Question

A capacitor of capacitance $C_0$ 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 $V$, Value of $C$ is :

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

Answer 1

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

The initial charge on the first capacitor,$Q_0=C_0{V}_0$

After first operation, let the common potential is $V_1$

So, $V_1=\frac{Q_0}{C_0+C}=\frac{C_0{V}_0}{C_0+C}=\frac{C_0}{C_0+C}{V}_0$

Similarly after n th operation, $V={\left(\frac{C_0}{C_0+C}\right)}^n{V}_0$

${\left(\frac{V}{V_0}\right)}^{\frac{1}{n}}=\frac{C_0}{C_0+C}$

$C+C_0=C_0{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}$

$C=C_0[{\left(\frac{V_0}{V}\right)}^{\frac{1}{n}}-1]$

Ans:$\left(B\right)$