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2. Electric Potential and Capacitance
hard
Two Identical capacttors $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ of equal capacitance are connected as shown in the circult. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $\mathrm{C}_{1}$ 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?.....$\%$

A
$75$
B
$0$
C
$50$
D
$25$
(NEET-2019)
Solution
$U_{\text {initial }}=\frac{1}{2} C V^{2},$ Loss $=\frac{C \cdot C}{2(C+C)}(V-0)^{2}=\frac{1}{4} C V^{2}$
$\%$ Loss $=\frac{\frac{1}{4} \mathrm{CV}^{2}}{\frac{1}{2} \mathrm{CV}^{2}} \times 100=50 \%$
Standard 12
Physics