Search any question & find its solution
Question:
Answered & Verified by Expert
Two identical capacitors $C_1$ and $C_2$ of equal capacitance are connected as shown in the circuit. Terminals a and $\mathrm{b}$ of the key $\mathrm{k}$ are connected to charge capacitor $\mathrm{C}_1$ using battery of emf $\mathrm{V}$ volt. Now, disconnecting a and $b$ the terminals $b$ and $c$ are connected. Due to this, what will be the percentage loss of energy?
Options:
Solution:
1436 Upvotes
Verified Answer
The correct answer is:
$50 \%$
When $\mathrm{C}_1$ is connected to voltage source, it is charged to a potential $V$ and this will be stored as a potential energy in the capacitor given by
$$
\mathrm{U}=\frac{1}{2} \mathrm{CV}^2
$$
When key is disconnected from battery and band c are connected, the charge will be transformed from the capacitor $\mathrm{C}_1$ to capacitor $\mathrm{C}_2$, then
The loss of energy due to redistribution of charge is given by
$$
\begin{aligned}
& \begin{aligned}
\Delta \mathrm{U} & =\frac{\mathrm{C}_1 \mathrm{C}_2}{2\left(\mathrm{C}_1+\mathrm{C}_2\right)}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2 \\
& =\frac{\mathrm{C} \times \mathrm{C}}{2(\mathrm{C}+\mathrm{C})}(\mathrm{V}-0)^2=\frac{1}{4} \mathrm{CV}^2 \quad\left[\because \mathrm{C}_1=\mathrm{C}_2\right]
\end{aligned} \\
& \therefore \text { Percentage loss }=\frac{\Delta \mathrm{U}}{\mathrm{U}} \times 100=\frac{\frac{1}{4} \mathrm{CV}^2}{\frac{1}{2} \mathrm{CV}^2} \times 100=50 \%
\end{aligned}
$$
$$
\mathrm{U}=\frac{1}{2} \mathrm{CV}^2
$$
When key is disconnected from battery and band c are connected, the charge will be transformed from the capacitor $\mathrm{C}_1$ to capacitor $\mathrm{C}_2$, then
The loss of energy due to redistribution of charge is given by
$$
\begin{aligned}
& \begin{aligned}
\Delta \mathrm{U} & =\frac{\mathrm{C}_1 \mathrm{C}_2}{2\left(\mathrm{C}_1+\mathrm{C}_2\right)}\left(\mathrm{V}_1-\mathrm{V}_2\right)^2 \\
& =\frac{\mathrm{C} \times \mathrm{C}}{2(\mathrm{C}+\mathrm{C})}(\mathrm{V}-0)^2=\frac{1}{4} \mathrm{CV}^2 \quad\left[\because \mathrm{C}_1=\mathrm{C}_2\right]
\end{aligned} \\
& \therefore \text { Percentage loss }=\frac{\Delta \mathrm{U}}{\mathrm{U}} \times 100=\frac{\frac{1}{4} \mathrm{CV}^2}{\frac{1}{2} \mathrm{CV}^2} \times 100=50 \%
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.