Question

- H/500 Question 2. A magnetic circuit in figure is made of silicon steel, for which the B-H relationship is given as B = 2(1 - e 500). The outer legs have 500 turns each. It is required to establish a flux of 3.6 mWb in the air-gap by applying equal currents to both windings. What is the current in each winding? 50 0.4 50 (All dimensions in centimeters)

Answer 1

6 7 I to so 0.4 5 .f 3 K 120 As are in both eequal the Currents and turns coils : they 9,+Q2 = produce same flun. 9g = 3.6m wb 3.6 or 9 = 2 = - 2 1.8m wb = 1-8810-3 wb Now dimensions for mean face path to ad = 50-5= uscm ab = 25-5 = 20cm b c = 50-5-0-4=44.6cm

The path badc has length l = 45 +20+20=85cm=0.85m The path befc has length lg = 45+20+20= 85cm = 0.85m the central core has length lz= 44.6cm = 0.446 m

flux density in core (leiht most) B,= 1.88 10-3 A 0.6 wb/m² 5x6x10-4 from the relation B= 2(1-e-H/500) 2 (i-e-/500) 0.6=2 -14/500 e I-0.3 -0.7 -4/500 = -0.3566 lage 0.7 H = soo x 0.3566 H = 178.3 AT/M . AT produced = Hal = 178.340.85 = 151.55 AT - flux density in Core (Right most) Q2 Bg 1-8X10-3 5X6X10-4 or 6 wb/m² Ac AT produced - 151.55 AT

4 Flux density in Central limb 3.68103 Bg = Ac 586610-4 = 1.2 wb/m2 from Relation 1.2 = 2(1-2-4/500) e-141500 = 0.4 -H/300 = ln 0.4 = -0.9163 H = 500x 0.9163=458.15 AT/M - AT required = 458.15 80.446 = 204.33 AT Now & Fluse density in air gap dg Bg = Ag 1-2 wblm? [ 89 = 3-6Kto = 3.6X10-3ub] Ag = SX6X109 m2 air gap and lg = = 0.4cm = 0.ooum - AT required in Bg. lg 1.280.004 47X10-7 3821.65 AT Mo

Total Ampere turns required = 151.55 +151.55 + 204.33+ 3821.65 4329.08 AT As both coils have same turns and currents - AT produced by each coil To tal AT 2 4329.08 2 = 2164.54 AT - Current in each winding AT 2164.54 11 = 4.33 A N 4.33 A ! each winding I= in