In the given circuit, charge Q 2 on the 2   μ F capacitor changes as C is varied from 1  …

In the given circuit, charge $Q_2$ on the $2 \mathrm{\mu }\mathrm{F}$ capacitor changes as C is varied from $1 \mathrm{\mu F}$ to $3 \mathrm{\mu }\mathrm{F}$. $Q_2$ as a function of 'C' is given properly by: (figures are drawn schematically and are not to scale)

Solution

$\because 1 \mu \mathrm{F} \& 2 \mu \mathrm{F}$ are in parallel.



$\therefore$ The equivalent capacitance of the series combination is, $C_{\mathrm{eq}}$ is$=\frac{3C}{C+3}$. So, the total charge supplied by the battery is, 
$\text{⇒}\mathit{\text{Q}}=C_{eq}E=\frac{3CE}{C+3}$.$\therefore$ The potential difference across the parallel combination of $1 \mu \mathrm{F}$ and $2\mu \mathrm{F}$ is, $∆V=\frac{Q}{3}=\frac{CE}{C+3}$. So charge on $2 \mu \mathrm{F}$ capacitor is, $Q_2=C_2∆V=\frac{2CE}{C+3}$$\Rightarrow \frac{Q_2}{2E}=\frac{C}{C+3}\Rightarrow \frac{Q_2}{2E}=\frac{C+3-3}{C+3}$

$\Rightarrow \frac{Q_2}{2E}=1-\frac{3}{C+3}$⇒ So the graph is a hyperbola. With downward curve line. i.e.

 

Asked in: JEE Main 2015 (04 Apr)

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