Capacitance Question 7
Question 7 - 2024 (30 Jan Shift 1)
A capacitor of capacitance $\mathrm{C}$ and potential V has energy E. It is connected to another capacitor of capacitance $2 \mathrm{C}$ and potential $2 \mathrm{~V}$. Then the loss of energy is $\frac{x}{3} E$, where $\mathrm{x}$ is
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Answer: (2)
Solution:
Energy loss $=\frac{1}{2} \frac{C_{1} C_{2}}{C_{1}+C_{2}}\left(V_{1}-V_{2}\right)^{2}$
$=\frac{2}{3} \cdot E$
$\therefore x=2$
