Two identical capacitors, have the same capacitance C. One of them is charged to potential {V_1} and the other to {V

Q: 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

c (c) 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 $V = \frac{{C{V_1} + C{V_2}}}{{2C}} = \frac{{{V_1} + {V_2}}}{2}$ Final energy of the system ${U_f} = \frac{1}{2}(2C){V^2} = \frac{1}{2}2C\,{\left( {\frac{{{V_1} + {V_2}}}{2}} \right)^2} = \frac{1}{4}C{({V_1} + {V_2})^2}$ Decrease in energy = ${U_i} - {U_f} = \frac{1}{4}C{({V_1} - {V_2})^2}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

IIT 2002, Diffcult

STD 11 Science
NEET
STD 12 - 2. Electric potential and capacitance
Physics

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