Electrostatics Ques 113
- Two identical capacitors, have the same capacitance $C$. One of them is charged to potential $V_{1}$ and the other to $V_{2}$. Likely charged plates are then connected. Then, the decrease in energy of the combined system is
(a) $\frac{1}{4} C\left(V_{1}^{2}-V_{2}^{2}\right)$
(b) $\frac{1}{4} C\left(V_{1}^{2}+V_{2}^{2}\right)$
(c) $\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2}$
(d) $\frac{1}{4} C\left(V_{1}+V_{2}\right)^{2}$
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Answer:
Correct Answer: 113.(c)
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
- $\Delta U=$ decrease in potential energy
$ \begin{aligned} & =U_{i}-U_{f} \\ & =\frac{1}{2} C\left(V_{1}^{2}+V_{2}^{2}\right)-\frac{1}{2}(2 C) \frac{V_{1}+V_{2}}{2} \\ & =\frac{1}{4} C\left(V_{1}-V_{2}\right)^{2} \end{aligned} $
BETA
