Determine self inductances of coils, Electrical Engineering

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

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Posted Date: 2/4/2013 6:17:37 AM | Location : United States







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