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 \( C_2 = 0.6\ \mu\text{F} \).
In a charged capacitor, the energy resides in
1. The positive charges
2. Both the positive and negative charges
3. The field between the plates
4. Around the edge of the capacitor plates
View Answer
In a charged capacitor, the electrostatic potential energy is stored in the electric field that exists between the plates of the capacitor. The energy density is given by \(u_E = \frac{1}{2}\epsilon_0 E^2\).
A capacitor of capacity \(C\) is connected with a battery of potential \(V\) in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential \(V\) again, the energy given by the battery will be
1. \(CV^2 / 4\)
2. \(CV^2 / 2\)
3. \(3CV^2 / 4\)
4. \(CV^2\)
View Answer
When the plate separation is halved, capacitance becomes \(2C\). Keeping charge constant at \(Q = CV\), to recharge it back to potential \(V\), the final charge is \(Q' = 2CV\). The charge flowing from the battery is \(\Delta Q = 2CV - CV = CV\). The energy supplied by the battery is \(W_b = \Delta Q \cdot V = CV^2\).
\(N\) 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 \(NV\).
A parallel plate capacitor is charged from a cell and then isolated from it. The separation between the plate is now increased
1. the force of attraction between the plates will decrease
2. the field in the region between the plates will change
3. the energy stored in the capacitor will increase
4. the potential difference between the plates will decreases
View Answer
Since the capacitor is isolated, its charge \(Q\) remains constant. When the separation \(d\) increases, the capacitance \(C = \frac{\epsilon_0 A}{d}\) decreases. Since \(U = \frac{Q^2}{2C}\), the stored energy increases due to the work done against the electrostatic attraction.
A parallel-plate capacitor has a plate area of \(0.3\text{ m}^2\) and a plate separation of \(0.1\text{ mm}\). If the charge on each plate has a magnitude of \(5 \times 10^{-6}\text{ C}\) then the force exerted by one plate on the other has a magnitude of about :
1. \(0\)
2. \(5\text{ N}\)
3. \(1 \times 10^4\text{ N}\)
4. \(9 \times 10^5\text{ N}\)
View Answer
The force of attraction between the plates of a parallel plate capacitor is given by \(F = \frac{Q^2}{2 \epsilon_0 A}\). Substituting the values: \(F = \frac{(5 \times 10^{-6})^2}{2 \times 8.85 \times 10^{-12} \times 0.3} \approx 4.7\text{ N}\), which is about \(5\text{ N}\).
A capacitor is connected to a battery. The electric energy stored in it is \(E\). If the separation between the plates is doubled, what will be the energy on the capacitor?
1. \(0.25 E\)
2. \(0.50 E\)
3. \(E\)
4. \(2 E\)
View Answer
The energy stored is \(E = \frac{1}{2} C V^2\). With the battery connected, potential \(V\) is constant. Doubling the separation halves the capacitance, so the new capacitance is \(C' = C/2\), and the stored energy becomes \(E' = E/2 = 0.50 E\).
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 \(V_c = \frac{Q_{\text{total}}}{C_{\text{total}}} = \frac{1500 + 100}{1 + 1} = 800\text{ V}\).