A capacitor of 2µF is charged as shown in the diagram.

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A capacitor of 2µF is charged as shown in the diagram. When the switch S is turned to position 2, the percentage of its stored energy dissipated is :

1. 0%
2. 20%
3. 75%
4. 80%

Correct Option: D

When S and 1 are connected
The 2µF capacitor gets charged. The potential difference across its plates will be V.
The potential energy stored in 2 µF capacitor

U_i =	1	CV² =	1	× 2 × V² = V²
	2		2

When S and 2 are connected
The 8µF capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is

∆U =	1	C₁C₂	(V₁ - V₂)²
	2	C₁ + C₂

Here, C₁ = 2µF, C₂ 8 µF, V₁= V, V₂ = 0

∴ ΔU =	1	×	2 × 8	(v - 0)² =	4	V²
	2		2 + 8		5

The percentage of the energy dissipated

=	ΔU	× 100 =	4/5 V²	× 100 = 80%
	Ui		V²