Determine self inductances of coils:

The number of turns in two coupled coils A and B are 600 and 1700 respectively. While a current of 6 A flows in coil B, the total flux in this coil is 0.8 m Wb and the flux linking the first coil is 0.5 m Wb. Determine self inductances of coils A and B, mutual inductance among the coils and coefficient of coupling.

Solution

N₁ = 600, N₂ = 1700, i₂ = 6 A, Φ₂ = 0.8 m Wb, Φ₂₁ = 0.5 m Wb

L  = N (φ₂/i₂ )= (1700 × 0.8 × 10^-3 /6) = 0.227 H

k = φ₂₁  / φ₂= (0.5 × 10 ^{- 3} )/ (0.8 × 10^-3  )= 0.625

Self inductance

 L= N ² μ A/ i

∴          L₁  =   (N₁) ² μ A/ l

L₂  =   (N₂) ² μ A/ l

∴          L₂/ L₁ =   (N₂) ² /(N₁) ²

∴          L₁ = L₂   (N₁) ² /(N₂) ²    = 0.227 ((600)²/(1700)² = 0.028 H

Mutual Inductance