Two identical capacitors, have the same capacitance C. One of the… - Vidyadip

MCQ

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

A
$\frac{1}{4}C\left( {V_1^2 - V_2^2} \right)$
B
$\frac{1}{4}C\left( {V_1^2 + V_2^2} \right)$
✓
$\frac{1}{4}C{\left( {{V_1} - {V_2}} \right)^2}$
D
$\frac{1}{4}C{\left( {{V_1} + {V_2}} \right)^2}$

✓

Answer

Correct option: C.

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

c
Initial energy of combined system

$U_{1}=\frac{1}{2} C V_{1}^{2}+\frac{1}{2} C V_{2}^{2}$

Final common potential, $V=\frac{V_{1}+V_{2}}{2}$

Final energy of system,

$U_{2}=2 \times \frac{1}{2}\left(\frac{V_{1}+V_{2}}{2}\right)^{2}$

Hence loss of energy $=U_{1}-U_{2}=\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2}$

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