Electrostatics Question 648

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

Options:

A) $ \frac{1}{4}C(V^{2} _{{{1}^{{}}}}-V^{2} _{2}) $

B) $ \frac{1}{4}C(V^{2} _{1}+V^{2} _{2}) $

C) $ \frac{1}{4}C{{(V _{1}-V _{2})}^{2}} $

D) $ \frac{1}{4}C{{(V _{1}+V _{2})}^{2}} $

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Answer:

Correct Answer: C

Solution:

[c] Initial energy of the system $ U _{i}=\frac{1}{2}CV^{2} _{1}+\frac{1}{2}CV^{2} _{2} $

When the capacitors are joined, common potential $ V=\frac{CV _{1}+CV _{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}{4}C{{(V _{1}-V _{2})}^{2}} $



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