n following FIG, the energy stored in c4 is 27J.To find:Obtain an expression for energy stores in a capacitor.  Calculate the TOTAL energy stores in the system.Solution:Now first let us consider,V(QR) = VThe energy stored in a capacitor is given by,U = 1/2 CV²The energy stored in C4 is given by,U4 = 1/2 C4V²27 = 1/2 (6 × 10^{-6}) V²         (given U4 = 27 J)V² = 9 × 10^{-6}Now consider the energy stored in capacitor C2U2 = 1/2 C2V²U2 = 1/2 (2 × 10^{-6}) × 9 × 10^{-6}U2 = 9 JNow consider the energy stored in capacitor C3U3 = 1/2 C3V²U3 = 1/2 (3 × 10^{-6}) × 9 × 10^{-6}U3 = 13.5 JNow consider,U234 = U2 + U3 + U4= 9 + 13.5 + 27∴ U234 = 49.5 JNow consider the capacitors,C' = C2 + C3 + C4= 2 + 3 + 6∴ C' = 11 ×10^{-6} CNow let us consider other formula for stored energy,U = Q²/2CSo, we have,U234 = Q²/2C'49.5 = Q²/2(11 × 10^{-6})Q² = 49.5 × 22 × 10^{-6}Now consider the energy stored in capacitor C1 U1 = Q²/2C1= 49.5 × 22 × 10^{-6} / 2 × 1 × 10^{-6}∴ U1 = 544.5 JThe total energy stored is,U_{Total} = U1 + U234= 544.5 + 49.5∴ U_{Total} = 594 JTherefore the total energy in the system is 594 J