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A battery of `20V` is connected to a capacitor of capacitance `10 muF`. Now battery is removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, the total energy stored in the two capacitors will beA. `250muJ`B. `500muJ`C. `750muJ`D. `1000muJ` |
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Answer» Correct Answer - D Since final charges are same, potential is same i.e., `C_(1)=C_(2)=10muF` `V=(10xx20+10xx0)/(10+10)=10V` `U_(f)=(1)/(2)(C_(1)+C_(2))V^(2)=(1)/(2)(10+10)(10)^(2)xx10^(-6)` `=1000muJ` |
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