JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Capacitance

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Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:

(A) $\frac{1}{4} C V^{2}$ ✓ Correct
(B) $2 CV^{2}$
(C) $\frac{1}{2} C V^{2}$
(D) $\frac{3}{4} C V^{2}$

Answer: (A)

Solution

Two identical capacitors of capacitance

C

charged to

V

and

2 V

are connected with negative ends together, then positive ends together — charges combine on the same-sign plates:

V_{co mm o n} = \frac{C V + C ( 2 V )}{2 C} = \frac{3 V}{2}

Energy before:

U_{i} = \frac{1}{2} C V^{2} + \frac{1}{2} C (2 V)^{2} = \frac{5}{2} C V^{2}

.
Energy after:

U_{f} = \frac{1}{2} (2 C) (\frac{3 V}{2})^{2} = \frac{9}{4} C V^{2}

.
Decrease in energy:

Δ U = \frac{5}{2} C V^{2} - \frac{9}{4} C V^{2} = \frac{1}{4} C V^{2}

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Capacitance

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