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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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 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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