Find the energy of the circuitSolutiontwo inductors in paral

Find the energy of the circuit

Solution

two inductors in parallel = Leq = 2*2/(2+2) =1H

two capacitors in parallel =Ceq = 1+1 = 2F

voltage across Leq= 3V = voltage across Ceq

energy stored in Ceq =1/2*Ceq*V^2 = 1/2*2*9 =9Joules

under dc conditions inductors are shortcircuited and capacitors are opened

theoritically current in the inductors would be infinite but if we consider a 1ohm resistance in series with source voltage then current in Leq= 3/1 =3A

energy stored in Leq= 1/2*L*I^2= 9/2

total energy= 9+9/2 =27/2 joules