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?.....$\%$
NEET 2019, Diffcult
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$U_{\text {initial }}=\frac{1}{2} C V^{2},$ Loss $=\frac{C \cdot C}{2(C+C)}(V-0)^{2}=\frac{1}{4} C V^{2}$
$\%$ Loss $=\frac{\frac{1}{4} \mathrm{CV}^{2}}{\frac{1}{2} \mathrm{CV}^{2}} \times 100=50 \%$
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