Two identical capacitors have same capacitance C. One of them is charged to the potential V and other to the potential 2V. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is
\(\frac{CV^2}{4}\)
- 2CV2
\(\frac{1}{2}CV^2\)
- \(\frac{3}{4}CV^2\)
The Correct Option is A
Solution and Explanation
The potential of the combined system of capacitors is determined by:
\[ V_c = \frac{q_{net}}{C_{net}} = \frac{CV + 2CV}{2C} = \frac{3V}{2} \]
Energy loss within the system is computed as:
\[ \text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(2V)^2 - \frac{1}{2}C \left(\frac{3V}{2}\right)^2 \]
The preceding expression simplifies to:
\[ \text{Loss of energy} = \frac{1}{2}CV^2 + \frac{1}{2}C(4V^2) - \frac{1}{2}C \left(\frac{9V^2}{4}\right) \]
\[ \text{Loss of energy} = \frac{CV^2}{2} + 2CV^2 - \frac{9CV^2}{8} \]
\[ \text{Loss of energy} = \frac{4CV^2}{8} + \frac{16CV^2}{8} - \frac{9CV^2}{8} \]
\[ \text{Loss of energy} = \frac{CV^2}{4} \]
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