Question

The energy stored in the system of capacitors shown in the circuit is (Capacitance of each capacitor is $C=0.1\mu F)$

• A. $355\mu J$
• B. $455\mu J$
• C. $555\mu J$
• D. $255\mu J$

Answer 1

The correct option is B $455\mu J$

Capacitance in the path AB is

$C_{AB}=C\parallel C=2C$

Capacitance in the path ABG is
$C_{AG}=\left(2C\right)seriesC$
$=\frac{2C\times C}{2C+C}=\frac{2}{3}C$
Capacitance between D and E is
$C_{DE}=\left(CseriesC\right)\parallel \left(\frac{2C}{3}\right)$
$C_{DE}=\frac{C}{2}+\frac{2C}{3}=\frac{7C}{6}$
Capacitance between D and F is
$C_{DF}=Cseries\frac{7C}{6}=\frac{C\times \left(\frac{7C}{6}\right)}{C+\left(\frac{7C}{6}\right)}$
$C_{DF}=\left(\frac{7C}{13}\right)$
Energy
$E=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{7C}{13}\right)(130{)}^2$
Substituting C = 0.1 \mu F we get,
$E=\frac{1}{2}\left(\frac{7\times 0.1\times {10}^{-6}}{13}\right)\times (130{)}^2=455\times {10}^{-6}=455\mu J$