A 6,mu F capacitor is charged from 10,V to 20 | vedclass

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(B) The energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.The increase in energy $\Delta E$ is given by $\Delta E = E_{Final} - E_{Initia

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$9 	imes 10^{-4}\,J$

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(B) The energy stored in a capacitor is given by $E = \frac{1}{2}CV^2$.The increase in energy $\Delta E$ is given by $\Delta E = E_{Final} - E_{Initial} = \frac{1}{2}C(V_{Final}^2 - V_{Initial}^2)$.Given: $C = 6\,\mu F = 6 	imes 10^{-6}\,F$,$V_{Initial} = 10\,V$,$V_{Final} = 20\,V$.Substituting the values:$\Delta E = \frac{1}{2} 	imes (6 	imes 10^{-6}) 	imes (20^2 - 10^2)$$\Delta E = 3 	imes 10^{-6} 	imes (400 - 100)$$\Delta E = 3 	imes 10^{-6} 	imes 300$$\Delta E = 900 	imes 10^{-6}\,J = 9 	imes 10^{-4}\,J$.

$A$ $6\,\mu F$ capacitor is charged from $10\,V$ to $20\,V$. The increase in energy will be:

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