Physics - Capacitor Question with Solution | TestHub

Equivalent capacitance for three capacitors ${\mathrm{C}}_1,{\mathrm{C}}_2$ and ${\mathrm{C}}_3)$ in series is given by $\begin{array}{l}\frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{{\mathrm{C}}_1}+\frac{1}{{\mathrm{C}}_2}+\frac{1}{{\mathrm{C}}_3}=\frac{{\mathrm{C}}_2{\mathrm{C}}_3+{\mathrm{C}}_3{\mathrm{C}}_1+{\mathrm{C}}_{\mathrm{C}}{\mathrm{C}}_2}{{\mathrm{C}}_{\mathrm{C}}\mathrm{C}{\mathrm{C}}_3}\\ \Rightarrow {\mathrm{C}}_{\mathrm{eq}}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2{\mathrm{C}}_3}{{\mathrm{C}}_1{\mathrm{C}}_2+{\mathrm{C}}_2{\mathrm{C}}_3+{\mathrm{C}}_3{\mathrm{C}}_1}\\ \Rightarrow {\mathrm{C}}_{\mathrm{eq}}=\frac{\mathrm{C}(2\mathrm{C})(3\mathrm{C})}{\mathrm{C}(2\mathrm{C})+(2\mathrm{C})(3\mathrm{C})+(3\mathrm{C})\mathrm{C}}=\frac{6}{11}\mathrm{C}\end{array}$ $\Rightarrow$ Charge on capacitors $\left({\mathrm{C}}_1,{\mathrm{C}}_2\&{\mathrm{C}}_3\right)$ in series $={\mathrm{C}}_{\mathrm{eq}}\mathrm{V}=\frac{6\mathrm{C}}{11}\mathrm{v}$ Charge on capacitor ${\mathrm{C}}_4={\mathrm{C}}_4\mathrm{V}=4\mathrm{CV}$ $\frac{\text{ Charge }{\mathrm{onC}}_2}{\text{ Charge }{\mathrm{onC}}_4}=\frac{\frac{6\mathrm{C}}{11}\mathrm{V}}{4\mathrm{CV}}=\frac{6}{11}\times \frac{1}{4}=\frac{3}{22}$