In the given electrical circuit, if the switch S is closed then the maximum energy stored in the inductors is:

Step-by-step solution

In figure below, if switch S is open then total energy stored in the capacitors, is suppose E₀ . Energy in IF capacitor, E₁= 1 2 C V^2= 1 2 C ( Q C )^2= 1 2 Q^2 C = 1 2 4^2 1 =8 ~J Similarly, E₂= 1 2 Q^2 C = 2^2 2 2 =1 ~J So, the total energy, E₀=E₁+E₂=8+1=9 ~J Now, switch's is closed then the common potential of Capacitors, aligned V_ common & = C₁ V₁+C₂ V₂ C₁+C₂ & = 1 4+2 1 1+2 = 6 3 =2 ~V aligned Hence, now the new arrangement of energy, and aligned & E₁= 1 2 C₁ V_ common ^2= 1 2 1 4=2 ~J & E₂= 1 2 C₂ V_ common