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A capacitance of C(0) is charged to a po...

A capacitance of `C_(0)` is charged to a potential `V_(0)` and then isolated. `A` small capacitor `C` is then charged from `C_(0)`, discharged and charged again, the process being repeated `n` times. Due to this, potential of the large capacitor is decreased to `V`. find the capacitance of the small capacitor. is it possible to remove charge on `C_(0)` completely in this way?

Text Solution

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The correct Answer is:
`C = C_(0) [((V_(0))/(V))^(1//n)-1] = 0.01078 muF, n = 20`,No
After first operation charge remain on `C_(0)` in
the ratio of capacity `q' = ((C_(0))/(C_(0)+C)) xxq`
`q' = ((C_(0))/(C_(0) +C)) C_(0)V_(0)`
& `V' = ((C_(0))/(C_(0)+C)) V_(0)`
After `n^(th)` operation `V = ((C_(0))/(C_(0)+C))^(n) V_(0)`
& `C=[((V_(0))/(V))^(1//n)-1] C_(0)`
By putting the given values we get `n = 20`
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