JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 29)
A capacitor of capacitance $$900 \mu \mathrm{F}$$ is charged by a $$100 \mathrm{~V}$$ battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as $$x \times 10^{-} { }^{2} \mathrm{~J}$$. The value of $$x$$ is _____________.
คำตอบ
225
คำอธิบาย
$${U_i} = {1 \over 2}C{V^2} = {1 \over 2} \times 900 \times {10^{ - 6}} \times {100^2} = 4.5$$ J
As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So
$${U_f} = {1 \over 2}2C{\left( {{V \over 2}} \right)^2} = {1 \over 2} \times 2 \times 900 \times {10^{ - 6}}{\left( {{{100} \over 2}} \right)^2}$$
$$ = {9 \over 4}$$ J = 2.25 J
So, loss in energy $$\Delta {U_{loss}} = {U_i} - {U_f}$$
= 2.25 J
= 225 $$\times$$ 10$$^{-2}$$ J