Question 177 of 180

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

ABCD

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Answer: (C)

∵1\mu F&2\mu F are in parallel.



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

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