Question

What is the energy stored in the capacitor (in $nJ$ ) between terminals $a$ and $b$ of the network (as shown in figure)? If capacitance of each capacitor is $1\mu F$ .

Answer 1

Answer: 12500

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


Charge on capacitor between terminals $a$ and $b$
$\frac{Q}{2}=\frac{CV}{2}$
Energy stored in that capacitor $=\frac{1}{2}\frac{{\left(\frac{Q}{2}\right)}^2}{C}$
$=\frac{Q^2}{8C}$
$=\frac{C{V}^2}{8}$
$=\frac{1\times 10^{-6}\times 10^2}{8}J$
$=12.5\mu J=12500nJ$