Two identical capacitors, have the same capacitance (C). One of them is charged to potential \(V_1\) and the other to \(V_2\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is :
1. \(\frac{1}{4} C (V_1^2 - V_2^2)\)
2. \(\frac{1}{4} C (V_1^2 + V_2^2)\)
3. \(\frac{1}{4} C (V_1 - V_2)^2\)
4. \(\frac{1}{4} C (V_1 + V_2)^2\)
View Answer
Initial energy \(U_i = \frac{1}{2}CV_1^2 + \frac{1}{2}CV_2^2\). Common potential after connection \(V = \frac{CV_1 + CV_2}{C + C} = \frac{V_1 + V_2}{2}\) for like-polarity connection. Final energy \(U_f = \frac{1}{2}(2C)V^2 = C \left(\frac{V_1 + V_2}{2}\right)^2 = \frac{C}{4}(V_1^2 + V_2^2 + 2V_1V_2)\) . Decrease in energy \(Delta U = U_i - U_f = \frac{1}{4}C(V_1^2 + V_2^2 - 2V_1V_2) = \frac{1}{4}C(V_1 - V_2)^2\).
The capacitance of a a parallel plate capacitor is (C) when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant (k). The capacitor is connected to a cell of emf (E), and the slab is taken out
1. charge (CE(k - 1)\) flows through the cell
2. energy (E^2C(k - 1)\) is absorbed by the cell.
3. the energy stored in the capacitor is reduced by (E^2C(k - 1)\)
4. the external agent has to do \frac{1}{2} E^2C(k - 1)\) amount of work to take the slab out.
View Answer
Concept: Work done by external agent in removing dielectric from capacitor connected to battery. Formula: (W_{\text{ext}} = \Delta U - W_{\text{cell}}\), where (W_{\text{cell}} = E\Delta Q\). Solution: Initial stored energy (U_1 = \frac{1}{2} kCE^2\) and charge (Q_1 = kCE\). Final stored energy (U_2 = \frac{1}{2} CE^2\) and charge (Q_2 = CE\). Change in stored energy (Delta U = U_2 - U_1 = -\frac{1}{2} CE^2(k-1)\). Charge returned to cell (Delta Q = Q_1 - Q_2 = CE(k-1)\). Work done by cell (W_{\text{cell}} = E (Q_2 - Q_1) = -E^2C(k-1)\). Work done by external agent (W_{\text{ext}} = \Delta U - W_{\text{cell}} = -\frac{1}{2} CE^2(k-1) - (-E^2C(k-1)) = \frac{1}{2} E^2C(k-1)\).
3 capacitors each of capacitance \(2\text{ }\mu\text{F}\) are to be connected to obtain an equivalent capacitance of \(3\text{ }\mu\text{F}\). Which of the following combination is possible?
1. All in series
2. All in parallel
3. 2 in parallel, 1 in series
4. 2 in series, 1 in parallel
View Answer
Connecting two \(2\text{ }\mu\text{F}\) capacitors in series gives \(1\text{ }\mu\text{F}\). Adding the third \(2\text{ }\mu\text{F}\) capacitor in parallel to this combination yields \(1 + 2 = 3\text{ }\mu\text{F}\).
Assertion (A): When two capacitors of capacitance \(300 \text{ pF}\) and \(600 \text{ pF}\) which can work upto maximum potential of \(4 \text{ kV}\) and \(3 \text{ kV}\) respectively, are connected in series, their combination can work upto maximum potential of \(7 \text{ kV}\).
Reason (R): In series combination, maximum working potential will be sum of maximum working potential of individual capacitors.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
For capacitors in series, the maximum charge \(Q_{\text{max}}\) the combination can hold is the minimum of individual \(C V_{\text{max}}\). Here, \(Q_{1, \text{max}}\ = 300 \text{ pF} \cdot 4 \text{ kV} = 1200 \text{ pC}\) and \(Q_{2, \text{max}}\ = 600 \text{ pF} \cdot 3 \text{ kV} = 1800 \text{ pC}\). So, \(Q_{\text{max}}\ = 1200 \text{ pC}\). Equivalent capacitance \(C_{\text{eq}}\ = (300 \cdot 600) / (300 + 600) = 200 \text{ pF}\). The maximum potential for the combination is \(V_{\text{max}}\ = Q_{\text{max}} / C_{\text{eq}}\ = 1200 \text{ pC} / 200 \text{ pF} = 6 \text{ kV}\). Thus, both Assertion (A) and Reason (R) are false.
Assertion (A): Two parallel plates having unequal charges have same capacitance as that of equal and opposite charges on same plates and same configuration.
Reason (R): Capacitance of system/ configuration is independent of charge on plates.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
A: True. The capacitance of a parallel plate capacitor, \(C = \frac{\epsilon_0 A}{d}\), is a geometric property and does not depend on the specific charge values on the plates, only their configuration.\nR: True. Capacitance is an intrinsic property dependent on geometry and dielectric, not on charge or potential.\n(R) correctly explains (A).
Assertion (A): Electrolytic capacitors have larger capacities.
Reason (R): Electrolytic capacitors have a positive and a negative terminal.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true; electrolytic capacitors offer high capacitance due to their very thin dielectric layer and large effective plate area. Reason (R) is also true, as all capacitors have two terminals. However, the presence of terminals doesn't explain why they have *larger* capacities, so (R) is not the correct explanation for (A).