Two inductors of 60 mH each are joined in parallel. The current passing through this combination is 2.2 ~A . The energy stored in this combination of inductors in joule is

Step-by-step solution

L ₁= L ₂= L =60 mH When two inductors are connected in parallel, their equivalent inductance is given by, aligned & 1 ~L _ eq = 1 ~L ₁ + 1 ~L ₂ & L _ eq = L 2 =30 mH & u _ B = 1 2 ~L _ eq I ^2 & u _ B = 1 2 30 10⁻³ 2.2 2.2 & u _ B =0.0726 ~J & aligned