Question:
Two capacitors of capacitances \(C\) and \(2C\) are charged to potential differences \(V\) and \(2V\), respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
Two capacitors of capacitances \(C\) and \(2C\) are charged to potential differences \(V\) and \(2V\), respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is :
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With opposite polarities joined, the net charge is the difference of the two charges. Find the common voltage, then use \(U = \tfrac12 C_{eq}V_f^2\).
Updated On: Nov 5, 2026
- Zero
- \(\dfrac{3}{2}CV^2\)
- \(\dfrac{9}{2}CV^2\)
- \(\dfrac{25}{6}CV^2\)
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The Correct Option is B
Solution and Explanation
Charges: \(q_1 = CV\), \(q_2 = (2C)(2V) = 4CV\).
Opposite plates are joined, so the net charge on the combination is \(q = q_2 - q_1 = 3CV\).
Equivalent capacitance in parallel: \(C_{eq} = C + 2C = 3C\).
Common potential: \(V_f = \dfrac{3CV}{3C} = V\).
Final energy: \(U_f = \tfrac12 (3C)V^2 = \tfrac32 CV^2\).
Answer: (2) \(\dfrac{3}{2}CV^2\)
Opposite plates are joined, so the net charge on the combination is \(q = q_2 - q_1 = 3CV\).
Equivalent capacitance in parallel: \(C_{eq} = C + 2C = 3C\).
Common potential: \(V_f = \dfrac{3CV}{3C} = V\).
Final energy: \(U_f = \tfrac12 (3C)V^2 = \tfrac32 CV^2\).
Answer: (2) \(\dfrac{3}{2}CV^2\)
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