Magnetic equivalent circuit question

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Homework Help Overview

The discussion revolves around drawing a magnetic equivalent circuit and understanding the associated reluctance calculations. Participants are exploring the relationships between magnetic and electrical components, particularly focusing on reluctance and dimensions in the context of magnetic circuits.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to clarify the dimensions for the cross-sectional area and the calculation of reluctance. Questions arise regarding the symbols used in the circuit and whether multiple current sources are necessary. There is also discussion about the number of reluctances and how they relate to the overall circuit.

Discussion Status

Participants are actively questioning the setup and dimensions of the magnetic equivalent circuit. Some guidance has been provided regarding the dimensions of the cross-section, while others are exploring the implications of different symbols and the number of reluctances in the circuit.

Contextual Notes

There is mention of differing symbols used in American and Danish contexts, which may affect understanding. Additionally, the problem involves finding all reluctances, with some participants noting discrepancies in the number of reluctances identified in their diagrams compared to the provided solution.

theone
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Homework Statement


The question is to draw a magnetic equivalent circuit for this
http://postimg.org/image/kek5fgr9z/

Homework Equations



The reluctance is given by
##R=\frac{l}{μ_oμ_rA_c}##

The Attempt at a Solution


I am confused about the dimensions of the circuit, and also about what dimensions to use for the cross section ##A_c##

For example, is the reluctance R1 given by

##R_1=\frac{\frac{60}{1000}}{μ_0 1500 (\frac{20}{1000} \frac{20}{1000})}##
 

Attachments

  • IMG_20150522_211737145.jpg
    IMG_20150522_211737145.jpg
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theone said:
The question is to draw a magnetic equivalent circuit for this

Draw an electric equivalent, substituting the (magnetic) bars by (7) resistances. N1 can be substituted by a current source, N2 by an amp-meter.
 
Hesch said:
Draw an electric equivalent, substituting the (magnetic) bars by (7) resistances. N1 can be substituted by a current source, N2 by an amp-meter.

but I did that in the attachment. I wanted to check if the dimensions I'm using for ##A_c## is correct
 
Sorry, I didn't see the attached.

Yes, Ac = 20 * 20 mm2.

What are these ±O symbols meant to be? And why are there two? Are two currents induced in the "transformer" at the same time?
 

What are these ±O symbols meant to be? And why are there two? Are two currents induced in the "transformer" at the same time?
the mmf from the wounded coil. I'm not sure what the correct symbol is
Shouldn't there be two? ##N_1i_1## on the left and ##N_2i_2## on the right
 
theone said:
the mmf from the wounded coil. I'm not sure what the correct symbol is
Shouldn't there be two? N1i1N_1i_1 on the left and N2i2N_2i_2 on the right

I don't know anything about american symbols ( I'm from Denmark ), but you must have different symbols for a current source/a voltage source. As suggested in #2, I'd use a current source because of the analogy between electrical current and magnetic flux.

I don't know if there must be one or two current sources, for example: Is N2 loaded/current supplied? I just know:
theone said:
The question is to draw a magnetic equivalent circuit for this
 
Hesch said:
I don't know anything about american symbols ( I'm from Denmark ), but you must have different symbols for a current source/a voltage source. As suggested in #2, I'd use a current source because of the analogy between electrical current and magnetic flux.

I don't know if there must be one or two current sources, for example: Is N2 loaded/current supplied? I just know:

I forgot to mention that in addition to drawing a magnetic equivalent, the question also asked to find all the reluctance's.

On my diagram there are 4 reluctance's. But the solution only lists three reluctance's : 0.212, 0.292, 0.159 MA/m.

The 0.159 MA/m I can get as ##R_4##,

##R_4=\frac{\frac{60}{1000}}{μ_0 1500 (\frac{10}{1000} \frac{20}{1000})} = 0.159 MA/m##

But I don't know how they got the other two.
 
I guess that they have reduced the circuit into 3 reluctances:

a) R1+R2+R2 ( in series)
b) R1+R3+R3
c) R4
 
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Hesch said:
I guess that they have reduced the circuit into 3 reluctances:

a) R1+R2+R2 ( in series)
b) R1+R3+R3
c) R4

thanks, I got it now
 

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