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Serial connection: two transformers

  1. Jun 17, 2009 #1
    Good afternoon!

    To build a serial connection of two transformers, four coils are used: Two of this coils have n1 turning, two of them have n2 turnings.

    The coils are built together in the way, you can see on this picture:
    http://img197.imageshack.us/img197/2698/transformer.jpg [Broken]

    The supply on the left side amounts 50V, the voltameter on the right side measures only 30V.

    What is the quotient n1/n2 ?


    My solution are
    1. n1/n2=3/1 and
    2. n1/n2=1/3 (clear that there must exist two solustions).

    Is this solution right?

    Thank you very much.
    Mark
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 17, 2009 #2

    berkeman

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    Welcome to the PF.

    I must be missing something. If both primary and secondary have n1 + n2 windings, it would seem that the turns ratio is 1:1. What am I missing?
     
    Last edited by a moderator: May 4, 2017
  4. Jun 17, 2009 #3

    berkeman

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    Hmmm. Maybe that they do not share the same core. Let me think about this...
     
  5. Jun 17, 2009 #4

    berkeman

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    I don't think your solutions work. How did you derive them? They don't work in the equation that I've written for the transfer function.
     
    Last edited by a moderator: May 4, 2017
  6. Jun 17, 2009 #5
    Ok, I try to explan my way to solute the problem:

    1. U1,left : U2,left= (n1 : n2)^2

    2. U2,left = U1,left * (n1 : n2)^(-2)
    =50V * ( (n1 : n2)^2 + 1 )^-1

    3. U1,left = 50V * ( (n2 : n1)^2 + 1 )^-1


    4. 30V = U1,right + U2, right = U1,left* (n2/n1) + U2left * (n1/n2)
    =50V * ( (n1 : n2)^2 + 1 )^-1 * (n2/n1) + 50V * ( (n1 : n2)^2 + 1 )^-1 * (n1/n2)


    After calculation:
    n1/n2 = 3 , n1/n2 = 1/3

    Mark
     
  7. Jun 17, 2009 #6
    Why do you think my solution is wrong?
    Which equation do you mean?
     
  8. Jun 17, 2009 #7

    berkeman

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    I'm sorry, I wasn't able to follow your derivation. You may be correct, but your answers do not fit in the equation that I derived (although I did it quickly, so I could be wrong).

    I started with equations for the two voltages on the left that add up to 50V. The voltages across the two coils in the primary will ratio with each coil inductance, and hence will ratio with the number of turns squared. That gives you one equation (EDIT -- well, actually two equations). Then the output secondary voltage of 30V is formed by the transformed voltages from the primary, via each respective turns ratio. That gives you another equation.

    When I combine those two equations, I get an equation that is not satisfied by a 3:1 turns ratio for n1 and n2. Again, I could be wrong.

    Is that the way you are approaching the problem?
     
  9. Jun 17, 2009 #8

    berkeman

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    Ooo, hey. I take it back. 3:1 DOES work in my equation. Sorry for the confusion. The equation I got in the end was:

    [tex]\frac{3}{10} = \frac{n_1 n_2}{{n_1}^2 + {n_2}^2}[/tex]
     
  10. Jun 18, 2009 #9

    Redbelly98

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    Aha, that's what I was missing when I first read the problem. As I understand it, this assumes a fixed coil length. If, instead, the turns-per-length were fixed, I think the voltage would simply ratio in proportion to the number of turns. Is that thinking correct, berkeman?
     
  11. Jun 18, 2009 #10

    berkeman

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    Given the same core size and material (it looks like they are the same transformers, with one just reversed), the magnetizing inductance varies with the number of turns squared. Since the AC impedance varies liinearly with inductance, the voltage division will go with the number of turns squared. So for the two voltages on the primary, you would get:

    [tex]v_1 = 50V \frac{{n_1}^2}{{n_1}^2 + {n_2}^2}[/tex]

    [tex]v_2 = 50V \frac{{n_2}^2}{{n_1}^2 + {n_2}^2}[/tex]

    And then you write the equation for the output voltage as a function of those two input voltages and the respective turns ratios.
     
  12. Jun 18, 2009 #11

    The Electrician

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    This is a very interesting problem.

    The reciprocity theorem says that the voltage transfer ratio in one direction should equal the current transfer ratio in the other direction.

    So, if you short the left side and apply a current to the right side, is the current in the left side in a 5:3 ratio with respect to the current in the right side?

    My intuition fails me here; I would have to do the math.

    A somewhat related problem would be to connect a 100 ohm resistor to the right hand side, and calculate the impedance seen at the left hand side.
     
  13. Jun 18, 2009 #12

    berkeman

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    The problem as given has an open circuit on the secondary (I think), so the primary current is only determined by the primary magnetizing inductance and the frequency. Your related problem is an interesting one! If I have time later today I'll try it. Have you worked out the answer already?
     
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