Four capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure…
Four capacitors each of capacity $2 \mu \mathrm{F}$ are connected as shown in the figure $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=10 \mathrm{~V}$. The energy stored in the system is
$40 \times 10^{-8} \mathrm{~J}$
$625 \times 10^{-6} \mathrm{~J}$
$6250 \times 10^{-7} \mathrm{~J}$
$400 \times 10^{-7} \mathrm{~J}$
Solution
The combination of parallel resistors is
$\mathrm{C}_{\|}=2 \mu \mathrm{F}+2 \mu \mathrm{F}=4 \mu \mathrm{F}$
To find the equivalent capacitance of the circuit, we consider series combination:
$\begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{2 \mu \mathrm{F}}+\frac{1}{4 \mu \mathrm{F}}+\frac{1}{2 \mu \mathrm{F}}=\frac{2+1+2}{4 \mu \mathrm{F}} \\ & \Rightarrow \mathrm{C}_{\mathrm{eq}}=\frac{4}{5} \mu \mathrm{F}\end{aligned}$
$\begin{aligned} & \text { Energy stored }=\frac{\mathrm{c}_{\mathrm{eq}} \mathrm{v}^2}{2}=\frac{4 \times 10^{-6} \mathrm{~F} \times(10)^2 \mathrm{~V}^2}{5 \times 2} \\ & =40 \times 10^{-6} \mathrm{~J}=400 \times 10^{-7} \mathrm{~J}\end{aligned}$
