Question

Magnetic Circuits - Part A - Calculate reluctances Using the information given in the introduction, calculate the reluctances for the equivalent circuit shown. Express your answers to three significant figures in A turns w View Available Hints) 00 AED Ivec ? R. R₂ R- W Learning Goal: To understand how magnetic structures can be analyzed by drawing an equivalent circuit, and to use the equivalent circuit to calculate magnetic fixes and coll currents. When analyzing magnetic structures, the geometry is often complex enough that using the fundamental rules can be very difficult without numerical methods. However, there are approximate methods that are often sufficient for engineering calculations. When the magnetic field is mostly contained within cores of high permeability and small air gaps, the structure can can be converted into an equivalent magnetic circuit. These circuits can then be analyzed using standard techniques for electric circuits The role of a voltage source is taken by cols wrapped around the core. The magnetomotive force is the analogue of the voltage. A coil with Nurs carrying a current i has a magnetomotive force F-Ni The cores and air gaps all have a reluctance, which plays a similar role to the resistance in an electrical circuit. The reluctance of a path is R A where is the mean length of the path is the magnetic permeability along the path, and is the cross-sectional area of the path. We use the mean length because the cross-dimensions of magnetic structures are often significant, so a single length is not well defined. The permeability of air is taken to be the permeability of free space H. = 4 x 10-7 W.For magnetic materials, the permeability is usually given as a relative permeability such that The magnetic core shown has dimensions d 8 cm and g 0.5 cm. The relative permeability of the core is 400. What are the reluctances in the equivalent magnetic circuit? The iron core has a cross sectional area of 52 cm and the effective area of the air gap due to fringing is 8.3 cm Submit Part B. Calculate the flux density Both cols have N.500 tums and the current isi - 26 A. What is the fux dersity in the ar? Express your answer to three signficant figures in T. ► View Available int() DAX! Voc т Submit

Answer 1

Part. A R = l i, Ri= ? HAC Milun length 3d l=3x(8x10-2) (d=8cm) l= 0.24 m (Mr=400) x 400 4 TU X10 + M = Mour 16TX 105 M = 16 11 X 105 Ac = 5.28104 m² R 0.24 16 TTX 10 ix5.2010 MAC Ri=9.18 x 105 A-T Wb R3 = ? ६ R3 of Ri han dimensions sana as R₂=R. Rz = 9.18x105 AT wb Rs = ? R2 = 2R+R' 2 R. d d- g8-0-5x10²= 3.75 X 10" for R m 2. for 9 = 0.5x10²m for R Ac 5.38 10 4 m² to Fringing) ( duc 8.3 810-4 m2 Ac R3 = 21 (air Moal) + MAL MACI 2 0.5x10 ² + 3 X 3.75 x 16 41 Xto7 x 400 x 5-2 x 10 U nX 107 X1 XG-3X104 3.87 x 10 47.94 xios -

50.8810 AT Wb Part B: N: 500 ie). 6 A mmt Ni 500 * 3-6 = 1300 Equivalent circuit US V ↑ Ž Rs v mml - - R2 Ri 1 % WWW 03 R. 1300 13 00 62 Flux density CLLY gap = B = Ac! 1300 - V + 1300 - Nodol analyses R2 RI 3 (Ri=R3) 5 ( 1300 - V) Rz V =) 2 ( 1300 - V) 50.8x10s 9.18 * 105 11.067 (1300) = v 14587.1 = 12.067 V 1192.31 A-T Ø : 1192.27 = 23.47X10S 50.8xios R2

B= Ac' 5 こ 23.47 X10 8.3 X 10 4 B= 0.283T Partei 0.3 cm Horo to dcercand ما d=8cm Ac'= 8.3 cm² Ac = 5.2 cm2 as 9 decrease R3 changes Rg (now) = 21 MAC MAC' m d-g 2 8-0-3 2 = 3.85 x 10² 2 0.38 0 R2 (new) = + «Όχιο .7x8.3x104 2x3.858 102 GTX107x400x5-2x10 4 2.946x 105 +28.763 x 105 31.7xios Rz (new) = 31-7X10S AT طلا Given Flux dinsitu US constant 02= BAC constant B = Ac

- S Sca 02 2 23.47 x 10 ابيا Ri Rs R2(New) v' 400 +4ooi Ø 2 x R2 (NCL) = 23.47X105x31-7X 105 v' = 744 using Nodal analyses 4001- V' 400i-v' R2(New) Ri R3 744 23,47 x 10 2 (4001 - 744) 9.18 X105 31.78105 2(4001 - 744) = 215.455 800i - 1488 215-455 3.13 A Answers 9.18 X 10 Part A R = R3 AT alb R2 = 50.88 103 AT طنا Part 13 : B=0.283T Partei i-13 A