Question

The energy stored in the capacitor (in $μJ$ ) between terminals $a$ and $b$ of the network (as shown in figure) is $\frac{n}{2}μJ$ . Fin the value of $n$ . If capacitance of each capacitor is $1μF$ .

Answer 1

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

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