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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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