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A capacitor of capacitance `6 muF` is charged to a potential of 150V. Its potential falls to 90V, when another uncharged capacitor is connected to it. Find the capacitance of the second capacitor and the amount of energy lost due to the connection.

Answer» Here, `C_(1) = 6 muF, V_(1) = 150 V, V_(2) = 0`
`V = 90 V, C_(2) = ?`
Common potential `V = (C_(1) V_(1) + C_(2) V_(2))/(C_(1) + C_(2))`
`90 = (6xx10^(-6)xx150+0)/(6xx10^(-6) + C_(2))`
`C_(2) = 4xx10^(-6) F = 4 muF`
Initial energy, `U_(1) = (1)/(2) C_(1) V_(1)^(2)`
`= (1)/(2)xx6xx10^(-6)xx(150)^(2) = 6.75xx10^(-2) J`
Final energy, `U_(2) = (1)/(2) (C_(1) + C_(2)) V^(2)`
`= (1)/(2) (6+4) xx 10^(-6) xx (90)^(2) = 4.05xx10^(-2) J`
Loss of energy on connecting two capacitors,
`DeltaU = U_(2) - U_(1) = (6.75xx10^(-2) - 4.05 xx10^(-2))`
`= 0.027 J`


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