Physics - Electrostatics Question with Solution | TestHub

$3\mathrm{ }\mathrm{\mu F}$&$1\mathrm{ }\mathrm{\mu F}$are in parallel hence${\mathrm{C}}_{\mathrm{net}}={\mathrm{C}}_1+{\mathrm{C}}_2$hence is equal to$4\mathrm{ }\mathrm{\mu F}$.

Now$2\mathrm{ }\mathrm{\mu F}$&$2\mathrm{ }\mathrm{\mu F}$are in series with this equivalent capacitor$\left(4\mathrm{ }\mathrm{\mu F}\right)$.Solving$2\mathrm{ }\mathrm{\mu F}$and$2\mathrm{ }\mathrm{\mu F}$are in series, we get$1\mathrm{ }\mathrm{\mu F}$.

$\therefore$${\mathrm{V}}_{4\mathrm{\mu F}}=\frac{\mathrm{V}\times {\mathrm{C}}_{1\mathrm{\mu F}}}{\left({\mathrm{C}}_{1\mathrm{\mu F}}+{\mathrm{C}}_{4\mathrm{\mu F}}\right)}$
$\Rightarrow {\mathrm{V}}_{4\mathrm{\mu F}}=\frac{10\times 1}{\left(1+4\right)}=2$
$\Rightarrow {\mathrm{V}}_{4\mathrm{\mu F}}=2 \mathrm{V}$

(potential difference across$4\mathrm{ }\mathrm{\mu F}$capacitor is calculated by voltage division rule)

$\therefore \mathrm{ }{\mathrm{V}}_{5\mathrm{\mu F}}=10 \mathrm{V}$

(Potential difference across$5\mathrm{ }\mathrm{\mu F}$is same as battery)

$\therefore \frac{{\mathrm{V}}_{1\mathrm{\mu F}}}{{\mathrm{V}}_{5\mathrm{\mu F}}}=\frac{2}{10}=\frac{1}{5}$