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Current and Impedance in a Transformer

  1. Apr 3, 2013 #1
    My textbook doesn't cover transformers in very much detail at all, so I only have a very vague idea as to how they work...

    1. The problem statement, all variables and given/known data

    The secondary voltage of an ignition transformer in a furnace is 13.9 kV. When the primary operates at an rms voltage of 120 V, the primary impedance is 22.6 Ω and the transformer is 91.1% efficient.
    a) What turns ratio is required?
    b) What is the current in the secondary?
    c) What is the impedance in the secondary?

    2. Relevant equations

    V2/V1 = N2/N1 (where N2/N1 is the turns ratio)
    V1 * I1 = V2 * I2
    V1/I1 = R/(N2/N1)^2

    3. The attempt at a solution

    I already solved a) correctly, which was just 13.9 kV/120 V (i.e. N2/N1). But besides that, I honestly don't know what else to do. I guess I could start by plugging in 22.6 ohms into the equation for impedance [i.e. I(rms) = V(rms)/Z], but I'm getting completely mixed up. I'm also not sure how efficiency affects the final result...
     
  2. jcsd
  3. Apr 3, 2013 #2
    Efficiency refers to the ratio of V*I power of secondary and primary circuits. So, if you find Iprimary you can find Isecondary and hence the impedance of the secondary circuit.
     
  4. Apr 3, 2013 #3
    Thanks! It doesn't say whether or not the secondary voltage (13.9 kV) is an rms voltage... should I assume that it is?
     
  5. Apr 3, 2013 #4
    Yes I think it should be assumed so.
     
  6. Apr 3, 2013 #5

    SammyS

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    That's a reasonable assumption, since it says that the primary voltage is rms .
     
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