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
N_{1} = 600, N_{2} = 1700, i_{2} = 6 A, Φ_{2} = 0.8 m Wb, Φ_{21} = 0.5 m Wb
L = N (φ_{2}/i_{2 })= (1700 × 0.8 × 10^{-3 }/6) = 0.227 H
k = φ_{21} / φ_{2}= (0.5 × 10 ^{- 3 })/ (0.8 × 10^{-3} )= 0.625
Self inductance
L= N ^{2} μ A/ i
∴ L_{1 } = (N_{1}) ^{2} μ A/ l
L_{2 } = (N_{2}) ^{2} μ A/ l
∴ L_{2}/ L_{1} = (N_{2}) ^{2} /(N_{1}) ^{2}
∴ L_{1} = L_{2} (N_{1}) ^{2} /(N_{2}) ^{2} = 0.227 ((600)^{2}/(1700)^{2} = 0.028 H
Mutual Inductance