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
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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] $
BETA
