Megnetic circuits: MMF and reluctance.
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The magnetic circuit shown in figure below is built up of solid iron cross section, 1x1cm2. Each air gap is 1 mm wide. For this type of air gap, the effective cross-section of the air may be taken as 1.2 x that of iron path to allow for fringing. Two different grades of iron are used. When a coil having 2000 turns is wound on part B, and a current of 0.5A is applied to it, the relative permeability of part A is measured as 500, and of part B as 1000.
a) Calculate the Reluctance of part A
b) Calculate the Reluctance of part B
Ra = 0.2 / [1 x 10^-4 x 4pi x 10^-7 x 500] = 3.18 x 10^6 At/W
Rb = 0.38 / [1 x 10^-4 x 4pi x 10^-7 x 1000] = 3.02 x 10^6 At/W
The above is a provided solution for parts (a) and (b). Please explain (comprehensively) how these equations were formed and particularly, where did the 0.38 come from when calculating the reluctance of part B?
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Solution Summary
The magnetic circuit with two air gaps is given and the reluctance or the circuit is calculated by calculating the series and parallel combinations of the reluctances.
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