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Capacitor in an AC Circuit

A capacitor o...

Question

A capacitor of capacitance $C_{0}$ 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 capacitor, $C_{0}$ is decreased to $V$ . Find the value of $C$ .

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Solution

$q_{0} = C_{0} V_{0}$

After charging $q_{1} = q_{0} (\frac{C_{0}}{C_{0} + C})$

After discharging and again charging first time,

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

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

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

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

After $n$ th charging

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

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

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

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

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