A capacitor of capacitance 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 the uncharged capacitor is connected to the positive plate and the othe

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Solution

Q.
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 the uncharged capacitor is connected to the positive plate and the other plate of the uncharged capacitor is connected to the 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 ________ . [2023]

Ans.
(225)

$C = 900 μ F$

$Hence Q = C V = 900 \times 10^{- 6} \times 100 = 9 \times 10^{- 2} = 90 mC$

Common potential will be developed across both capacitors by KVL.

Total charge on left plates of capacitors should be conserved.

$∴$ $90 mC + 0 = 2 C V_{0}$

$C V_{0} = 45 mC$

$Heat dissipated = U_{i} - U_{f}$ [Change in energy stored in the capacitors]

$= \frac{1}{2} \frac{{(90 mC)}^{2}}{900 μ F} - 2 \times \frac{1}{2} \frac{{(45 mC)}^{2}}{900 μ F} [U = \frac{Q^{2}}{2 C}]$

$= \frac{1}{2 \times 900 \times 10^{- 6}} (8100 - 4050) \times 10^{- 6} = 2.25 Joule$

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