Q. Two coils with self inductances 1H and 4H have a mutual inductance of 1H. The RMS value of current flowing in the two coils is 4A and 1A respectively. Find the coupling factor

Ans.

Given

                  Mutual inductance M = 1H

                       L₁ = 1 H

                       L₂ = 4 H

Given     I₁ = 4 Amp

                      I₂ = 1 Amp

 The total magnetic energy stored

                W = 1/2L₁I₁² + 1/2L₂I₂² ±1/2MI₁I₂

                W = {1/2 ×1 × 16 } + {1/2 × 4 × 1}± {1/2 × 1 × 4 × 1}

                W = 12 or 8 watt