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 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.An LC circuit with negligible resistance containing a 20 mH inductor and a 50 $\mu$F capacitor with an initial charge of 10 mC is closed at t = 0. Then the minimum time taken (in $\mu$s) for the total energy to be shared equally between the inductor and capacitor is (A) $375\pi$ (B) $125\pi$ (C) $500\pi$ (D) $250\pi$

   ›Reveal solutionSolution

   Energy is shared equally when $q = q_0/\sqrt2$, i.e. $\omega t = \pi/4$; with $\omega = 1000\ \text{rad/s}$ this gives $t = 250\pi\ \mu\text{s}$.

   In an ideal LC circuit the charge oscillates as $q = q_0\cos(\omega t)$, with angular frequency

   $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(20\times 10^{-3})(50\times 10^{-6})}} = \frac{1}{\sqrt{10^{-6}}} = 1000\ \text{rad/s}.$

   The capacitor energy is $U_C = \dfrac{q^2}{2C}$ and the total energy is $U = \dfrac{q_0^2}{2C}$. For the energy to be shared equally between capacitor and inductor, $U_C = \tfrac12 U$:

   $\frac{q^2}{2C} = \frac12\cdot\frac{q_0^2}{2C} \;\Rightarrow\; q^2 = \frac{q_0^2}{2} \;\Rightarrow\; q = \frac{q_0}{\sqrt2}.$ …

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2. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.In the given circuit, if the potential difference between the points A and B is 90 V and the resistance of the voltmeter is 6000 $\Omega$, then the reading of the voltmeter is (A) 40 V (B) 50 V (C) 30 V (D) 45 V

   ›Reveal solutionSolution

   The $6000\ \Omega$ voltmeter in parallel with the $2000\ \Omega$ resistor gives $1500\ \Omega$; with $1200\ \Omega$ in series across $90\ \mathrm{V}$, the p.d. across the parallel section — which is what the meter shows — is $\mathbf{50\ V}$. Option (B).

   The concept: a voltmeter is part of the circuit

   An ideal voltmeter has infinite resistance and draws no current, so it reports the p.d. that existed before it was connected. A real voltmeter has a large but finite resistance, so when you place it across a component:

   • it forms a parallel combination with that component;
   • the parallel resistance is always less than either branch, so the section's share of the total voltage drops;
   • the meter faithfully reports this new, reduced p.d. — the meter is not lying, the circuit has genuinely changed. This is the loading effect, and it is why a good voltmeter is designed with resistance far larger than the resistances it is used across.

   This question is designed so that the loading is deliberately noticeable ($6000\ \Omega$ is only $3\times$ the $2000\ \Omega$ resistor).

   Step-by-step

   1. Combine the voltmeter with the resistor it is across (parallel):

   $\frac{1}{R_p} = \frac{1}{2000} + \frac{1}{6000} = \frac{3+1}{6000} = \frac{4}{6000} \;\Rightarrow\; R_p = 1500\ \Omega$

   2. Total resistance between A and B (this parallel section is in series with the $1200\ \Omega$ resistor):

   $R_{AB} = 1200 + 1500 = 2700\ \Omega$

   3. Current drawn from the supply: $I = \frac{V_{AB}}{R_{AB}} = \frac{90}{2700} = \frac{1}{30}\ \mathrm{A}$ …

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3. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The quality factor of a series LCR circuit with L = 0.12 H, C = 480 nF and R = 25 $\Omega$ connected to a 220 V variable frequency supply is (A) 576 (B) 1152 (C) 60 (D) 20

   ›Reveal solutionSolution

   The quality factor ($Q$) of a series LCR circuit measures the sharpness of its resonance. It is calculated using the inductance ($L$), capacitance ($C$), and resistance ($R$) of the circuit. For the given values, the quality factor is $\boxed{20}$.

   Concept and Intuition

   A series LCR circuit consists of an inductor ($L$), a capacitor ($C$), and a resistor ($R$) connected in series. When an alternating current (AC) voltage is applied to such a circuit, its impedance (total opposition to current flow) varies with the frequency of the supply.

   At a specific frequency, known as the resonant frequency ($\omega_0$), the inductive reactance ($X_L = \omega L$) becomes equal in magnitude to the capacitive reactance ($X_C = \frac{1}{\omega C}$). At this point, they cancel each other out, and the circuit's impedance is purely resistive and at its minimum value ($Z = R$). This leads to a maximum current flow for a given voltage, a phenomenon called resonance.

   The quality factor ($Q$) of a series LCR circuit is a dimensionless parameter that quantifies the sharpness of this resonance.

   • A high $Q$ factor indicates a very sharp and narrow resonance curve, meaning the circuit is highly selective to frequencies close to its resonant frequency. Such a circuit stores a large amount of energy compared to the energy it dissipates per cycle. This is desirable in applications like radio tuning, where you want to pick up a specific frequency very precisely.
   • A low $Q$ factor indicates a broad and flat resonance curve, meaning the circuit responds to a wider range of frequencies. It dissipates a significant amount of energy compared to what it stores.

   Physically, the quality factor can be understood as the ratio of the energy stored in the circuit (in the inductor's magnetic field or capacitor's electric field) to the energy dissipated per cycle by the resistor.

   The quality factor $Q$ for a series LCR circuit is given by:

   $Q = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$

   where $\omega_0 = \frac{1}{\sqrt{LC}}$ is the resonant angular frequency.

   The supply voltage (220 V in this case) is irrelevant for calculating the quality factor itself, as $Q$ is an intrinsic property of the LCR components.

   Step-by-step Derivation

   1. Identify the given parameters:

      We are given the following values for the series LCR circuit:

      • Inductance, $L = 0.12 \text{ H}$
      • Capacitance, $C = 480 \text{ nF}$
      • Resistance, $R = 25 \text{ } \Omega$

   2. Convert units to SI base units: …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The frequency band of standard amplitude modulated broadcast is (A) $896 - 901$ MHz (B) $88 - 108$ MHz (C) $540 - 1600$ kHz (D) $3.7 - 4.2$ GHz

   ›Reveal solutionSolution

   Standard AM broadcast uses medium-wave frequencies in the kHz range, specifically 540–1600 kHz, which corresponds to option (C).

   The question asks about the frequency band for standard amplitude modulated (AM) broadcast — the kind you tune into on an AM radio. This is a factual recall point from communication systems, but it helps to understand why these frequencies are chosen.

   AM broadcast works by varying the amplitude of a carrier wave to encode audio information. The carrier frequency must be high enough to carry audio (which goes up to about 5 kHz for AM) but low enough to travel long distances via ground wave propagation. Ground waves follow the Earth’s curvature and are most effective at lower frequencies — typically below a few MHz. That’s why AM broadcast uses the medium-wave (MW) band, which lies in the kHz range, not the MHz or GHz range.

   Now let’s check each option:

   1. Option (A): $896 - 901$ MHz — This is in the UHF range, used for mobile phones and some TV broadcasts, not AM radio. Too high for ground wave propagation.

   2. Option (B): $88 - 108$ MHz — This is the FM broadcast band. FM uses frequency modulation, not amplitude modulation, and operates at VHF frequencies. So this is incorrect for AM. …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If L and C are inductance and capacitance respectively, then the dimensional formula of $(LC)^{-\frac{1}{2}}$ is (A) $[M^0L^0T^{-1}]$ (B) $[M^1L^1T^{-1}]$ (C) $[M^0L^1T^1]$ (D) $[M^0L^0T^{-2}]$

   ›Reveal solutionSolution

   The quantity $(LC)^{-1/2}$ has the dimensions of frequency (angular frequency), which is $[M^0 L^0 T^{-1}]$. The correct option is (A).

   ---

   The key insight is that $L$ and $C$ together determine the natural frequency of an LC circuit. In physics, the product $LC$ appears in the expression for the angular frequency $\omega = 1/\sqrt{LC}$. Since $\omega$ is a rate (radians per second), its dimension is simply inverse time. So we expect $(LC)^{-1/2}$ to have dimensions $[T^{-1}]$.

   Let’s verify this formally using dimensional analysis.

   1. Recall the dimensions of inductance $L$. Inductance is defined by $V = L \, \frac{dI}{dt}$. Voltage $V$ has dimensions $[M L^2 T^{-3} A^{-1}]$ (since $V = \text{energy/charge}$ and energy $= ML^2T^{-2}$, charge $= AT$). Current $I$ has dimensions $[A]$, so $dI/dt$ has $[A T^{-1}]$. Hence:

   $[L] = \frac{[V]}{[dI/dt]} = \frac{M L^2 T^{-3} A^{-1}}{A T^{-1}} = M L^2 T^{-2} A^{-2}.$

   2. Recall the dimensions of capacitance $C$. Capacitance is defined by $Q = CV$, where charge $Q$ has dimensions $[A T]$. So:

   $[C] = \frac{[Q]}{[V]} = \frac{A T}{M L^2 T^{-3} A^{-1}} = M^{-1} L^{-2} T^{4} A^{2}.$

   3. Multiply $L$ and $C$.

   $[L C] = (M L^2 T^{-2} A^{-2}) \cdot (M^{-1} L^{-2} T^{4} A^{2}) = M^{0} L^{0} T^{2} A^{0} = [T^2].$

   Notice that all mass, length, and current dimensions cancel perfectly — leaving only time squared.

   4. Take the square root and then the reciprocal. …

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6. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.A storage battery of emf 10 V and internal resistance 1 $\Omega$ is being charged by a 100 V dc supply using a series resistor of 17 $\Omega$. The terminal voltage of the battery during charging is (A) 25 V (B) 30 V (C) 20 V (D) 15 V

   ›Reveal solutionSolution

   The charging current is $5$ A, giving a terminal voltage of $10 + 5(1) = 15$ V.

   The $100\,\text{V}$ supply drives current against the battery's emf through the total resistance (series resistor + internal resistance).

   Charging current:

   $I = \frac{V_{\text{supply}} - \varepsilon}{R + r} = \frac{100 - 10}{17 + 1} = \frac{90}{18} = 5\,\text{A}$ …

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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.Three capacitors of capacitances $10\,\mu\mathrm{F}$, $5\,\mu\mathrm{F}$ and $20\,\mu\mathrm{F}$ are connected in series with a $14\,\mathrm{V}$ dc supply. The charge on $5\,\mu\mathrm{F}$ capacitor is (A) $20\,\mu\mathrm{C}$ (B) $40\,\mu\mathrm{C}$ (C) $70\,\mu\mathrm{C}$ (D) $2.8\,\mu\mathrm{C}$

   ›Reveal solutionSolution

   In a series capacitor network, each capacitor stores the same charge. The equivalent capacitance is found via the reciprocal sum, and the common charge equals that equivalent capacitance times the applied voltage. Here the charge on the 5 µF capacitor is 40 µC.

   Concept & Intuition

   When capacitors are connected in series to a DC source, the same amount of charge flows onto each capacitor’s plates (because the current is the same through each, and charge accumulates at the same rate). The total voltage divides inversely with capacitance. So instead of computing individual voltages, we can find the equivalent series capacitance and multiply by the total voltage — that product gives the common charge on every capacitor.

   ---

   1. Find the equivalent series capacitance For capacitors in series, the reciprocal of the equivalent capacitance is the sum of the reciprocals:

   $\frac{1}{C_{\text{eq}}} = \frac{1}{10} + \frac{1}{5} + \frac{1}{20}$

   (all capacitances in µF). Compute:

   $\frac{1}{10} = 0.1,\quad \frac{1}{5} = 0.2,\quad \frac{1}{20} = 0.05$

   Sum = $0.1 + 0.2 + 0.05 = 0.35$.

   Thus:

   $C_{\text{eq}} = \frac{1}{0.35} = \frac{100}{35} = \frac{20}{7}\,\mu\mathrm{F}.$

   2. Determine the common charge The total voltage across the series combination is $V = 14\,\mathrm{V}$. The charge stored on the equivalent capacitor (and hence on each individual capacitor) is:

   $Q = C_{\text{eq}} \cdot V = \frac{20}{7}\,\mu\mathrm{F} \times 14\,\mathrm{V} = 20 \times 2 = 40\,\mu\mathrm{C}.$

   3. Interpret the result …

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8. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.Resonance phenomenon is exhibited by a circuit only if (A) L and R are present in the circuit (B) C and R are present in the circuit (C) R and Z are present in the circuit (D) L and C are present in the circuit

   ›Reveal solutionSolution

   Resonance in an AC circuit requires an energy exchange between electric and magnetic fields, which demands both inductance and capacitance. The answer is (D).

   Why resonance needs both L and C

   Resonance is fundamentally about energy oscillation. In a mechanical system, a swing exchanges kinetic and potential energy at its natural frequency. The electrical analog requires two energy-storage elements that can trade energy back and forth:

   • An inductor stores energy in its magnetic field ($U_L = \frac{1}{2}LI^2$)
   • A capacitor stores energy in its electric field ($U_C = \frac{1}{2}\frac{Q^2}{C}$)

   At resonance, these two elements exchange energy at a characteristic frequency where their reactive effects cancel. The inductor's inductive reactance $X_L = \omega L$ exactly balances the capacitor's capacitive reactance $X_C = \frac{1}{\omega C}$, making the circuit behave purely resistively at that instant.

   Without both elements, there is no energy exchange and no resonance frequency.

   Examining each option

   1. Option (A): L and R

      A resistor dissipates energy as heat; it does not store it. An RL circuit has no second storage element to exchange energy with the inductor. The impedance varies with frequency, but there is no resonance peak—just a monotonic change. No resonance occurs.

   2. Option (B): C and R

      Similarly, an RC circuit has only one energy-storage element (the capacitor). The resistor cannot participate in energy oscillation. The impedance decreases with frequency, but again there is no resonant frequency where reactances cancel. No resonance.

   3. Option (C): R and Z

      This option is physically unclear. $Z$ denotes impedance, which is a property of the entire circuit, not a component. You cannot "have impedance present" as a circuit element. This is not a meaningful statement.

   4. Option (D): L and C

      An LC circuit (even with resistance present) exhibits resonance. The resonance condition is: …

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9. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.The quality factor of a series LCR resonant circuit is 75. If the resistance is decreased by 50% and the inductance is increased by 100%, then the quality factor of the circuit is (A) $150$ (B) $150\sqrt{2}$ (C) $300$ (D) $300\sqrt{2}$

   ›Reveal solutionSolution

   The quality factor $Q$ of a series LCR circuit is $Q = \frac{1}{R}\sqrt{\frac{L}{C}}$. Changing $R$ and $L$ as described multiplies $Q$ by a factor of $2\sqrt{2}$, so the new $Q$ is $75 \times 2\sqrt{2} = 150\sqrt{2}$. The correct option is (B).

   Concept & Intuition

   The quality factor $Q$ measures how underdamped an RLC circuit is — essentially, how sharply it resonates. For a series LCR circuit, $Q$ depends on resistance, inductance, and capacitance in a specific way:

   $Q = \frac{1}{R}\sqrt{\frac{L}{C}}.$

   Notice that $Q$ is inversely proportional to $R$ and proportional to $\sqrt{L}$. So if we change $R$ and $L$ while keeping $C$ fixed, we can directly compute the new $Q$ by multiplying the original by the ratio of changes. The problem gives percentage changes, which translate to simple multiplicative factors.

   Step-by-step solution

   1. Write the original quality factor. For a series LCR circuit,

   $Q_0 = \frac{1}{R}\sqrt{\frac{L}{C}}.$

   We are told $Q_0 = 75$.

   2. Interpret the changes.

      • Resistance is decreased by 50%: new resistance $R' = R - 0.5R = 0.5R$.
      • Inductance is increased by 100%: new inductance $L' = L + 1.0L = 2L$.
      • Capacitance is not mentioned, so it remains unchanged: $C' = C$.

   3. Write the new quality factor.

      Substitute the new values into the formula:

   $Q' = \frac{1}{R'}\sqrt{\frac{L'}{C'}} = \frac{1}{0.5R}\sqrt{\frac{2L}{C}}.$

   4. Simplify the expression. $$$$

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10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.For the displacement current through the plates of a parallel plate capacitor of capacitance $30\,\mu\mathrm{F}$ to be $150\,\mu\mathrm{A}$, the potential difference across the plates of the capacitor has to vary at the rate of (A) $10\ \mathrm{Vs}^{-1}$ (B) $5\ \mathrm{Vs}^{-1}$ (C) $15\ \mathrm{Vs}^{-1}$ (D) $20\ \mathrm{Vs}^{-1}$

    ›Reveal solutionSolution

    The displacement current between capacitor plates equals the conduction current in the wires, which is $C \frac{dV}{dt}$. Setting $C \frac{dV}{dt} = 150\,\mu\mathrm{A}$ with $C = 30\,\mu\mathrm{F}$ gives $\frac{dV}{dt} = 5\ \mathrm{V/s}$, so the correct option is (B).

    The key idea is that displacement current is not a separate physical current — it’s the term Maxwell added to make the total current continuous in circuits with capacitors. For a parallel-plate capacitor, the displacement current between the plates exactly equals the conduction current in the connecting wires. That conduction current, in turn, is related to how fast the voltage across the capacitor changes: $I = C \frac{dV}{dt}$. So the problem reduces to a simple rate calculation.

    1. Recall the definition of displacement current For a parallel-plate capacitor, the displacement current $I_d$ through the region between the plates is given by

    $I_d = \varepsilon_0 \frac{d\Phi_E}{dt},$

    where $\Phi_E$ is the electric flux. For a uniform field between the plates, $\Phi_E = E A$, and $E = V/d$, so $\Phi_E = \frac{V}{d} A$. But the capacitance is $C = \varepsilon_0 A / d$, so

    $I_d = \varepsilon_0 \frac{d}{dt}\left(\frac{V A}{d}\right) = \varepsilon_0 \frac{A}{d} \frac{dV}{dt} = C \frac{dV}{dt}.$

    This shows that the displacement current equals $C$ times the rate of change of voltage.

    2. Set the displacement current equal to the given value …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.A potentiometer balances at 44 cm when a cell of internal resistance $1~\Omega$ is in the secondary circuit. To obtain the balancing point at 40 cm, the resistance to be connected parallel to cell is (A) $20~\Omega$ (B) $10~\Omega$ (C) $30~\Omega$ (D) $5~\Omega$

    ›Reveal solutionSolution

    The key idea is that the balancing length is proportional to the terminal voltage of the cell. By shunting the cell with a parallel resistor, we reduce its terminal voltage to match the new balancing length. The required parallel resistance is 10 Ω.

    ---

    Concept and Intuition

    In a potentiometer, the balancing length is directly proportional to the potential difference across the cell being measured (when the cell is in the secondary circuit).

    If the cell has internal resistance, its terminal voltage is less than its emf when current flows. By connecting a resistor in parallel with the cell, we effectively lower the total external resistance, which changes the current drawn from the cell and thus its terminal voltage.

    The problem gives two balancing lengths: one for the cell alone, and one for the cell with a parallel resistor. The ratio of these lengths equals the ratio of the corresponding terminal voltages.

    ---

    Step-by-step solution

    1. Relate balancing length to terminal voltage For a potentiometer, the balancing length $l$ is proportional to the potential difference across the cell’s terminals:

    $V \propto l$

    So,

    $\frac{V_2}{V_1} = \frac{l_2}{l_1}$

    Here, $l_1 = 44\ \text{cm}$ (cell alone), $l_2 = 40\ \text{cm}$ (cell with parallel resistor).

    2. Express terminal voltages Let the emf of the cell be $E$ and its internal resistance $r = 1\ \Omega$.

       • Case 1 (cell alone): The secondary circuit is just the cell and the potentiometer wire. The current through the cell is $I_1 = \frac{E}{R + r}$, where $R$ is the resistance of the potentiometer wire segment up to the balance point. But since the potentiometer draws negligible current at balance, the terminal voltage is simply $E$ (no load). Wait — careful: In a standard potentiometer setup, the cell in the secondary circuit is not under load because the galvanometer draws no current at balance. So the terminal voltage equals the emf $E$.

         However, the problem states the cell has internal resistance, implying that when we connect a parallel resistor, current does flow through the cell, causing a voltage drop across its internal resistance. So for the first case (no parallel resistor), the cell is essentially open-circuit, so $V_1 = E$.

       • Case 2 (with parallel resistor $R_p$): The cell now supplies current to the parallel resistor. The total external resistance is $R_p$. The current from the cell is

         $I_2 = \frac{E}{r + R_p}$ …

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12. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.A series LCR circuit with inductance 20 mH, capacitance 50 $\mu$F and resistance 25 $\Omega$ is connected to a variable frequency supply of 220 V. The frequency of the source at which resonance occurs is (A) 1000 Hz (B) 159 Hz (C) 59 Hz (D) 50 Hz

    ›Reveal solutionSolution

    Resonant frequency $f=1/(2\pi\sqrt{LC}) \approx 159$ Hz.

    Concept: At resonance in a series LCR circuit, $X_L = X_C$, so $\omega_0 = 1/\sqrt{LC}$ and $f_0 = 1/(2\pi\sqrt{LC})$. The resistance and supply voltage do not affect the resonant frequency.

    Calculation:

    • $L = 20\text{ mH} = 0.02\text{ H}$
    • $C = 50\ \mu\text{F} = 50\times10^{-6}\text{ F}$
    • $LC = 0.02 \times 50\times10^{-6} = 1\times10^{-6}\ \text{s}^2$ …

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