EE 255 - Chapter 1b

EE 255 - Chapter 1b - 9 425 A t 163 A t/m 2.60 m c H l = =...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 9 425 A t 163 A t/m 2.60 m c H l = = = F From the magnetization curve, 0.15 T B = and the total flux in the core is ( )( )( ) TOT 0.15 T 0.15 m 0.15 m 0.0033 Wb BA = = = The relative permeability of the core can be found from the reluctance as follows: A l r TOT TOT = = F R Solving for r yields ( )( ) ( ) ( ) ( )( ) TOT-7 TOT 0.0033 Wb 2.6 m 714 425 A t 4 10 H/m 0.15 m 0.15 m r l A = = = F The assumption that r = 1000 is not very good here. It is not very good in general. 1-13. A core with three legs is shown in Figure P1-10. Its depth is 8 cm, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure 1-10 c . Answer the following questions about this core: (a) What current is required to produce a flux density of 0.5 T in the central leg of the core? (b) What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part (a) ? (c) What are the reluctances of the central and right legs of the core under the conditions in part (a) ? (d) What are the reluctances of the central and right legs of the core under the conditions in part (b) ? (e) What conclusion can you make about reluctances in real magnetic cores? 10 S OLUTION The magnetization curve for this core is shown below: (a) A flux density of 0.5 T in the central core corresponds to a total flux of ( )( )( ) TOT 0.5 T 0.08 m 0.08 m 0.0032 Wb BA = = = By symmetry, the flux in each of the two outer legs must be 1 2 0.0016 Wb = = , and the flux density in the other legs must be ( )( ) 1 2 0.0016 Wb 0.25 T 0.08 m 0.08 m B B = = = The magnetizing intensity H required to produce a flux density of 0.25 T can be found from Figure 1-10 c . It is 50 At/m. Similarly, the magnetizing intensity H required to produce a flux density of 0.50 T is 70 At/m. Therefore, the total MMF needed is TOT center center outer outer H l H l = + F ( )( ) ( )( ) TOT 70 A t/m 0.24 m 50 A t/m 0.72 m 52.8 A t = + = F and the required current is TOT 52.8 A t 0.13 A 400 t i N = = = F (b) A flux density of 1.0 T in the central core corresponds to a total flux of ( )( )( ) TOT 1.0 T 0.08 m 0.08 m 0.0064 Wb BA = = = By symmetry, the flux in each of the two outer legs must be 1 2 0.0032 Wb = = , and the flux density in the other legs must be ( )( ) 1 2 0.0032 Wb 0.50 T 0.08 m 0.08 m B B = = = 11 The magnetizing intensity H required to produce a flux density of 0.50 T can be found from Figure 1-10 c . It is 70 At/m. Similarly, the magnetizing intensity H required to produce a flux density of 1.00 T is about 160 At/m. Therefore, the total MMF needed is TOT center center outer outer H I H I = + F ( )( ) ( )( ) TOT 160 A t/m 0.24 m 70 A t/m 0.72 m 88.8 A t = + = F and the required current is TOT 88.8 A t 0.22 A 400 t i N = = = This current is less not twice the current in part...
View Full Document

Page1 / 14

EE 255 - Chapter 1b - 9 425 A t 163 A t/m 2.60 m c H l = =...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online