Question

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and 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

• A. $\frac{1}{4}C(V_1^2-V_2^2)$
• B. $\frac{1}{4}C(V_1+V_2{)}^2$
• C. $\frac{1}{4}C(V_1^2+V_2^2)$
• D. $\frac{1}{4}C(V_1-V_2{)}^2$

Answer 1

The correct option is D $\frac{1}{4}C(V_1-V_2{)}^2$

Initial energy $\left(E_i\right)=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2$

Let, $V$ be the common potential By law of conservation of charge total initial charge $=$ total final charge
$q_1+q_2=q+q$ ( final charges will be same due to same voltage and same capacitance )

$C{V}_1+C{V}_2=CV+CV$

$V=\frac{V_1+V_2}{2}$

Final energy $\left(E_f\right)=\frac{1}{2}(C+C)\left(V^2\right)$

$=\frac{1}{2}\left(2C\right){\left(\frac{V_1+V_2}{2}\right)}^2$

$=\frac{1}{4}C(V_1+V_2{)}^2$

$\therefore$ Energy loss $=E_i-E_f$

$=\frac{1}{2}C{V}_1^2+\frac{1}{2}C{V}_2^2-\frac{1}{4}C(V_1+V_2{)}^2$

$=\frac{1}{4}C(V_1-V_2{)}^2$

Thus, energy loss is $\frac{1}{4}C(V_1-V_2{)}^2$
Hence, option (c) is correct.