QJEEM23Phc8a2: JEE Main 2023 Physics - Class 12 JEEM23Phc8a2 Solution: A capacitor of capacitance 900 μ F is charged by a 100   V …

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QJEEM23Phc8a2: JEE Main 2023 Physics - Class 12 JEEM23Phc8a2 Solution: A capacitor of capacitance 900 μ F is charged by a 100   V …

A capacitor of capacitance $900 μ F$ is charged by a $100 V$ 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 $x \times 10^{-}^{2} J$ . The value of $x$ is _____________.

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Accepted Answer

$U_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 900 \times 10^{- 6} \times 100^{2} = 4.5$ J

As the other capacitor is identical therefore charge is equally divided and potential difference across the capacitors becomes half. So

$U_{f} = \frac{1}{2} 2 C {(\frac{V}{2})}^{2} = \frac{1}{2} \times 2 \times 900 \times 10^{- 6} {(\frac{100}{2})}^{2}$

$= \frac{9}{4}$ J = 2.25 J

So, loss in energy $Δ U_{l o s s} = U_{i} - U_{f}$

= 2.25 J

= 225 $\times$ 10 $^{- 2}$ J

QJEEM23Phc8a2: JEE Main 2023 Physics - Class 12 JEEM23Phc8a2 Solution: A capacitor of capacitance 900 μ F is charged by a 100 &nbsp; V …

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QJEEM23Phc8a2: JEE Main 2023 Physics - Class 12 JEEM23Phc8a2 Solution: A capacitor of capacitance 900 μ F is charged by a 100 V …