Q.At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Resonance in AC Circuits
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.
--- …
The doorway of the detector contains a coil that, together with a capacitor, forms an LC circuit driven near its resonant frequency f0=2πLC1. At resonance the circuit carries a large, sharply-tuned current.
When a person carrying metal walks through, the metal alters the effective inductance of the coil, shifting the circuit away from resonance. This produces a marked change in the circuit's impedance and hence in its current. The electronics sense this change and trigg …
A metal detector is a coil–capacitor (LC) circuit tuned to resonance. Metal carried through the doorway changes the coil's inductance, throwing the circuit off resonance; the resulting sharp change in impedance and current is sensed electronically and sounds the alarm.
The archway you walk through is not a magnet — it is an AC circuit deliberately operated at resonance, exploiting how sensitively a resonant circuit responds to a small change in its components.
1. A tuned LC circuit
Built into the doorway is a coil (inductance L) connected with a capacitor (capacitance C) and driven by an oscillating source. Such a series circuit has a resonant frequency
f0=2πLC1
at which the inductive and capacitive reactances cancel, the impedance drops to its minimum Z=R, and the current is at its sharp maximum. The circuit is set to run at (or very near) this resonant frequency.
2. Effect of a metal object
When a person carrying metal passes through the coil, the metal changes the magnetic environment of the coil and hence its effective inductance L. Because f0 depends on L, the circuit is pushed away from resonance.
3. A large, detectable change …
Method: Electromagnetic Induction (Principle of Eddy Currents)
This is a concept-based reasoning method — no calculation is needed. The answer follows from understanding how changing magnetic fields induce currents in conductors.
Step 1 — Identify the core physical principle
The metal detector works on electromagnetic induction, specifically the production of eddy currents in a metal object.
Step 2 — Describe the setup
- The doorway contains coils that carry an alternating current (AC).
- This AC produces a changing magnetic field in the space around the doorway.
Step 3 — What happens when metal enters the field
- When a person carrying a metal object (conductor) walks through, the changing magnetic field induces circulating currents inside the metal.
- These are called eddy currents (by Faraday’s Law of Induction).
Step 4 — How the detector senses the metal
- The eddy currents themselves produce a secondary magnetic field.
- This secondary field is detected by receiver coils in the doorway. …
Here are the common mistakes students make on this question, along with the correct reasoning to avoid them.
Mistake #1: Saying it works on "Magnetic Effect of Current"
- The error: Students think the detector creates a magnetic field and if metal is present, it "attracts" the metal or somehow completes a circuit.
- Why it's wrong: The metal detector does not rely on magnetic attraction. It relies on a changing magnetic field inducing a current in the metal object.
- How to avoid: Remember: Static magnetic fields don't trigger the alarm. The field must be changing (alternating) to induce anything.
Mistake #2: Confusing it with "Electromagnetic Induction" in a transformer
- The error: Students write "mutual induction between two coils" but forget to mention the metal object acts as a secondary coil.
- Why it's wrong: In a transformer, both coils are fixed. Here, the metal object is the "secondary" — it has no wire attached.
- How to avoid: Say: "The metal object acts as a secondary coil in which eddy currents are induced."
Mistake #3: Forgetting to mention Eddy Currents
- The error: Students stop at "electromagnetic induction" without specifying that eddy currents are induced in the metal.
- Why it's wrong: The induced current in the metal is not a simple current in a wire — it's a loop of current (eddy current) inside the metal itself.
- How to avoid: Always include the phrase: "Eddy currents are induced in the metal object."
Mistake #4: Not explaining the detection mechanism
- The error: Students describe induction but don't explain how the detector knows metal is present.
- Why it's wrong: The induced eddy current in the metal creates its own magnetic field, which then induces a current back in the detector's coil — this change is detected.
- How to avoid: Add this step: "The eddy currents produce a secondary magnetic field, which induces a current in the detector's receiver coil, triggering the alarm."
Mistake #5: Writing the principle as "Lenz's Law" instead of "Electromagnetic Induction"
- The error: Students write "Lenz's Law" as the principle. …
Showing the 12 most recent of 30 on this concept.
- CBSE 2026Set 55/1/11 markMCQQ.In a series LCR circuit, the voltage across the resistor, capacitor and inductor is 10 V each. If the capacitor is short circuited, the voltage across the inductor will be (A) 10 V (B) 52 V (C) 25 V (D) 102 V
›Reveal solutionSolution
In a series LCR circuit, when each component drops 10 V, the source voltage is 10 V (since VC and VL cancel) and the equal drops imply XL=XC=R. Shorting the capacitor leaves an RL circuit of impedance R2, so the current becomes I′=R210 and the inductor voltage is VL′=I′XL=210=52 V. The answer is (B).
Concept and intuition
The problem gives a series LCR circuit where the voltage across each element — resistor, capacitor, and inductor — is 10 V. That’s a strong clue: in a series circuit, the current is the same through all components, but the voltages are not in phase. The resistor voltage is in phase with current, the inductor voltage leads by 90°, and the capacitor voltage lags by 90°. So the three 10 V readings are phasor magnitudes, not simple arithmetic sums.
The key insight: if the capacitor is shorted, the circuit becomes a simple RL series circuit. The source voltage remains the same (it’s fixed by the supply), but the impedance changes. We need to find the new inductor voltage.
Step-by-step solution
1. Find the source voltage from the initial LCR condition.
In a series LCR circuit, the phasor sum of voltages across R, L, and C equals the source voltage Vs. Since VL and VC are opposite in phase (180° apart), they subtract. Given VR=VL=VC=10 V:
Vs=VR2+(VL−VC)2=102+(10−10)2=10 V
So the source supplies only 10 V. This makes sense: the inductor and capacitor voltages cancel exactly, so the source only “sees” the resistor drop.
TipThis cancellation is the hallmark of resonance in a series LCR circuit — at resonance, XL=XC, and the impedance is purely resistive. Here, VL=VC implies XL=XC, so the circuit is at resonance.
2. Determine the relationship between R and XL (or XC).
At resonance, the current is I=Vs/R=10/R. The voltage across the inductor is VL=IXL=(10/R)XL=10 V. Therefore:
R10XL=10⇒XL=R
So the inductive reactance equals the resistance. Similarly, XC=R as well. …
- CBSE 2026Set V11 markMCQQ.Power factor of a series LCR circuit is maximum when :(a) XL=XC(b) XC=0(c) XL>XC(d) XL<XC
›Reveal solutionSolution
- CBSE 2026Set ANNUAL1 markQ.Write True or False: The quality factor is ω_r L / R.
›Reveal solutionSolution
True — for a series resonant circuit, Q = ω_r L / R.
The quality factor (Q-factor) of a series resonant LCR circuit measures the sharpness of resonance. It is defined as the ratio of the inductive reactance at resonance to the resistance:
Q = ω_r L / R = (1/R)√(L/C),
…
- CBSE 2026Set SEM31 markMCQQ.The condition of getting maximum current in an LCR series circuit is(a) X_L = 0(b) X_C = 0(c) X_L = X_C(d) R = X_L − X_C
›Reveal solutionSolution
A series LCR circuit carries maximum current at resonance, where the inductive and capacitive reactances are equal (X_L = X_C), leaving impedance Z = R minimum. Option (c).
Step 1 — impedance of a series LCR circuit: Z = √(R² + (X_L − X_C)²), from NCERT/CBSE Class 12 Physics, Alternating Current.
…
- CBSE 2025Set D1 markMCQQ.In resonance condition, the frequency of L-C circuit is (A) (1/2π)√(1/LC) (B) 2π√(1/LC) (C) 2π√(LC) (D) (1/2π)√(LC)
›Reveal solutionSolution
At resonance the inductive and capacitive reactances are equal, giving the natural frequency f = 1/(2π√(LC)).
Resonance in an L-C (or series L-C-R) circuit occurs when the inductive reactance equals the capacitive reactance:
XL=XC ⇒ ωL=ωC1
Solving for the angular frequency,
…
- CBSE 2025Set ANNUAL1 markMCQQ.A series LCR circuit fed by an ac source with angular frequency ω acts as a purely resistive circuit, when(a) ωL > 1/ωC(b) ωL < 1/ωC(c) ωL = 1/ωC(d) ω³L = 1/ωC²
›Reveal solutionSolution
A series LCR circuit behaves as purely resistive at resonance, when the inductive and capacitive reactances cancel.
The impedance of a series LCR circuit is
Z=R2+(ωL−ωC1)2 …
- CBSE 2025Set ANNUAL1 markMCQQ.When LCR series circuit is at resonance then the phase angle (phi) between current and voltage is –(a) pi/2(b) pi(c) 2 pi(d) 0
›Reveal solutionSolution
At resonance in a series LCR circuit, current and voltage are exactly in phase.
In a series LCR circuit, the phase angle ϕ between the applied voltage and current is given by tanϕ=RXL−XC, where XL=ωL and XC=ωC1.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The power delivered by the AC source of a circuit becomes maximum when(i) wL = wC(ii) wL = 1/(wC)(iii) wL = -(1/(wC))^2(iv) wL = sqrt(wC)
›Reveal solutionSolution
Maximum power occurs at resonance, wL = 1/(wC).
In a series LCR circuit the impedance is Z=R2+(XL−XC)2 with XL=ωL and XC=1/ωC. Power P=VrmsIrmscosϕ is greatest when Z is minimum (Z = R) and the current is in phase with the voltage. …
- CBSE 2024Set A11 markMCQQ.The resonance phenomenon is exhibited by a circuit only if following components are present(a) L and R(b) R and C(c) L and C(d) None of the above
›Reveal solutionSolution
- CBSE 2024Set ANNUAL1 markMCQQ.In a series LCR circuit, resonant frequency depends on which of the following -(a) LCR(b) CL(c) LC1(d) RC1
›Reveal solutionSolution
Resonance in a series LCR circuit occurs when inductive and capacitive reactances are equal.
…
- CBSE 2024Set ANNUAL1 markMCQQ.For a series L-C-R circuit at resonance, the relation among inductance (L), capacitance (C) and frequency (ω) is(a) ω = LC(b) ω = 1/LC(c) ω = √(L/C)(d) ω = 1/√(LC)
›Reveal solutionSolution
At resonance the inductive and capacitive reactances of a series LCR circuit are equal, giving ω = 1/√(LC).
In a series L-C-R circuit driven by an AC source of angular frequency ω, the total reactance is X = X_L − X_C = ωL − 1/(ωC). The circuit is said to be at resonance when this net reactance is zero, i.e. the impedance is purely resistive (Z = R) and is minimum, so the current is maximum.
Setting X_L = X_C:
ωL = 1/(ωC)
ω² = 1/(LC)
ω = 1/√(LC)
…
- CBSE 2024Set ANNUAL1 markMCQQ.What is the value of resonant frequency ω0 of a series LCR circuit ?(a) LC(b) 1 / LC(c) √LC(d) 1 / √LC
›Reveal solutionSolution
Resonance in a series LCR circuit occurs when inductive and capacitive reactances cancel, giving ω0 = 1/√(LC).
In a series LCR circuit, the impedance is Z = √[R² + (XL − XC)²], with XL = ωL and XC = 1/(ωC).
…
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