A capacitor of capacitance C₀ is charged to a potential V₀ and then isolated. A small capacitor C is then charged from C₀, discharged and charged again; the process being repeated n times. Due to this, the potential of the larger capacitor is decreased to V. The value of C is

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$C_{0} [{(\frac{V}{V_{0}})}^{n} + 1]$

$C_{0} {[(\frac{V}{V_{0}}) - 1]}^{n}$

$C_{0} {(\frac{V_{0}}{V})}^{1 / n}$

$C_{0} [{(\frac{V_{0}}{V})}^{1 / n} - 1]$

answer is B.

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Detailed Solution

Potential of larger capacitor after the first charging is

$V_{1} = \frac{C_{0} V_{0}}{(C + C_{0})}$

After second charging, potential is

$V_{2} = \frac{C_{0} V_{1}}{(C + C_{0})} = {(\frac{C_{0}}{C + C_{0}})}^{2} V_{0}$

After n^th charging, potential is

$V_{n} = {(\frac{C_{0}}{C + C_{0}})}^{n} V_{0}$

But $V_{n} = V$

So $C = C_{0} [{(\frac{V_{0}}{V})}^{1 / n} - 1]$

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