JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Capacitance

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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:

(A) $\frac{3}{2} C V^{2}$ ✓ Correct
(B) $\frac{9}{2} C V^{2}$
(C) $\frac{25}{6} C V^{2}$
(D)zero

Answer: (A)

Solution

Before connection the stored charges are

Q_{1} = C V

and

Q_{2} = (2 C) (2 V) = 4 C V

. Joining the positive plate of one to the negative plate of the other makes the charges oppose, so the net shared charge is

4 C V - C V = 3 C V

spread over a total capacitance

C + 2 C = 3 C

, giving a common potential

V^{'} = \frac{3 C V}{3 C} = V

. The final stored energy is

\frac{1}{2} (3 C) V^{'2} = \frac{1}{2} (3 C) V^{2} = \frac{3}{2} C V^{2}

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Capacitance

Solution

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