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 $1\,μF$ .

Answer 1

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


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