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Question

A capacitor C is fully charged with voltage V0. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor of capacitance C/2. The energy loss in the process after the charge is distributed between the two capacitors is:


A

12CV02

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B

14CV02

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C

13CV02

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D

16CV02

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Solution

The correct option is D

16CV02


Step 1. Given data:

Capacitance of fully charged capacitor is C and voltage is V0.

Capacitance of uncharged capacitor is C2

Step 2. Solving for energy loss:

Since the two capacitors are connected in parallel, using Cp=C1+C2

The capacity of new combination is Cf=C+C2=3C2

Since the total charge (Q=CV) is constant, we have

CfVf=CV0

3C2Vf=CV0

Vf=23V0

The initial energy is U0=12CV02

The energy of new configuration is, Uf=123C22V032

Uf=1223CV02

The Energy loss is, U0-Uf=1-2312CV02

U0-Uf=13.12CV02=16CV02

Hence, option D is correct


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