Electrostatic Potential and Capacitance - Result Question 23
25. Two identical capacitors $C_1$ and $C_2$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_1$ using battery of emf 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?
[NEET Odisha 2019]
(a) $25 %$
(b) $75 %$
(c) $0 %$
(d) $50 %$
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Answer:
Correct Answer: 25. (d)
Solution:
- (d)
$ =0.44 \times 10^{-10} C^{2} N^{-1} m^{-2} $
$U_i=\frac{1}{2} C V^{2}$
On switching key at point $c$
$ \begin{aligned} & \frac{q_0-q}{C}=\frac{q}{C} \\ & 2 q=q_0 \end{aligned} $
$ \begin{aligned} & q=(\frac{q_0}{2}) \\ & U_f=\frac{1}{2}(\frac{q_0}{2})^{2} \times \frac{1}{C}+\frac{1}{2}(\frac{q_0}{2})^{2} \times \frac{1}{C} \\ & U_f=\frac{1}{4} C V^{2} \\ & \text{ loss }=50 % \end{aligned} $
