In the circuit shown in the figure below, the potential difference across the 4.5 $\mathrm{\mu F}$

 

 

 

 

 

 

 

capacitor is 

In the given circuit, the net capacitance of 3$\mathrm{\mu F}$ and 6$\mathrm{\mu F}$ capacitors being connected in

${\mathrm{\mu }}_0 {\mathrm{\mu }}_1 \mathrm{Ni}$

$\mathrm{C}=3\mathrm{\mu F}+6\mathrm{\mu F}=9\mathrm{\mu F}$

$\mathrm{Now}, 4.5 \mathrm{\mu F} \mathrm{and} 9 \mathrm{\mu F} \mathrm{capacitors} \mathrm{are} \mathrm{in} \mathrm{series}, \mathrm{so} \mathrm{the} \mathrm{total} \mathrm{apacity} \mathrm{of} \mathrm{apacitor},$$\therefore \mathrm{C}=\frac{9\times 4.5}{9+4.5}=\frac{9\times 4.5}{13.5}3 \mathrm{\mu F}$

$$\begin{gathered} \mathrm{Charge}, \mathrm{q}=\mathrm{CV}=3\times 12=36 \mathrm{\mu F} \\ (\mathrm{given}, \mathrm{V}=12\mathrm{V}) \\ \therefore \mathrm{Potential} \mathrm{difference} \mathrm{across} 4.5 \mathrm{\mu F}, \\ {\mathrm{V}}^{\text{'}}=\frac{\mathrm{q}}{4.5} =\frac{36}{4.5}=8\mathrm{V} \end{gathered}$$

parallel,