Two identical capacitors, have the same capacitance $C$. One of them is charged to potential $V_{1}$ and the…

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
  1. $\frac{1}{4} C\left(V_{1}^{2}-V_{2}^{2}\right)$
  2. $\frac{1}{4} C\left(V_{1}^{2}+V_{2}^{2}\right)$
  3. $\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2}$
  4. $\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$

Solution

Initial energy of the system $U_{i}=\frac{1}{2} C V_{1}^{2}+\frac{1}{2} C V_{2}^{2}$
When the capacitors are joined, common potential
$V=\frac{C V_{1}+C V_{2}}{2 C}=\frac{V_{1}+V_{2}}{2}$
Final energy of the system
$U_{f}=\frac{1}{2}(2 C) V^{2}=\frac{1}{2} 2 C\left(\frac{V_{1}+V_{2}}{2}\right)^{2}=\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$
Decreased in energy $=U_{i}-U_{f}=\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2}$ ,

Asked in: JEE Mains - Capacitance - Test 3

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