### NEET Solved Paper 2017 Question 3

##### Question: A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

#### Options:

A) Increases by a factor of 2

B) Increases by a factor of 4

C) Decreases by a factor of 2

D) Remains the same

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

- Charge on capacitor q = CV when it is connected with another uncharged capacitor.

${{V} _{c}}=\frac{{{q} _{1}}+{{q} _{2}}}{{{C} _{1}}+{{C} _{2}}}=\frac{q+0}{C+C}$ ${{V} _{c}}=\frac{V}{2}$

Initial energy ${{U} _{i}}=\frac{1}{2}C{{V}^{2}}$

Final energy ${{U} _{f}}=\frac{1}{2}C{{\left( \frac{V}{2} \right)}^{2}}+\frac{1}{2}C{{\left( \frac{V}{2} \right)}^{2}}$ $=\frac{C{{V}^{2}}}{4}$

Loss of energy $={{U} _{i}}-{{U} _{f}}$ $=\frac{C{{V}^{2}}}{4}$

i.e. decreases by a factor [b] $