Question

Capacitance of each capacitor in the circuit is $1\mu F$. What is the energy stored in the capacitor (in $\mu J$ ) between terminals ' $a$ ' and ' $b$ ' of the network (as shown in figure)?

Answer 1

Answer: 12.5

Redrawing network, we get a balanced Wheatstone's network.


$\Rightarrow C_{eq}=C$
Charge on capacitor between terminals $a$ and $b$,
$\frac{Q}{2}=\frac{CV}{2}$
Energy stored in that capacitor
$=\frac{1}{2}\frac{{\left(\frac{Q}{2}\right)}^2}{C}$
$=\frac{Q^2}{8C}$
$=\frac{C{V}^2}{8}$
$=\frac{1\times 10^{-6}\times 10^2}{8}J$
$=12.5\mu J$