Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A capacitor of capacitance $C_0$ is charge to potential $V_0$ and then isolated . A small capacitor C is charged from $C_0$, discharge and then charged again; this process is repeated n times. Due to this potential of $C_0$ decreases to r. Value of C is :

$(A)\;C_0 \bigg( \frac{V_0}{V} \bigg)^{1/n}\\ (B)\;C_0 \bigg[\bigg(\frac{V}{V_0}\bigg )^n +1\bigg] \\ (C)\; C_0 \bigg[\bigg(\frac{V_0}{v}\bigg)^{1/n}-1 \bigg] \\ (D)\;C_0 \bigg[\bigg(\frac{V}{V_0}\bigg)-1 \bigg]^n $

Can you answer this question?

1 Answer

0 votes
Potential of C on 1st charge
$V_1 =\large\frac{C_0V_0}{(C+C_0)}$
On second charging ,
$V_2 =\large\frac{C_0V_1}{(C+C_0)}$
$\qquad= \bigg( \large\frac{C_0}{C+C_0}\bigg)^2$$ V_0$
Similarly after $n^{th}$ charging,
$V_n= \bigg( \large\frac{C_0}{C+C_0}\bigg)^n$$ V_0$
=>$ C= C_0 \bigg[\bigg(\frac{V_0}{v}\bigg)^{1/n}-1 \bigg]$
Hence C is the correct answer.
answered Jan 15, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App