Q.A series LCR circuit connected to a variable frequency 230 V source as shown below (a series loop of AC source E, resistor R, capacitor C, and inductor L), where L = 5.0 H, C = 80 μF and R = 40 Ω. (Given ω = 50 rad s^-1).

Showing the 12 most recent of 30 on this concept.

1. CBSE 2026Set 55/1/11 markMCQ
   Q.In a series LCR circuit, the voltage across the resistor, capacitor and inductor is $10\ \text{V}$ each. If the capacitor is short circuited, the voltage across the inductor will be (A) $10\ \text{V}$ (B) $5\sqrt{2}\ \text{V}$ (C) $\dfrac{5}{\sqrt{2}}\ \text{V}$ (D) $10\sqrt{2}\ \text{V}$

   ›Reveal solutionSolution

   In a series LCR circuit, when each component drops 10 V, the source voltage is 10 V (since $V_C$ and $V_L$ cancel) and the equal drops imply $X_L = X_C = R$. Shorting the capacitor leaves an RL circuit of impedance $R\sqrt{2}$, so the current becomes $I' = \dfrac{10}{R\sqrt{2}}$ and the inductor voltage is $V_L' = I' X_L = \dfrac{10}{\sqrt{2}} = 5\sqrt{2}\ \text{V}$. The answer is (B).

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   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.

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   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 $V_s$. Since $V_L$ and $V_C$ are opposite in phase (180° apart), they subtract. Given $V_R = V_L = V_C = 10\ \text{V}$:

   $V_s = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{10^2 + (10 - 10)^2} = 10\ \text{V}$

   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.

   Tip

   This cancellation is the hallmark of resonance in a series LCR circuit — at resonance, $X_L = X_C$, and the impedance is purely resistive. Here, $V_L = V_C$ implies $X_L = X_C$, so the circuit is at resonance.

   2. Determine the relationship between $R$ and $X_L$ (or $X_C$).

   At resonance, the current is $I = V_s / R = 10 / R$. The voltage across the inductor is $V_L = I X_L = (10/R) X_L = 10\ \text{V}$. Therefore:

   $\frac{10}{R} X_L = 10 \quad \Rightarrow \quad X_L = R$

   So the inductive reactance equals the resistance. Similarly, $X_C = R$ as well. …

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2. CBSE 2026Set V11 markMCQ
   Q.Power factor of a series LCR circuit is maximum when :

   (a) $X_L = X_C$

   (b) $X_C = 0$

   (c) $X_L > X_C$

   (d) $X_L < X_C$

   ›Reveal solutionSolution

   (a) $X_L = X_C$ …

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3. CBSE 2026Set ANNUAL1 mark
   Q.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),

   …

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4. CBSE 2026Set SEM31 markMCQ
   Q.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.

   …

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5. CBSE 2025Set D1 markMCQ
   Q.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:

   $X_L = X_C \ \Rightarrow\ \omega L = \frac{1}{\omega C}$

   Solving for the angular frequency,

   …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.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

   $Z=\sqrt{R^2+\left(\omega L-\dfrac{1}{\omega C}\right)^2}$ …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.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 $\phi$ between the applied voltage and current is given by $\tan\phi = \dfrac{X_L - X_C}{R}$, where $X_L = \omega L$ and $X_C = \dfrac{1}{\omega C}$.

   …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.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 $Z = \sqrt{R^2 + (X_L - X_C)^2}$ with $X_L = \omega L$ and $X_C = 1/\omega C$. Power $P = V_{rms} I_{rms}\cos\phi$ is greatest when Z is minimum (Z = R) and the current is in phase with the voltage. …

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9. CBSE 2024Set A11 markMCQ
   Q.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

   (c) L and C …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.In a series LCR circuit, resonant frequency depends on which of the following -

    (a) $\dfrac{R}{LC}$

    (b) $\dfrac{L}{C}$

    (c) $\dfrac{1}{\sqrt{LC}}$

    (d) $\dfrac{1}{\sqrt{RC}}$

    ›Reveal solutionSolution

    Resonance in a series LCR circuit occurs when inductive and capacitive reactances are equal.

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.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)

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.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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