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Q.
What is the energy stored in the capacitor between terminals $a$ and $b$ of the network shown in the figure? (Capacitance of each capacitor $C = 1 \,\mu F$)
AIIMSAIIMS 2009Electrostatic Potential and Capacitance
Solution:
Network is redrawn as shown in figure.
This is a balanced Wheatstone's network.
Equivalent capacitance$C_{e q}=C$
Charge on capacitor between the terminals a and $b$ $\frac{Q}{2}=\frac{C V}{2}$
Energy stored in that capacitor
$=\frac{1\left(\frac{Q}{2}\right)^{2}}{2 C}=\frac{Q^{2}}{8 C}=\frac{C^{2} V^{2}}{8 C}=\frac{C V^{2}}{8}$
Given: $C=1 \mu F, V=10 V$
$=\frac{\left(1 \times 10^{-6}\right) \times 10^{2}}{8}$
$=\frac{100}{8} \times 10^{-6}=12.5 \mu J$