Question 61 markMCQSection A

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

Alternating Current
Series LCR circuit with capacitor short-circuited

Options

(A)10 V10 \text{ V}
(B)52 V5\sqrt{2} \text{ V}
(C)52 V\frac{5}{\sqrt{2}} \text{ V}
(D)102 V10\sqrt{2} \text{ V}

Official answer

Correct option: B — 52 V5\sqrt{2} \text{ V}.


  • With VR=VC=VL=10 VV_R = V_C = V_L = 10 \text{ V}, the circuit is at resonance, so the source voltage is V=VR2+(VL−VC)2=10 VV = \sqrt{V_R^2 + (V_L - V_C)^2} = 10 \text{ V}, and XL=XC=RX_L = X_C = R.
  • Shorting the capacitor leaves a series R–L across the same 10 V source. New impedance Z′=R2+XL2=R2Z' = \sqrt{R^2 + X_L^2} = R\sqrt{2}, so current I′=10/(R2)I' = 10/(R\sqrt{2}).
  • Voltage across inductor: VL′=I′⋅XL=(10/R2)⋅R=10/2V_L' = I'\cdot X_L = (10/R\sqrt{2})\cdot R = 10/\sqrt{2} = 52 V5\sqrt{2} \text{ V}.
Answer keywords:series LCR circuitresonancephasor voltagesimpedancecapacitor short circuitedvoltage across inductorX_L equals X_C5√2 volts

Marking scheme

  1. 1.1 mark: correct option B (52 V5\sqrt{2} \text{ V}).
  2. 2.Reasoning credited: recognising resonance (R=XL=XCR = X_L = X_C, source 10 V), new impedance R2R\sqrt{2}, and VL=10/2=52 VV_L = 10/\sqrt{2} = 5\sqrt{2} \text{ V}.

Hint

Equal VR,VC,VLV_R, V_C, V_L means resonance, so R=XL=XCR = X_L = X_C and source = 10 V. After shorting C, use Z=R2+XL2Z = \sqrt{R^2 + X_L^2} with XL=RX_L = R to get the new current, then VL=I⋅XLV_L = I\cdot X_L.

Quick oral answer

Equal R, C and L voltages mean resonance, so the source is 10 V and XL=RX_L = R; shorting the capacitor makes the impedance R2R\sqrt{2}, the current 10/R210/R\sqrt{2}, and the inductor voltage 10/2=52 V10/\sqrt{2} = 5\sqrt{2} \text{ V}.

Common mistakes

  1. 1.Assuming the current stays the same after shorting the capacitor instead of recomputing it with the new impedance.
  2. 2.Forgetting that the source voltage (10 V) is fixed and only the circuit configuration changes.
  3. 3.Dividing by 2 instead of 2\sqrt{2}, giving 5/2 V5/\sqrt{2} \text{ V} instead of 52 V5\sqrt{2} \text{ V}.

Analysis and explanation

This is a multi-step phasor problem that rewards understanding of resonance and impedance.


Step 1 — Interpret the original state

  • Equal voltages VR=VC=VL=10 VV_R = V_C = V_L = 10 \text{ V} (at the same current) mean R=XC=XLR = X_C = X_L. Since XL=XCX_L = X_C, the circuit is at resonance.
  • The applied (source) rms voltage is V=VR2+(VL−VC)2=102+0=10 VV = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{10^2 + 0} = 10 \text{ V}.

Step 2 — Short the capacitor

  • Removing C leaves a series R–L circuit across the unchanged 10 V source, with XL=RX_L = R.
  • New impedance Z′=R2+XL2=R2+R2=R2Z' = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + R^2} = R\sqrt{2}.
  • New current I′=V/Z′=10/(R2)I' = V/Z' = 10/(R\sqrt{2}).

Step 3 — Voltage across the inductor

  • VL′=I′⋅XL=(10/(R2))⋅R=10/2=52≈7.07 VV_L' = I'\cdot X_L = (10/(R\sqrt{2}))\cdot R = 10/\sqrt{2} = 5\sqrt{2} \approx 7.07 \text{ V}.

Why the distractors are wrong

  • A (10 V): assumes the inductor voltage is unchanged, ignoring that shorting C raises impedance and lowers the current.
  • C (5/2 V5/\sqrt{2} \text{ V}): results from dividing by 2 instead of √2 in the current.
  • D (102 V10\sqrt{2} \text{ V}): would require the current to rise, but impedance increases, so current and VLV_L fall below 10 V.

Exam trap

  • The source voltage (10 V) stays fixed; only the circuit changes. Students often keep the old current instead of recomputing it with the new impedance.

Interesting facts

At resonance the inductor and capacitor voltages can each individually exceed the source voltage — here each is 10 V while the source is also 10 V — because they are 180° out of phase and cancel.

This voltage magnification at resonance (the circuit Q-factor) is exactly what tunes a radio: a series LCR circuit selects one station by resonating at its frequency.

Question details

Q6 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Alternating Current
Topic
Series LCR circuit with capacitor short-circuited
Bloom's level
apply
Syllabus topic
Unit IV: Electromagnetic Induction & AC — Alternating Current: Series LCR circuit, resonance, phasor voltages
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do we know the original circuit is at resonance?

The voltages across the inductor and capacitor are equal (10 V each) at the same current, so XL=XCX_L = X_C — the defining condition of series resonance. At resonance the reactive voltages cancel and the source voltage equals VR=10 VV_R = 10 \text{ V}.

Why does the inductor voltage drop from 10 V to 52 V5\sqrt{2} \text{ V} after shorting the capacitor?

Shorting the capacitor changes the impedance from R (at resonance) to R2+XL2=R2\sqrt{R^2 + X_L^2} = R\sqrt{2}, so the current falls by a factor √2. The inductor voltage VL=I⋅XLV_L = I\cdot X_L therefore becomes 10/2=52≈7.07 V10/\sqrt{2} = 5\sqrt{2} \approx 7.07 \text{ V}.