Question

What is the energy stored in the capacitor between terminals $a$ and $b$ of the network shown in the figure? (Capacitance of each capacitor $C=1\mu F$)

• A. $12.5\mu J$
• B. Zero
• C. $25\mu J$
• D. $50\mu J$

Answer 1

Answer: (A)

Network is redrawn as shown in figure.

This is a balanced Wheatstone's network.
Equivalent capacitance$C_{eq}=C$
Charge on capacitor between the terminals a and $b$ $\frac{Q}{2}=\frac{CV}{2}$
Energy stored in that capacitor

$=\frac{1{\left(\frac{Q}{2}\right)}^2}{2C}=\frac{Q^2}{8C}=\frac{C^2{V}^2}{8C}=\frac{C{V}^2}{8}$
Given: $C=1\mu F,V=10V$
$=\frac{\left(1\times 10^{-6}\right)\times 10^2}{8}$
$=\frac{100}{8}\times 10^{-6}=12.5\mu J$