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Magnetic equivalent circuit question

  1. May 22, 2015 #1
    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:

  2. jcsd
  3. May 23, 2015 #2

    Hesch

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    Draw an electric equivalent, substituting the (magnetic) bars by (7) resistances. N1 can be substituted by a current source, N2 by an amp-meter.
     
  4. May 23, 2015 #3
    but I did that in the attachment. I wanted to check if the dimensions I'm using for ##A_c## is correct
     
  5. May 23, 2015 #4

    Hesch

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    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?
     
  6. May 23, 2015 #5
    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
     
  7. May 23, 2015 #6

    Hesch

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    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:
     
  8. May 23, 2015 #7
    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.
     
  9. May 23, 2015 #8

    Hesch

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

    a) R1+R2+R2 ( in series)
    b) R1+R3+R3
    c) R4
     
  10. May 23, 2015 #9
    thanks, I got it now
     
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