Physics - Capacitor Question with Solution | TestHub

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