Physics - Capacitor Question with Solution | TestHub

The combination of parallel resistors is $${\mathrm{C}}_{\Vert }=2\mu \mathrm{F}+2\mu \mathrm{F}=4\mu \mathrm{F}$$ To find the equivalent capacitance of the circuit, we consider series combination: $\begin{array}{rl}& \frac{1}{{\mathrm{C}}_{\mathrm{eq}}}=\frac{1}{2\mu \mathrm{F}}+\frac{1}{4\mu \mathrm{F}}+\frac{1}{2\mu \mathrm{F}}=\frac{2+1+2}{4\mu \mathrm{F}}\\ & \Rightarrow {\mathrm{C}}_{\mathrm{eq}}=\frac{4}{5}\mu \mathrm{F}\end{array}$ $\begin{array}{rl}& \text{ Energy stored }=\frac{{\mathrm{c}}_{\mathrm{eq}}{\mathrm{v}}^2}{2}=\frac{4\times 10^{-6}\mathrm{F}\times (10{)}^2{\mathrm{V}}^2}{5\times 2}\\ & =40\times 10^{-6}\mathrm{J}=400\times 10^{-7}\mathrm{J}\end{array}$