1 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

Q: 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

**Correct option: B — $5\sqrt{2} \text{ V}$.** - With $V_R = V_C = V_L = 10 \text{ V}$, the circuit is at **resonance**, so the source voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2} = 10 \text{ V}$, and $X_L = X_C = R$. - Shorting the capacitor leaves a series R–L across the same 10 V source. New impedance $Z' = \sqrt{R^2 + X_L^2} = R\sqrt{2}$, so current $I' = 10/(R\sqrt{2})$. - Voltage across inductor: $V_L' = I'\cdot X_L = (10/R\sqrt{2})\cdot R = 10/\sqrt{2}$ = **$5\sqrt{2} \text{ V}$**.

Alternating Current

Series LCR circuit with capacitor short-circuited

Options

(A)

10 \text{ V}

(B)

5\sqrt{2} \text{ V}

(C)

\frac{5}{\sqrt{2}} \text{ V}

(D)

10\sqrt{2} \text{ V}

Official Answer

Correct option: B — $5\sqrt{2} \text{ V}$ .

With $V_R = V_C = V_L = 10 \text{ V}$ , the circuit is at resonance, so the source voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2} = 10 \text{ V}$ , and $X_L = X_C = R$ .
Shorting the capacitor leaves a series R–L across the same 10 V source. New impedance $Z' = \sqrt{R^2 + X_L^2} = R\sqrt{2}$ , so current $I' = 10/(R\sqrt{2})$ .
Voltage across inductor: $V_L' = I'\cdot X_L = (10/R\sqrt{2})\cdot R = 10/\sqrt{2}$ = $5\sqrt{2} \text{ V}$ .

series LCR circuitresonancephasor voltagesimpedancecapacitor short circuitedvoltage across inductorX_L equals X_C5√2 volts

Marking Scheme

11 mark: correct option B ( $5\sqrt{2} \text{ V}$ ).
2Reasoning credited: recognising resonance ( $R = X_L = X_C$ , source 10 V), new impedance $R\sqrt{2}$ , and $V_L = 10/\sqrt{2} = 5\sqrt{2} \text{ V}$ .

Hint

Equal $V_R, V_C, V_L$ means resonance, so $R = X_L = X_C$ and source = 10 V. After shorting C, use $Z = \sqrt{R^2 + X_L^2}$ with $X_L = R$ to get the new current, then $V_L = I\cdot X_L$ .

Quick Oral Answer

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

Analysis & Explanation

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

Step 1 — Interpret the original state

Equal voltages $V_R = V_C = V_L = 10 \text{ V}$ (at the same current) mean $R = X_C = X_L$ . Since $X_L = X_C$ , the circuit is at resonance.
The applied (source) rms voltage is $V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{10^2 + 0} = 10 \text{ V}$ .

Step 2 — Short the capacitor

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

Step 3 — Voltage across the inductor

$V_L' = I'\cdot X_L = (10/(R\sqrt{2}))\cdot R = 10/\sqrt{2} = 5\sqrt{2} \approx 7.07 \text{ V}$ .

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 ( $5/\sqrt{2} \text{ V}$ ): results from dividing by 2 instead of √2 in the current.
D ( $10\sqrt{2} \text{ V}$ ): would require the current to rise, but impedance increases, so current and $V_L$ 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.

Common Mistakes

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

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.

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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 $X_L = X_C$ — the defining condition of series resonance. At resonance the reactive voltages cancel and the source voltage equals $V_R = 10 \text{ V}$ .

Why does the inductor voltage drop from 10 V to $5\sqrt{2} \text{ V}$ after shorting the capacitor?

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