Electrostatics Question 614
Question: Two identical capacitors, have the same capacitance C. One of them is charged to potential $ V _{1} $ and the other to $ V _{2} $ . The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is [IIT-JEE (Screening) 2002]
Options:
A) $ \frac{1}{4}C(V _{1}^{2}-V _{2}^{2}) $
B) $ \frac{1}{4}C(V _{1}^{2}+V _{2}^{2}) $
C) $ \frac{1}{4}C{{( V _{1}-V _{2} )}^{2}} $
D) $ \frac{1}{4}C{{( V _{1}+V _{2} )}^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Initial energy of the system
$ U _{i}=\frac{1}{2}CV _{1}^{2}+\frac{1}{2}CV _{2}^{2} $
When the capacitors are joined, common potential $ V=\frac{C V_{1}+C V_{2}}{2C}=\frac{V_{1}+V_{2}}{2} $
Final energy of the system
$ U _{f}=\frac{1}{2}(2C)V^{2}=\frac{1}{2}2C{{( \frac{V _{1}+V _{2}}{2} )}^{2}}=\frac{1}{4}C{{(V _{1}+V _{2})}^{2}} $
Decrease in energy = $ U _{i}-U _{f}=\frac{1}{2}C{{(V _{1}-V _{2})}^{2}} $
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