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
L_{1} = 1 H
L_{2} = 4 H
Given I_{1} = 4 Amp
I_{2} = 1 Amp
The total magnetic energy stored
W = 1/2L_{1}I_{1}^{2} + 1/2L_{2}I_{2}^{2} ±1/2MI_{1}I_{2}
W = {1/2 ×1 × 16 } + {1/2 × 4 × 1}± {1/2 × 1 × 4 × 1}
W = 12 or 8 watt