Key Fact
Resonance in LCR
At resonance frequency ω₀ = 1/√(LC): X_L = X_C, Z = R minimum, I maximum. Quality factor Q = ω₀L/R = 1/(ω₀ R C).
-- NCERT Class 12 Physics, Ch. 7, p. 189Resonance LCR
8 MCQs2 revision cards9-step worked example
Source: NCERT Electromagnetic Induction and Alternating CurrentsPYQ coverage: NEET 2021, 2022, 2023, 2026Official key: NTA-verifiedLast updated: 27 Sep 2026
Resonance LCR, explained for NEET
The resonance formula is where marks leak. Two errors dominate: dropping the square root, so 1/(2π·LC) is written instead of 1/(2π√(LC)); and dropping the 2π, so ω₀ is handed in when f₀ was asked. Both produce answers that sit on the option list — the distractors are built from exactly these slips.
Resonance in a series LCR circuit is the condition X_L = X_C. Since X_L rises with frequency and X_C falls, there is one frequency where they cancel. At that frequency the reactive term in Z = √(R² + (X_L − X_C)²) vanishes, so the impedance collapses to its minimum value Z = R, and the current reaches its maximum I = V/R. The source voltage and current are in phase there: φ = 0. NCERT Class 12 Physics Chapter 7 states the resonance condition and derives the frequency on page 190.
Setting ωL = 1/(ωC) gives ω₀ = 1/√(LC), and dividing by 2π gives f₀ = 1/(2π√(LC)). Note what is absent: R does not appear. Changing the resistance changes how sharp the resonance is and how large the peak current is, but not where the peak sits.
NEET asks this as a three-link chain — find f₀, then ω₀, then Z, then I. Each link carries its own factor-of-2π or √2 opportunity, and the errors compound, so a final current can be off by a factor of four with every individual step looking reasonable. Work one substitution per line and check the order of magnitude before moving on.
One more consequence worth holding: at resonance the individual voltages across L and across C are equal in magnitude and opposite in phase, so they cancel in the sum even though each can separately exceed the source voltage.
Watch-out: if a question gives you R along with L and C and asks for the resonant frequency, R is there as bait. Do not try to fit it in.
Can you answer these Resonance LCR MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In a series LCR circuit at resonance, the impedance of the circuit is equal to
Show answer and why every option is right or wrong
Answer: C. At resonance X_L = X_C, so the reactive term (X_L − X_C) in the impedance expression vanishes and Z reduces to R alone. NCERT Class 12 Physics Chapter 7 gives this result on page 189.
Why A is wrong: A is wrong because it retains an inductive reactance term that has been cancelled by the equal capacitive reactance at resonance; the reactances subtract, they do not survive in the square root.
Why B is wrong: B is wrong because the reactances enter the impedance as a difference, not a sum — the whole point of resonance is that X_L − X_C = 0.
Why D is wrong: D is wrong because the resistance is always present and never cancels; only the reactive part vanishes. A zero impedance would mean infinite current.
MCQ 2Easy RecallPractice
The angular resonant frequency of a series LCR circuit is given by
Show answer and why every option is right or wrong
Answer: B. Setting ωL = 1/(ωC) gives ω² = 1/(LC), so ω₀ = 1/√(LC). NCERT Class 12 Physics Chapter 7 derives this on page 189.
Why A is wrong: A is wrong because it omits the square root; 1/(LC) is ω₀², not ω₀ itself. This is the single most common slip on this formula.
Why C is wrong: C is wrong because it is the ordinary frequency f₀ in hertz, not the angular frequency ω₀ in rad s⁻¹. The two differ by the factor 2π.
Why D is wrong: D is wrong because it inverts the dependence — increasing L or C must lower the resonant frequency, not raise it.
MCQ 3Easy RecallPractice
At the resonant frequency of a series LCR circuit, the phase angle between the source voltage and the current is
Show answer and why every option is right or wrong
Answer: D. With tan φ = (X_L − X_C)/R and X_L = X_C at resonance, the numerator is zero, so φ = 0 and the circuit behaves as purely resistive. NCERT Class 12 Physics Chapter 7, page 189.
Why A is wrong: A is wrong because a 90° lead would require a purely capacitive circuit; here the capacitive reactance is exactly cancelled by the inductive reactance.
Why B is wrong: B is wrong because 45° would require |X_L − X_C| = R, which is not the resonance condition — at resonance the difference is zero regardless of R.
Why C is wrong: C is wrong because a 90° lag would require a purely inductive circuit; the resistance is present and the reactances have cancelled.
MCQ 4Direct ApplicationPractice
A series LCR circuit has L = 2.0 H and C = 8.0 × 10⁻⁶ F. Its angular resonant frequency is closest to
Show answer and why every option is right or wrong
Answer: A. LC = 2.0 × 8.0 × 10⁻⁶ = 1.6 × 10⁻⁵ s², so √(LC) = 4.0 × 10⁻³ s and ω₀ = 1/(4.0 × 10⁻³) = 2.5 × 10² rad s⁻¹. NCERT Class 12 Physics Chapter 7, page 189.
Why B is wrong: B is wrong because it drops the square root and computes 1/(LC) = 6.25 × 10⁴, which has units of s⁻² and is ω₀², not ω₀.
Why C is wrong: C is wrong because it divides the correct ω₀ by 2π, converting to f₀ = 40 Hz while the question asked for the angular frequency.
Why D is wrong: D is wrong because 6.3 × 10¹ rad s⁻¹ is not produced by misplacing the 2π: 1/(2π√(LC)) = 1/(2π × 4.0 × 10⁻³) ≈ 4.0 × 10¹, which is option C. ω₀ = 1/√(LC) = 1/(4.0 × 10⁻³) = 2.5 × 10² rad s⁻¹.
MCQ 5Direct ApplicationPractice
A series LCR circuit contains L = 0.50 H, C = 2.0 × 10⁻⁶ F and R = 1.0 × 10² Ω. The resonant frequency f₀ is closest to
Show answer and why every option is right or wrong
Answer: B. LC = 1.0 × 10⁻⁶ s², √(LC) = 1.0 × 10⁻³ s, so f₀ = 1/(2π × 1.0 × 10⁻³) ≈ 1.6 × 10² Hz. The resistance plays no part in locating the resonance. NCERT Class 12 Physics Chapter 7, page 189.
Why A is wrong: A is wrong because it reports ω₀ = 1/√(LC) = 1.0 × 10³ rad s⁻¹ and labels it as f₀; the 2π divisor has been dropped.
Why C is wrong: C is wrong because it drops the square root: 1/(LC) = 1.0 × 10⁶ has units s⁻² and cannot be a frequency.
Why D is wrong: D is wrong because 5.0 × 10² Hz = 1/(2√(LC)) drops the π from the 2π divisor while keeping the 2. The resistance plays no part in locating the resonance.
MCQ 6Concept TrapPractice
In a series LCR circuit driven at its resonant frequency, the resistance R is now replaced by a smaller resistance while L and C are unchanged. Compared with before, the resonant frequency and the peak current respectively
Show answer and why every option is right or wrong
Answer: C. The resonance condition X_L = X_C involves only L and C, so f₀ is unchanged; at resonance Z = R, so a smaller R gives a larger current I = V/R. NCERT Class 12 Physics Chapter 7, page 189.
Why A is wrong: A is wrong on the current: at resonance the impedance equals R exactly, so halving R doubles the current even though the frequency is fixed.
Why B is wrong: B is wrong because it makes the resonant frequency depend on R; the condition ωL = 1/(ωC) contains no resistance at all.
Why D is wrong: D is wrong for the same reason as B — R does not appear in ω₀ = 1/√(LC), so it cannot shift the resonant frequency in either direction.
MCQ 7Direct ApplicationPractice
A series LCR circuit with L = 1.0 H, C = 1.0 × 10⁻⁶ F and R = 5.0 × 10¹ Ω is driven by an AC source of RMS voltage 1.0 × 10² V at the resonant frequency. The RMS current in the circuit is closest to
Show answer and why every option is right or wrong
Answer: A. At resonance Z = R = 5.0 × 10¹ Ω, so I_rms = V_rms/R = 1.0 × 10²/5.0 × 10¹ = 2.0 A. The values of L and C only fix where resonance occurs; they do not enter the current at resonance. NCERT Class 12 Physics Chapter 7, page 189.
Why B is wrong: B is wrong because it divides by a reactance computed away from resonance instead of recognising that the reactive terms cancel, leaving Z = R.
Why C is wrong: C is wrong because it applies a spurious √2 factor, treating the given RMS voltage as though it needed conversion; the voltage is already RMS and the current asked for is RMS.
Why D is wrong: D is wrong because it uses an impedance four times too large, the signature of compounding a 2π slip in ω₀ with a reactance evaluated at the wrong frequency.
MCQ 8Concept TrapPractice
A series LCR circuit is at resonance. The capacitance is then changed to one-quarter of its original value, with L and R held fixed and the source frequency readjusted to the new resonance. The new resonant frequency is
Show answer and why every option is right or wrong
Answer: D. Since f₀ ∝ 1/√(LC), reducing C by a factor of 4 reduces √(LC) by a factor of 2, so f₀ doubles. NCERT Class 12 Physics Chapter 7, page 189.
Why A is wrong: A is wrong because it applies the change to C directly with the wrong sense of proportionality — f₀ varies inversely with √C, so a smaller C raises the frequency.
Why B is wrong: B is wrong because it has the right factor of 2 but the wrong direction; reducing the capacitance increases the resonant frequency.
Why C is wrong: C is wrong because it ignores the square root and scales f₀ by the full factor of 4 rather than by √4 = 2.
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Resonance LCR: quick recall before you leave
How do you solve a Resonance LCR question? A worked example
- 1
Given
L = 0.40 H, C = 1.0 × 10⁻⁵ F, R = 2.0 × 10¹ Ω, source V_rms = 6.0 × 10¹ V.
- 2
Required
The resonant frequency f₀, and the RMS current in the circuit when the source is set to that frequency.
- 3
Concept
Resonance occurs where the inductive and capacitive reactances are equal. There the reactive contribution to the impedance vanishes, leaving Z = R, which is the minimum possible impedance for this circuit — hence the maximum current.
- 4
Formula
f₀ = 1/(2π√(LC)); at resonance Z = R, so I_rms = V_rms/R.
- 5
Substitution
LC = (0.40)(1.0 × 10⁻⁵) = 4.0 × 10⁻⁶ s²
f₀ = 1/(2π√(4.0 × 10⁻⁶)) - 6
Calculation
√(4.0 × 10⁻⁶) = 2.0 × 10⁻³ s
2π × 2.0 × 10⁻³ = 1.2566 × 10⁻²
f₀ = 1/(1.2566 × 10⁻²) = 79.58 Hz → 8.0 × 10¹ Hz (2 sig figs)
At this frequency Z = R = 2.0 × 10¹ Ω, so
I_rms = (6.0 × 10¹)/(2.0 × 10¹) = 3.0 A
The factor 2π is a mathematical constant and the 2 inside √(LC) does not arise from a measurement; neither constrains the significant-figure count. The two sig figs come from the given L, C, R and V, each quoted to two. - 7
Final answer
f₀ = 8.0 × 10¹ Hz; I_rms = 3.0 A.
- 8
Common trap
This problem is a chain — √(LC), then the 2π division, then the current — and each link can be broken independently. Dropping the square root gives 1/(2π × 4.0 × 10⁻⁶) ≈ 4.0 × 10⁴ Hz. Dropping the 2π gives 5.0 × 10² Hz, which is ω₀ wearing a hertz label. Both wrong values sit comfortably among plausible options. The defence is one substitution per line with an order-of-magnitude glance after each: a 0.40 H inductor with a 10 µF capacitor resonating at 40 kHz should look wrong before the arithmetic is even checked. A second bait is R: it is needed for the current but plays no part in f₀, and the habit of using every given number puts it into the frequency calculation.
- 9
Similar NEET-style question
A series LCR circuit has L = 0.10 H and C = 4.0 × 10⁻⁶ F, with R = 5.0 × 10¹ Ω, connected to a 1.0 × 10² V RMS supply. Find the resonant frequency and the RMS current at resonance. *(Expected: f₀ ≈ 2.5 × 10² Hz, I_rms = 2.0 A.)*
What to remember before solving Resonance LCR questions
Which Resonance LCR formulas do you need for NEET?
2 formulas — click to collapse
LCR series impedance
Effective resistance of series LCR. Phase angle phi between V and I.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Z | impedance | Ω |
| R | resistance | Ω |
| X_L, X_C | reactances | Ω |
| phi | phase angle | rad |
Valid when
- Series LCR
- AC steady state
- Single frequency source
Resonance frequency LCR
Frequency at which X_L = X_C, Z = R minimum, current maximum.
| Symbol | Quantity | SI Unit |
|---|---|---|
| f0 | resonance freq | Hz |
| L | inductance | H |
| C | capacitance | F |
Valid when
- Series LCR
- Steady AC
Where do students lose marks on Resonance LCR?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Negative Marking
Multi-step LCR resonance problem (compute resonance frequency, then current, then impedance). Each sub-step has factor-of-2π or sqrt-of-2 error opportunities. Final answer can be off by 4× cumulatively.
When it triggers
LCR question asks for current/voltage/power at resonance with multi-step computation.
How to avoid
Slow down. Compute f_0 = 1/(2π√(LC)) carefully. Then ω_0 = 2πf_0. Then Z = R, I = V/R. Each line one substitution; check unit/order at each step. Answers off by powers of 2π or √2 are deliberate distractors.
Root cause: formula misuse
Correction
ω_0 = 1/√(LC); f_0 = 1/(2π√(LC)). Common errors: drop the sqrt (treat 1/(LC) as ω_0²), or drop the 2π (give ω_0 instead of f_0).
More in Electromagnetic Induction and Alternating Currents: 2 exam traps and mistakes · 9 formulas · 5 question patterns from its other lessons.
Resonance LCR questions from past NEET papers
4 questions from NEET 2021, 2022, 2023, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
12.2 A
25.5 A
311.0 A
422.0 A
NTA Answer: Option 1(final)
NEET 2023
115.9 kHz
21.59 rad/s
31.59 kHz
415.9 rad/s
NTA Answer: Option 3(final)
NEET 2022
1ν = 100 Hz; ν₀ = 100/π Hz
2ν₀ = ν = 50 Hz
3ν₀ = ν = 50/π Hz
4ν₀ = 50/π Hz, ν = 50 Hz
NTA Answer: Option 3(final)
NEET 2021
142 rad/s and 58 rad/s
225 rad/s and 75 rad/s
350 rad/s and 25 rad/s
446 rad/s and 54 rad/s
NTA Answer: Option 4(final)
All 20 past-paper questions from Electromagnetic Induction and Alternating Currents →
How does NEET ask about Resonance LCR?
1 recurring pattern from past papers — click to collapse
Series LCR resonance frequency f0 = 1/(2π sqrt(LC)). At resonance Z=R, current max.
Multi StepMedium
Common distractors
forgets square root
Uses 1/(2π LC) without sqrt
conflates X_L X_C formulas
Substitutes wrong reactance formula
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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