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 larger capacitor is decreased to $V$. The value of $C$ is

• A. $C_0{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}$
• B. $C_0[{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}-1]$
• C. $C_0{[\left(\frac{V}{V_0}\right)-1]}^n$
• D. $C_0[{\left(\frac{V}{V_0}\right)}^n+1]$

Answer 1

The correct option is B $C_0[{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}-1]$

If a capacitor of capacitance $C$ is connected to a capacitor of capacitance $C_0$ then the potential $V_0$ will change to some common potential $V_1$

$⟹V_1=\frac{C_0{V}_0}{C_0+C}$
When the capacitor $C$ is charged again after discharging the common potential on both the capacitors will be
$V_2=\frac{C_0{V}_1}{C_0+C}$
$⟹V_2={\left(\frac{C_0}{C_0+C}\right)}^2{V}_0$
After continuing the same process of charging and discharging for n times, The potential $V_n$ obtained will be
$V_n={\left(\frac{C_0}{C_0+C}\right)}^n{V}_0$
Its given that final potential is equal to $V$
$\therefore V=V_n$
So, the value of capcitance $C$ will be
$C=C_0[{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}-1]$