Q.A charged $30\ \mu\text{F}$ capacitor is connected to a $27\ \text{mH}$ inductor. What is the angular frequency of free oscillations of the circuit?

Showing the 12 most recent of 36 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.To have resonance in a LCR series circuit, the impedance (Z) value (A) $Z > R$ (B) $Z = R$ (C) $Z < R$ (D) $Z = 0$

   ›Reveal solutionSolution

   At series resonance the reactances cancel exactly, so the impedance equals the pure resistance: $Z=R$.

   Concept and Intuition

   In a series LCR circuit, impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}$. Resonance is defined precisely as the condition where the inductive and capacitive reactances are equal in magnitude and opposite in phase effect, so they cancel each other in the impedance expression, leaving only the resistive part.

   Step-by-Step Solution

   1. Impedance of a series LCR circuit: $Z = \sqrt{R^2+(X_L-X_C)^2}$.
   2. At resonance, by definition, $X_L = X_C$.
   3. Substituting: $Z = \sqrt{R^2+0^2} = R$.

   Common Mistakes …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In the given circuit, the readings of voltmeters $V_1$ and $V_2$ are 300 V each. The reading of voltmeter $V_3$ and ammeter A are respectively: [FIGURE - a series circuit with an inductor L, a capacitor C, and a resistor $R = 100\ \Omega$ connected in series; a voltmeter $V_1$ is connected across L, $V_2$ across C, and $V_3$ across R; an ammeter A is in series with the circuit; the circuit is powered by an AC source of 220 V, 50 Hz] (A) 100 V, 2.0 A (B) 150 V, 2.2 A (C) 220 V, 2.0 A (D) 220 V, 2.2 A

   ›Reveal solutionSolution

   Equal $V_L$ and $V_C$ readings cancel in the series-LCR phasor sum, so the resistor takes the full 220 V source voltage, giving $I = 2.2$ A.

   Concept and Intuition

   In a series LCR circuit, the voltages across L and C are always 180° out of phase with each other (they point in opposite directions on the phasor diagram), while the resistor voltage is in phase with the current and perpendicular to both. The source voltage is the phasor sum:

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

   When $V_L = V_C$ (as given, both 300 V), the reactive terms cancel completely, leaving $V = V_R$. This is exactly the condition of resonance-like behaviour — even though $V_L$ and $V_C$ individually can be much larger than the source voltage, they cancel each other out, and the resistor "sees" the entire applied EMF.

   Step-by-Step Solution

   1. Given $V_1 = V_L = 300$ V, $V_2 = V_C = 300$ V, source $= 220$ V (rms).
   2. Phasor relation: $V_{source} = \sqrt{(V_L-V_C)^2 + V_R^2}$.
   3. Since $V_L = V_C$, $(V_L - V_C) = 0$, so $V_{source} = V_R$.
   4. Hence $V_3 = V_R = 220$ V (matching the given source voltage, consistent — this is the resonance condition). …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.To have dissipative power in a LCR series circuit to be half (A) current amplitude $= 2 \times$ (Maximum current amplitude) (B) current amplitude $= \dfrac{\text{(Maximum current amplitude)}}{2}$ (C) current amplitude $= (\text{Maximum current amplitude})^{\frac{1}{2}}$ (D) current amplitude $= \dfrac{\text{(Maximum current amplitude)}}{\sqrt{2}}$

   ›Reveal solutionSolution

   This tests the half-power point condition in a driven LCR series circuit, central to defining bandwidth and quality factor. Answer: current amplitude $= \dfrac{I_{0,max}}{\sqrt2}$.

   Concept and Intuition

   In a series LCR circuit driven at varying frequency, the current amplitude $I_0(\omega)$ peaks at resonance ($I_{0,max}$) and falls off away from resonance. Since the power dissipated depends on the square of the current amplitude, halving the power does not mean halving the current — it means reducing the current amplitude by a factor of $\sqrt2$ (because squaring $1/\sqrt2$ gives $1/2$). This is exactly the condition used to define the 'half-power frequencies' that set the resonance bandwidth.

   Step-by-Step Solution

   1. Power dissipated for a given current amplitude: $P \propto I_0^2$ (with everything else, e.g. the resistance, held fixed for the comparison).
   2. Maximum power (at resonance): $P_{max} \propto I_{0,max}^2$.
   3. We want $P = \dfrac{P_{max}}{2}$: $I_0^2 = \dfrac{I_{0,max}^2}{2}$.
   4. Taking the square root: $I_0 = \dfrac{I_{0,max}}{\sqrt2}$.

   Common Mistakes …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.In the figure shown, three ac voltmeters are connected. At resonance, [FIGURE: an ac source connected in series to a resistor R, an inductor L, and a capacitor C; voltmeter $V_1$ is connected across R; voltmeter $V_2$ is connected across the series combination of L and C; voltmeter $V_3$ is connected across the same L-C combination on the source side] (A) $V_2 = 0$ (B) $V_1 = 0$ (C) $V_3 = 0$ (D) $V_3 = V_2 \neq 0$

   ›Reveal solutionSolution

   This is a series RLC resonance question read from the figure: $V_1$ is across R, $V_2$ is across the L–C series combination, and $V_3$ is across the whole branch. At resonance $X_L=X_C$, so the L and C voltages cancel exactly — meaning $V_2=0$.

   Concept and Intuition

   In a series RLC circuit, the current $I$ is common to all elements. The voltage across the inductor leads the current by $90°$, while the voltage across the capacitor lags it by $90°$ — so $V_L$ and $V_C$ are exactly $180°$ out of phase with each other. At resonance, $X_L = X_C$, so $V_L = IX_L$ and $V_C = IX_C$ have equal magnitude too. Two phasors of equal magnitude, opposite in phase, sum to zero.

   Step-by-Step Solution

   1. At resonance, $\omega L = \dfrac{1}{\omega C} \Rightarrow X_L = X_C$.
   2. $V_L = IX_L$ and $V_C = IX_C$ are equal in magnitude.
   3. Since $V_L$ leads $I$ by $90°$ and $V_C$ lags $I$ by $90°$, they are anti-phase to each other; their phasor sum $V_2 = V_L + V_C = 0$. …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In a series LCR circuit the resistance, inductive reactance and capacitive reactance are in the ratio 3 : 8 : 4. If the potential difference across the resistor is 12 V, then the emf of the ac source used in the circuit is (A) 48 V (B) 16 V (C) 180 V (D) 20 V

   ›Reveal solutionSolution

   Recognizing the 3-4-5 impedance triangle hidden in the given ratio gives an emf of 20 V.

   Concept and Intuition

   In a series LCR circuit, $V_R=IR$, $V_L=IX_L$, $V_C=IX_C$ all share the same current $I$, so the given ratio $R:X_L:X_C=3:8:4$ is also the ratio $V_R:V_L:V_C$. The net reactive voltage is $V_L-V_C$ (they oppose each other, 180° out of phase), and the source emf is the phasor sum $\varepsilon=\sqrt{V_R^2+(V_L-V_C)^2}$.

   Step-by-Step Solution

   1. Let $R=3k$, $X_L=8k$, $X_C=4k$ for some common constant $k$.
   2. Current $I=\dfrac{V_R}{R}=\dfrac{12}{3k}=\dfrac{4}{k}$.
   3. Net reactance $X=X_L-X_C=4k$, so the net reactive voltage $=IX=\dfrac{4}{k}\times4k=16$ V.
   4. Impedance $Z=\sqrt{R^2+X^2}=\sqrt{(3k)^2+(4k)^2}=5k$ — a 3-4-5 triangle. …

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6. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A series LCR circuit with $R = 40\ \Omega$, $L = 3\ H$ and $C = 20\ \mu F$ is connected to a 400 V ac supply with variable frequency. When the frequency of supply equals to natural frequency of the circuit, the average power transferred to the circuit in one complete cycle is (A) 200 W (B) 4000 W (C) 6000 W (D) 800 W

   ›Reveal solutionSolution

   Tests power dissipation in a series LCR circuit at resonance, where the reactive elements cancel and the circuit behaves as a pure resistor.

   Concept and Intuition

   At resonance, the inductor's reactance $X_L=\omega L$ exactly equals the capacitor's reactance $X_C=\dfrac{1}{\omega C}$, and since they act with opposite phase in a series circuit, their effects cancel completely. The circuit's impedance collapses to just $R$, current and voltage are in phase, and the power dissipated is the same as it would be for a plain resistor connected directly to the same supply — no need to even compute $L$ or $C$ numerically for the power itself (they only matter for finding the resonant frequency, not the power at it).

   Step-by-Step Solution

   1. Resonance condition: $\omega_0 = \dfrac{1}{\sqrt{LC}}$, at which $X_L = X_C$.
   2. Impedance at resonance: $Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{R^2+0} = R = 40\,\Omega$.
   3. At resonance, current and voltage are in phase (power factor $\cos\phi=1$), so average power $= V_{rms}I_{rms}\cos\phi = V_{rms}I_{rms} = \dfrac{V_{rms}^2}{Z} = \dfrac{V_{rms}^2}{R}$. …

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7. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Match the following, if $X_L$ and $X_C$ are inductive and capacitive reactances respectively List I | List II A. $X_L = X_C$ | I. Current in phase with voltage B. $X_L < X_C$ | II. Current lags behind voltage C. $X_L > X_C$ | III. Current leads voltage (A) A – II, B – III, C – I (B) A – III, B – II, C – I (C) A – III, B – I, C – II (D) A – I, B – III, C – II

   ›Reveal solutionSolution

   In a series LCR circuit, comparing $X_L$ and $X_C$ tells you whether the circuit behaves inductively (current lags) capacitively (current leads) or resistively at resonance (current in phase); this gives A–I, B–III, C–II.

   Concept and Intuition

   In an AC series LCR circuit, the phase angle $\phi$ between current and voltage is given by $\tan\phi=\dfrac{X_L-X_C}{R}$.

   • If $X_L=X_C$, $\phi=0$: this is resonance, and the circuit is purely resistive — current and voltage are in phase.
   • If $X_L<X_C$, $\phi<0$: the circuit is net capacitive, and in a capacitive circuit current leads the applied voltage.
   • If $X_L>X_C$, $\phi>0$: the circuit is net inductive, and current lags behind the voltage.

   Step-by-Step Solution

   1. Recall $\tan\phi=(X_L-X_C)/R$.
   2. $X_L=X_C\Rightarrow\phi=0\Rightarrow$ current in phase with voltage $\Rightarrow$ matches item I.
   3. $X_L<X_C\Rightarrow\phi<0\Rightarrow$ voltage lags current, i.e. current leads voltage $\Rightarrow$ item III. …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The resonant frequency of an LC circuit is $f_0$. If a dielectric slab of constant 16 is inserted completely between the plates of the capacitor, then the resonant frequency is (A) $\dfrac{f_0}{2}$ (B) $2f_0$ (C) $\dfrac{f_0}{4}$ (D) $4f_0$

   ›Reveal solutionSolution

   This tests how the resonant frequency of an LC circuit changes when the capacitor's dielectric is changed. Answer: $\dfrac{f_0}{4}$.

   Concept and Intuition

   The resonant frequency of an LC circuit depends on both $L$ and $C$ as $f_0\propto\dfrac{1}{\sqrt{LC}}$. Filling the capacitor's gap with a dielectric multiplies its capacitance by the dielectric constant $K$ (with $L$ unaffected), so the new frequency scales as $\dfrac{1}{\sqrt{K}}$ times the original.

   Step-by-Step Solution

   1. $f_0=\dfrac{1}{2\pi\sqrt{LC}}$.
   2. Dielectric constant $K=16$ inserted completely: $C'=16C$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.In an LCR series circuit, if the potential differences across inductor, capacitor and resistor are 60 V, 30 V and 40 V respectively, then the ac voltage applied to the circuit is (A) 50 V (B) 70 V (C) 130 V (D) 60 V

   ›Reveal solutionSolution

   In a series LCR circuit the voltages across L, C, R combine via a phasor (not scalar) sum; the applied voltage works out to 50 V.

   Concept and Intuition

   In a series LCR AC circuit, the same current flows through all elements, but the voltage across the resistor is in phase with the current, while the voltage across the inductor leads by $90^\circ$ and across the capacitor lags by $90^\circ$. So $V_L$ and $V_C$ are antiparallel to each other and both perpendicular to $V_R$ on a phasor diagram. The net applied voltage is the phasor sum: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.

   Step-by-Step Solution

   1. Given: $V_L = 60$ V, $V_C = 30$ V, $V_R = 40$ V.
   2. Net reactive voltage: $V_L - V_C = 60 - 30 = 30$ V. …

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10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.For better tuning of a series LCR circuit in a communication system, the preferred combination is (A) $R = 20\ \Omega;\ L = 15\ \mathrm{H};\ C = 35\ \mu\mathrm{F}$ (B) $R = 15\ \Omega;\ L = 40\ \mathrm{H};\ C = 20\ \mu\mathrm{F}$ (C) $R = 25\ \Omega;\ L = 15\ \mathrm{H};\ C = 45\ \mu\mathrm{F}$ (D) $R = 15\ \Omega;\ L = 20\ \mathrm{H};\ C = 45\ \mu\mathrm{F}$

    ›Reveal solutionSolution

    Tuning sharpness in a series LCR circuit is governed by its quality factor $Q=\dfrac{1}{R}\sqrt{L/C}$; the combination with the smallest $R$, largest $L$, and smallest $C$ gives by far the largest $Q$, hence the best tuning.

    Concept and Intuition

    A series LCR circuit used for tuning (e.g. selecting one radio station out of many nearby frequencies) needs a sharp resonance peak — a large response at the resonant frequency and rapidly falling response away from it. The sharpness of resonance is quantified by the quality factor $Q=\dfrac{\omega_0 L}{R}=\dfrac{1}{R}\sqrt{\dfrac{L}{C}}$. Physically: low resistance $R$ means less energy dissipated per cycle (less damping); a large $L$ (relative to $C$) means the reactive energy sloshing between the inductor and capacitor is large compared to what's lost in $R$ each cycle. All of these push $Q$ up, meaning "better tuning" corresponds to small $R$, large $L$, and small $C$ together — not any one factor alone, which is why we must compute $Q$ for each option rather than eyeball individual values.

    Step-by-Step Solution

    1. Formula: $Q = \dfrac{1}{R}\sqrt{\dfrac{L}{C}}$.
    2. Option (A): $R=20,\ L=15,\ C=35\,\mu\mathrm{F}$: $L/C = 15/35\times10^{-6}=4.286\times10^{5}$; $\sqrt{\cdot}\approx654.7$; $Q\approx654.7/20\approx32.7$.
    3. Option (B): $R=15,\ L=40,\ C=20\,\mu\mathrm{F}$: $L/C=40/20\times10^{-6}=2\times10^{6}$; $\sqrt{\cdot}\approx1414.2$; $Q\approx1414.2/15\approx94.3$.
    4. Option (C): $R=25,\ L=15,\ C=45\,\mu\mathrm{F}$: $L/C=15/45\times10^{-6}=3.333\times10^{5}$; $\sqrt{\cdot}\approx577.4$; $Q\approx577.4/25\approx23.1$. …

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11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.If a resistor of resistance $4\ \Omega$, a capacitor of capacitive reactance $6\ \Omega$ and an inductor of inductive reactance $9\ \Omega$ are connected in series with an ac source, then the impedance of the circuit is (A) $19\ \Omega$ (B) $11\ \Omega$ (C) $7\ \Omega$ (D) $5\ \Omega$

    ›Reveal solutionSolution

    For a series R-L-C circuit, only the net reactance (the difference between inductive and capacitive reactance) combines with resistance in quadrature to give the impedance.

    Concept and Intuition

    In a series AC circuit, the voltage across the inductor and the voltage across the capacitor are 180° out of phase with each other (both being 90° out of phase with the current, but in opposite senses). So their reactances partially cancel rather than add: the net reactance is $X = X_L - X_C$. The resistor's voltage is in phase with the current, 90° out of phase with the net reactive voltage, so resistance and net reactance combine as the two legs of a right triangle to give the impedance.

    Step-by-Step Solution

    1. Given: $R = 4\,\Omega$, $X_C = 6\,\Omega$, $X_L = 9\,\Omega$.
    2. Net reactance: $X = X_L - X_C = 9 - 6 = 3\,\Omega$. …

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12. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.A $100\Omega$ resistor, a $50\mu F$ capacitor and an inductor are connected in series to an ac source of frequency 50 Hz. If the circuit is in resonance, then the impedance of the circuit is (A) $100\Omega$ (B) $200\Omega$ (C) $180\Omega$ (D) $250\Omega$

    ›Reveal solutionSolution

    At resonance in a series RLC circuit, the reactive parts cancel and the circuit behaves purely resistively, so impedance equals R.

    Concept and Intuition

    In a series RLC circuit, impedance is $Z=\sqrt{R^2+(X_L-X_C)^2}$. At resonance, $X_L=X_C$ by definition, so the reactive term vanishes entirely, leaving only R.

    Step-by-Step Solution

    1. At resonance: $X_L=X_C$.
    2. $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{R^2+0}=R$. …

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