# Magnetic equivalent circuit question

1. May 22, 2015

### theone

1. The problem statement, all variables and given/known data
The question is to draw a magnetic equivalent circuit for this
http://postimg.org/image/kek5fgr9z/

2. Relevant equations

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

3. 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

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

#### Attached Files:

• ###### IMG_20150522_211737145.jpg
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2. May 23, 2015

### Hesch

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

3. May 23, 2015

### theone

but I did that in the attachment. I wanted to check if the dimensions I'm using for $A_c$ is correct

4. May 23, 2015

### Hesch

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?

5. May 23, 2015

### theone

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

6. May 23, 2015

### Hesch

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:

7. May 23, 2015

### theone

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.

8. May 23, 2015

### Hesch

I guess that they have reduced the circuit into 3 reluctances:

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

9. May 23, 2015

### theone

thanks, I got it now