In the given circuit, if the potential difference between A and B is 80 V, then the equivalent capacitance between A and B and the charge on $10\mathrm{\mu F}$ capacitor respectively, are

According to given circuit diagram.

Parallel equivalent capacitance,

${\text{C'}}_{\text{eq}}{\text{\hspace{0.17em}=C}}_{\text{1}}{\text{\hspace{0.17em}+\hspace{0.17em}C}}_{\text{2}}{\text{\hspace{0.17em}+C}}_{\text{2}}$

$\therefore {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}C'}}_{\text{eq}}\text{\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}9\hspace{0.17em}+\hspace{0.17em}10\hspace{0.17em}+\hspace{0.17em}5\hspace{0.17em}= 24\hspace{0.17em}μF}$

and series equivalent capacitance,$\frac{\text{1}}{{\text{C}}_{\text{eq}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1}}{{\text{C}}_{\text{a}}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{1}}{{\text{C}}_{\text{b}}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{1}}{{\text{C}}_{\text{c}}}$

$\Rightarrow \frac{\text{1}}{{\text{C}}_{\text{eq}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{1}}{\text{12}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{1}}{\text{24}}\text{\hspace{0.17em}+\hspace{0.17em}}\frac{\text{1}}{\text{8}}\Rightarrow \frac{\text{1}}{{\text{C}}_{\text{eq}}}=\frac{\text{2\hspace{0.17em}+\hspace{0.17em}1\hspace{0.17em}+\hspace{0.17em}3}}{\text{24}}$

$\Rightarrow \frac{\text{1}}{{\text{C}}_{\text{eq}}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{24}}{\text{6}}\text{μF\hspace{0.17em}=\hspace{0.17em}4μF}$

Since,$\text{C\hspace{0.17em}=\hspace{0.17em}}\frac{\text{Q}}{\text{V}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Q\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}CV}$

${\text{=\hspace{0.17em}4\hspace{0.17em}×\hspace{0.17em}10}}^{\text{-6}}\text{\hspace{0.17em}×\hspace{0.17em}80\hspace{0.17em}=\hspace{0.17em}320μC}$

Now, 9V + 10V + 5V = 320

$\Rightarrow \text{\hspace{0.17em}\hspace{0.17em}V=}\frac{\text{40}}{\text{3}}{\text{Volt.\hspace{0.17em}So,\hspace{0.17em}\hspace{0.17em}q}}_{\text{10μF}}\text{=10×}\frac{\text{40}}{\text{3}}\text{=133.3μC}$