A 6muF capacitor charged from 10 volts t...

A `6muF` capacitor charged from 10 volts to 20 volts. Increase in energy will be -

`18 xx10^(-4) J`

`9 xx10^(-4) J`

`4.5 xx10^(-4) J`

`9 xx10^(-6) J`

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The correct Answer is:

Increase in energy `=1/2C(V_2^2-V_1^2)`
`=1/2 xx6xx10^(-6)xx(20^2 -10^2)`
`=9 xx10^(-4) J`

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