Capacitance Question 7
Question 7 - 2024 (30 Jan Shift 1)
A capacitor of capacitance $C$ and potential V has energy E. It is connected to another capacitor of capacitance $2 C$ and potential $2 V$. Then the loss of energy is $\frac{x}{3} E$, where $x$ is
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Answer: (2)
Solution:
Energy loss $=\frac{1}{2} \frac{C _1 C _2}{C _1+C _2}\left(V _1-V _2\right)^{2}$
$=\frac{2}{3} \cdot E$
$\therefore x=2$
