Capacitors

Practice Questions:

Question 31: moderate

Two capacitors of capacitances \( C_1 \) and \( C_2 \) are connected in series across \( 200\text{V} \) power supply. The potential drop across \( C_1 \) is \( 120\text{V} \). A capacitor of capacitance \( 2\ \mu\text{F} \) is connected in parallel with \( C_1 \) and the potential drop across \( C_2 \) becomes \( 160\text{V} \). What are the values of \( C_1 \) and \( C_2 \) in \( \mu\text{F} \)?

1. \( 0.4 \text{ & } 0.6 \)

2. \( 0.6 \text{ & } 0.4 \)

3. \( 0.8 \text{ & } 1.2 \)

4. \( 1.2 \text{ & } 0.8 \)

View Answer

Initially, \( C_1(120) = C_2(80) \Rightarrow C_2 = 1.5 C_1 \). Finally, with parallel connection: \( (C_1 + 2)(40) = C_2(160) \). Substituting \( C_2 = 1.5 C_1 \) gives \( 40 C_1 + 80 = 240 C_1 \Rightarrow C_1 = 0.4\ \mu\text{F} \) and \( C_2 = 0.6\ \mu\text{F} \).

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Question 32: moderate

\(N\) identical capacitors are joined in parallel and the combination is charged to a potential \(V\). Now if they are separated and then joined in series then energy of combination will :

1. remain same and potential difference will also remain same

2. remain same and potential difference will become \(NV\)

3. increase \(N\) times and potential difference will become \(NV\)

4. increase \(N\) time and potential difference will remains same

View Answer

When connected in series, the charges do not change, so the total energy stored remains the same: \(U = N \times \left(\frac{1}{2} C V^2\right)\). However, the individual potential differences add up, so the total potential difference becomes \(NV\).

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Question 33: moderate

The plates of a parallel plate capacitor are charged up to 100 volt. A \(2\text{ mm}\) thick dielectric plate is inserted between the plates, then to maintain the same potential difference, the distance between the capacitor plates is increased by \(1.6\text{ mm}\). The dielectric constant of the plate is :

1. 5

2. 1.25

3. 4

4. 2.5

View Answer

The shift in plate distance to maintain constant potential is given by \(\Delta d = t\left(1 - \frac{1}{K}\right)\) where \(t\) is thickness and \(K\) is dielectric constant. Substituting the values: \(1.6 = 2\left(1 - \frac{1}{K}\right)⇒ 0.8 = 1 - \frac{1}{K} ⇒ K = 5\).

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Question 34: moderate

Condenser A has a capacity of \(15 \mu\text{F}\) when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity \(1 \mu\text{F}\) with air between the plates. Both are charged separately by a battery of \(100\text{ V}\). After charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is :

1. \(400\text{ V}\)

2. \(800\text{ V}\)

3. \(1200\text{ V}\)

4. \(1600\text{ V}\)

View Answer

Charge on A is \(15 \mu\text{F} \times 100\text{ V} = 1500 \mu\text{C}\) and on B is \(1 \mu\text{F} \times 100\text{ V} = 100 \mu\text{C}\). When dielectric of A is removed, its capacitance becomes \(1 \mu\text{F}\). The common potential is \(V_c = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{1500 + 100}{1 + 1} = 800\text{ V}\).

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Question 35: moderate

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of 30° with each other. When suspended in a liquid of density \(0.8\text{ g cm}^{-3}\), the angle remains the same. If density of the material of the sphere is \(1.6\text{ g cm}^{-3}\), the dielectric constant of the liquid is :

1. 1

2. 4

3. 3

4. 2

View Answer

When the angle remains unchanged in a liquid, the dielectric constant is given by \(K = \frac{\rho}{\rho - \sigma}\), where \(\rho\) is density of sphere and \(\sigma\) is density of liquid. Thus, \(K = \frac{1.6}{1.6 - 0.8} = \frac{1.6}{0.8} = 2\).

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Question 36: moderate

The electric field between the plates of a parallel-plate capacitor of capacitance \(2.0~\mu\text{F}\) drops to one third of its initial value in \(4.4~\mu\text{s}\) when the plates are connected by a thin wire. Find the resistance of the wire.

1. \(0.5~\Omega\)

2. \(0.1~\Omega\)

3. \(2~\Omega\)

4. \(1~\Omega\)

View Answer

The electric field in a discharging capacitor drops as \(E = E_0 e^{-t/RC}\). Given \(E = E_0/3\), we have \(RC = \frac{t}{\ln 3}\). Solving for \(R = \frac{4.4 \times 10^{-6}}{2.0 \times 10^{-6} \times 1.1} = 2~\Omega\).

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Question 37: moderate

A \(5.0~\mu\text{F}\) capacitor having a charge of \(20~\mu\text{C}\) is discharged through a wire of resistance \(5.0~\Omega\). Find the heat dissipated in the wire between 25 to 50 \(\mu\text{s}\) after the connections are made.

1. \(40\left(\frac{1}{e^2} - \frac{1}{e^4}\right)~\mu\text{J}\)

2. \(40\left(\frac{1}{e} - \frac{1}{e^2}\right)~\mu\text{J}\)

3. \(40\left(\frac{1}{e^2} - \frac{1}{e^3}\right)~\mu\text{J}\)

4. None of these

View Answer

The remaining energy in the capacitor is \(U(t) = \frac{q_0^2}{2C}e^{-2t/tau}\), where \(tau = RC = 25~\mu\text{s}\). The heat dissipated is \(H = U(t_1) - U(t_2) = \frac{q_0^2}{2C}\left(e^{-2} - e^{-4}\right)\) where \(frac{q_0^2}{2C} = 40~\mu\text{J}\).

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Question 38: moderate

Let C be the capacitance of a capacitor discharging through a resistor R. Suppose \(t_1\) is the time taken for the energy stored in the capacitor to reduce to half its initial value and \(t_2\) is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio \(t_1/t_2\) will be :

1. 2

2. 1

3. 1/2

4. ⇒1/4

View Answer

Energy is \(U \propto q^2 \propto e^{-2t/RC}\), so \(e^{-2t_1/RC} = 1/2 ⇒ t_1 = \frac{RC\ln 2}{2}\). Charge is \(q \propto e^{-t/RC}\), so \(e^{-t_2/RC} = 1/4 t_2 = 2RC\ln 2\). Thus, the ratio \(t_1/t_2 = 1/4\).

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Question 39: moderate

A capacitor of capacity \(C_0\) is connected to a battery of emf \(V_0\). When steady state is attained a dielectric slab of dielectric constant (K) is slowly introduced in the capacitor. Mark the Correct statement(s), in final steady state :

1. Magnitude of induced charge on the each surface of slab is \(C_0V_0(K - 1)\).

2. Net electric force due to induced charges on the plate is zero.

3. Force of attraction between plates of capacitor is \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\).

4. Net field due to induced charges in dielectric slab is \(\frac{8V_0(K-1)^2}{K \varepsilon_0 A}\)

View Answer

Final charge on plates \(Q = KC_0V_0\). Induced charge \(q_{\text{ind}} = Q(1 - 1/K) = KC_0V_0(1 - 1/K) = C_0V_0(K-1)\). This is correct. (C) Force of attraction between plates \(F = \frac{1}{2} C'V_0^2 / d = \frac{1}{2} (KC_0) V_0^2 / d\). Since \(C_0 = varepsilon_0 A / d\), \(F = \frac{1}{2} K (\varepsilon_0 A / d) V_0^2 / d = \frac{K \varepsilon_0 A V_0^2}{2d^2}\). This can be rewritten as \(\frac{K(C_0V_0)^2}{2 \varepsilon_0 A}\). Both A and C are correct statements. Option A is chosen as the primary answer.

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Question 40: moderate

A capacitor of capacitance \(C\) is connected to a battery of emf \(\varepsilon\) at \(t = 0\) through a resistance \(R\). Find the maximum rate at which energy is stored in the capacitor. When does the rate has this maximum value ?

1. \(\frac{\varepsilon^2}{4R}\)

2. \(\frac{\varepsilon^2}{2R}\)

3. \(RC\)

4. \(CR \ln 2\)

View Answer

The energy stored rate is \(P = \frac{\varepsilon^2}{R} e^{-t/RC}(1-e^{-t/RC})\). This is maximum when \(e^{-t/RC} = \frac{1}{2}\). The maximum rate is \(P_{max} = \frac{\varepsilon^2}{4R}\). This occurs at \(t = RC \ln 2\).

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