NEET 2019
- Exam
- NEET
- Year
- 2019
- Subject
- Physics
- Chapter
- Capacitance
NEET 2019 · Physics · Capacitance
Two identical capacitors C_1 and C_2 of equal capacitance are...
Two identical capacitors and of equal capacitance are connected as shown in the circuit. Terminals a and of the key are connected to charge capacitor using battery of emf volt. Now, disconnecting a and the terminals and are connected. Due to this, what will be the percentage loss of energy?
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
When is connected to voltage source, it is charged to a potential \mathrm{U}=\frac{1}{2} \mathrm{CV}^
\mathrm{C}_1\mathrm{C}_2
\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\frac{\frac{1}{4} \mathrm{CV}^2}{\frac{1}{2} \mathrm{CV}^2} \times$100=50 \%
\end{aligned}
\mathrm{C}_1\mathrm{C}_2\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\frac{\frac{1}{4} \mathrm{CV}^2}{\frac{1}{2} \mathrm{CV}^2} \times$100=50 \%
\end{aligned}