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

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


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