Question

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

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

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