How would i solve this? I have the answers and solution but it isnt a good exp\'

Question

How would i solve this? I have the answers and solution but it isnt a good exp'anation. Please give me step by step instructions!

Part 2: Tutorial Problems 1. Consider the following magnetic structure: 40 mm 250 mm 25 mm 45 mm Imm 1 mm 300 mm 25 mm The normal magnetisation characteristic of the core material is H(Am) 100 200 280400 6001000 1500 2500 B0.51 0.98 1.20 1.37 1.5 6 1.731.78 Determine the flus density in the centre limb and the necessary mmf for a winding on the centre limb: (a) For a flux density of 1.2 T in each air gap. (b) For a flux density of 1.2 T in one air gap when the other is closed with a magnetic material of the same permeability as the core material Page 2 of

Explanation / Answer

The flux the central limb divides at point B equally into 2 paths (say) ABCD and ABEF(LET CENTRAL PATH AB).

we may consider either path say ABEF and calculate mmf required for it. The same mmf will also send flux thru the other parallel path ABCD.

a) given flux density B of air gap = 1.2 T

H= B/ u0 = 1.2/(4 pi x10^-7) =0.954 x 10^6 AT/m

length of air gap = 1mm

mmf required for air gap= H x l = 0.954x 10^6 x10^-3 = 954 AT

Flux density in the outer limb EF is same as that of airgap = 1.2 T, H= 280 AT/m (from table)

mmf required for limb= H xl =280 x500x10^-3 =140AT

total mmf required for central limb = 954+140+954+140 = 2188 AT

H for central limb= 2188/(300 x10^-3) = 7293 AT/m

by extrapolation, flux density = B= 4.5T

b) if only one side is airgap,

total mmf required for central limb = 954+140+140 = 1234 AT

H for central limb= 1234/(300 x10^-3) = 4133 AT/m

by extrapolation, flux density = B= 1.9T

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