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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 $

1 Answer

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

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