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Two capacitor `C_(1)=2muF and C_(2)=6muF` are in series order connected is parallel to a third capacitor `C_(3)=4muF.` This combination is connected to 2 V battery. In charging these capacitor energy consumed by the battery is A. `2xx10^(-6)J`B. `11xx10^(6)J`C. `(32)/(3)xx10^(-6)J`D. `(16)/(3)xx10^(-6)J`

Answer» Correct Answer - B
`"Total capacity, C"=(C_(1)C_(2))/(C_(1)+C_(2))+C_(3)=(2xx6)/(2+6)+4`
`=(12)/(8)+4`
`C=(44)/(8)4muF=(44)/(8)xx10^(-6)F`
`because" Energy consumed by battery, U"=(1)/(2)CV^(2)`
`rArr" "U=(1)/(2)xx(44)/(8)xx10^(-6)xx(2)^(2)=11xx10^(-6)J`


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