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A 600 pF capacitor is charged by a 200 V...

A `600 pF` capacitor is charged by a `200 V` supply. It is then disconnected from the supply and is connected to another uncharged `600 pF` capacitor. What is the common potential in `V` and energy lost in `J` after reconnection?

A

`100, 6xx10^(-6)`

B

`200, 6xx10^(-5)`

C

`200, 5xx10^(-6)`

D

`100, 6xx10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
A
`V_(C)=(C_(1)V_(1))/(C_(1)+C_(2))=100 V`
Energy lost `=U_(i)-U_(f)`
`=1/2C_(1)V_(1)^(2)-1/2(C_(1)+C_(2))V_(c)^(2)`
`=6xx10^(-6)J`.
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Knowledge Check

  • A 4muF capacitor is charged by a 200V battery. It is then disconnected from the supply and is connected to another uncharged 2muF capacitor. During this process, Loss of energy (inJ) is

    A
    Zero
    B
    `5.33xx10^(-2)`
    C
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    D
    `2.67xx10^(-2)`

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