=LC1.
Steps.
- L=16 mH=16×10−3 H, C=10 μF=10×10−6 F. …
- GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQQ.At resonance, the value of the power factor is ______.(a) infinity(b) 1(c) 0(d) 0.5
›Reveal solutionSolution
At resonance, the reactive parts cancel and the circuit behaves as pure resistance, giving unity power factor.
Power factor is cosϕ=ZR, where Z=R2+(XL−XC)2.
…
- GUJCET 2021Set 151 markMCQQ.For LCR ac series circuits, L=25 mH, R=3Ω, C=62.5μF. What is the frequency of the sources at which resonance occurs? (A) 127.39 Hz (B) 35.40 Hz (C) 100 Hz (D) 21 Hz
›Reveal solutionSolution
Series resonance frequency is f0=2πLC1.
Concept:
LC=(25×10−3)(62.5×10−6)=1.5625×10−6,LC=1.25×10−3. …
- GUJCET 2021Set 151 markMCQQ.For a series LCR circuit with L=2 H, C=18μF and R=10Ω. What is the value Q-factor of this circuit? (A) 22.22 (B) 55.55 (C) 44.44 (D) 33.33
›Reveal solutionSolution
The quality factor of a series LCR circuit is Q=R1CL.
Concept. Q=R1CL.
Solution. L=2 H, C=18×10−6 F, R=10Ω. …
- GUJCET 2020Set 071 markMCQQ.A sine voltage having maximum value of 283 V & frequency of 50 Hz is applied to LCR series connection where R=3Ω, L=25.48 mH & C=796μF. Then impedence is ______ at resonance condition. (A) 3Ω (B) 5Ω (C) 15Ω (D) 4Ω
›Reveal solutionSolution
At resonance the reactances cancel and impedance equals the resistance.
Concept: For a series LCR circuit, Z=R2+(XL−XC)2. At resonance XL=XC, …
- GUJCET 2019Set 131 markMCQQ.In L-C-R, A.C. series circuit, L = 9H, R = 10Ω & C=100μF. Hence Q-factor of the circuit is ......... (A) 30 (B) 35 (C) 45 (D) 25
›Reveal solutionSolution
The quality factor Q=R1CL=30.
Concept: For a series L-C-R resonant circuit the quality factor is Q=Rω0L=R1CL.
Steps: …
- GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQQ.For L-C-R A.C. circuit resonance frequency is 600 Hz and frequencies at half power points are 550 Hz and 650 Hz. What will be the Q-factor?(a) 1/6(b) 1/3(c) 6(d) 3
›Reveal solutionSolution
Q equals resonance frequency divided by the bandwidth between half-power points: 600/(650-550) = 6.
The half-power (3 dB) points are at 550 Hz and 650 Hz, so the bandwidth is: …
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