Physics - Capacitor Question with Solution | TestHub

$\begin{array}{ll}\therefore & {\mathrm{C}}_1=3\mu \mathrm{F}\text{ and }{\mathrm{C}}_2=2\mu \mathrm{F}\\ & {\mathrm{C}}_{\text{series }}=\frac{{\mathrm{C}}_1{\mathrm{C}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{6}{5}\mu \mathrm{F}\\ & \text{ Also, }\mathrm{Q}=\mathrm{CV}\\ \mathrm{Q}& ={\mathrm{C}}_{\text{series }}\times \mathrm{V}\\ & =\frac{6}{5}\times 100=120\mu \mathrm{C}\end{array}$ $\mathrm{Q}$ will be the same across both the capacitors as they are in series. $\therefore$ Potential across capacitors, $\begin{array}{rl}& {\mathrm{V}}_1=\frac{\mathrm{Q}}{{\mathrm{C}}_1}=\frac{120}{3}=40\mathrm{V}\\ & {\mathrm{V}}_2=\frac{\mathrm{Q}}{{\mathrm{C}}_2}=\frac{120}{2}=60\mathrm{V}\\ & \therefore {\mathrm{V}}_2:{\mathrm{V}}_1=60:40=3:2\end{array}$