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Transformers Under Load and No Load

  1. May 2, 2015 #1
    1. The problem statement, all variables and given/known data
    I have a step down transformer (6 : 1)
    Copper Loss of 110W at full Load
    Primary Coil: I = 3.33A R = 0.35 Ohms Power Loss = 4W
    Secondary Coil: I = 20A R = 0.265 Power Loss = 106W
    Power Drawn Over the Primary = 75W

    9kVA Transformer
    2. Relevant equations

    The Equations i have are:

    E1/E2 = N1/N2 = I2/I1

    Power Loss = I^2 x R

    3. The attempt at a solution

    • No Load Current is 5% less than Full Load Current
    • Primary Copper Loss is less than 1/400 of full load copper loss
    Do i need to calculate the flux? Or Calculate the other Eddy Currents or Magnetic Flux?

    Your help will be appreciated

  2. jcsd
  3. May 2, 2015 #2


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    That is at full load. At no load, I2=0A, resulting in a copper loss = 0W

    Again at full load. I think, that your 5% comes from that the no load current is 5% of the full load current. This no load current keeps up the flux and delivers power to the iron-losses, which are the same at full or no load.

    That must be at no load and is the sum of primary copper losses and iron losses. You don't have to calculate the flux. Iron losses = 75W - ( primary copper losses ).
    Last edited: May 2, 2015
  4. May 2, 2015 #3
    So calculating the Load and No Load Current.
    My Full Load Current would be the 3.33A on the primary side since there is minimal current across the secondary side.

    My Load current, Would this involve inserting a resistor for load or would i use the Coil Resistance for both Primary and Secondary side?

    Thanks for your help!
  5. May 2, 2015 #4


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    At full load the currents are: I1 = 3.33A, I2 = 20A.

    You can just assume, that a resistor ( or other load) has been inserted. The value is not of interest in your calculations as for the transformer.

    Primary and secondary coil resistances are not "loads".

    I don't know what you are asked to calculate, but maybe you need to know cos φ and V1, V2.

    Edit: Sorry, I've seen it now (V1, V2): 9 kVA transformer.
    Last edited: May 2, 2015
  6. May 2, 2015 #5
    Hi Hesch,

    I have been asked to calculate the Non Load Current and Load Current for the transformer.

    Can i find this out using the formula's i have provided above since there is no voltage stated?

  7. May 2, 2015 #6


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    Well, I assumed that the load currents were given by the exercise (problem statement): 3.33A, 20A. ?

    If not: Please attach the exercise.
  8. May 2, 2015 #7
    Everything i have given to you i what i have been given and calculated!
    I have already completed these steps

    A 6/1 step down transformer has a full load secondary current of 20A and is rated at 9kVA. The copper loss at full load is 110W. The primary winding has a resistance of 0.35Ω.
    a) Find the resistance of the secondary coil and the power loss in the secondary coil.
    b) 75W of power is drawn by the primary when the secondary is open circuit. Calculate the efficiency at the rated load if the power factor is 0.75.

    But i need to do this
    Find if The no-load primary current is ≈ 5% less than the full-load primary current.
    Find if The primary copper loss on no-load is less than ≈ 1/400 of that on full-load.

    There is no mention of voltage but i have the Turns, Amps, Resistance on the windings (R1 = 0.35, R2 = 0.265) , Primary Power Drawn on Open Circuit(75w), Power Loss of Primary (4w) & Secondary (106w), Cooper Loss at Full Load (110w)

  9. May 2, 2015 #8
    So if my Full load current or Closed Circuit current is
    I1 = 3.33A
    I2 = 20A

    Can i use the 75w that is drawn under a no load condition since it is open circuit, So If i was to use the 75w / 0.35 Ohms = 214A --> sqrt(214) = 14.127A this give me the 14.127A / 6 = 2.355A you can see within the simulation. This has given me the simulation value i have been given. Also within the simulation, i have used a current source since i have no voltage. Would this be correct?

    Attached Files:

  10. May 2, 2015 #9


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    If the rated power = 9 kVA, and the full load (rated) current, I2 = 20A, V2 must be 450V.

    You have solved a) correct. So now you know, that full load primary copper losses = 4W.

    R1 ≈ 4W / I12 = 0,36Ω

    Now there is no load. There are no copper losses in the secondary as I2 = 0. The 75W loss is due to primary copper losses + iron losses.

    Iron losses = 75W - 4W = 71W.

    Riron ≈ (6*V2)2 / 71W = 103kΩ

    There it was: Power factor = cos φ.

    But there is a problem: We don't know if the phase ( 41.4° ) is due to the load or the reactances in the transformer? So I really don't know what to do, concerning the efficiency. The efficiency must be calculated as active outputpower/inputpower, and copper losses are calculated by absolute currents. That's why a short-circuit test of a transformer is done: Then you know about reactance in the transformer.
    Last edited: May 2, 2015
  11. May 2, 2015 #10
    Hi Hersch,

    Thanks for that, The efficiency is not an issue as i do not need to calculate that for this example but maybe required if he decides they want to know more.

    Ok So i got the voltage part since i had reverted the KvA formula. My V1 = 2700V, V2 = 450V

    Now the Iron Losses have been worked out, I have the voltages. Using Ohms law i should be able to use that Resistance to calculate the current, Its gives me a far lower value if i am looking for a 5% value of Non Loaded - Loaded Current.

    I really do appreciate your help Hersch. Helping me make more sense of it than i have been taught.

  12. May 2, 2015 #11


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    No, you cannot do this, using only resistors. A model of a transformer also includes inductance, and you don't know the values:

  13. May 2, 2015 #12
    Thanks for the reply,

    So the Unloaded current cannot be calculated this way. If we have no circuit on the secondary side, we are only considering the Primary Side Resistance from the windings (0.35 ohms). I am sorry if i am missing the obvious.

    Thanks again!
  14. May 2, 2015 #13


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    In the model, Ip = Is + Ie. But you don't know Ie as you don't know the value of Xm. Thus, not knowing Ip, you don't know the primary copper losses.
  15. May 3, 2015 #14


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    At no load the primary copper losses will drop to a tiny fraction of their full-load value, so almost zero here.

    I think it's reasonable to assume the quoted power factor refers to the pf of the secondary's load.

    The question involves a low-power power transformer, so transformer reactances are probably neglected for these calculations.
  16. May 3, 2015 #15
    Hi Nascent, Thanks for the reply.

    So i have my primary Closed Circuit Current which is the 3.33A i had calculated earlier since i was able to calculate the secondary current. Is that correct?
    Via the simulation, to get the calculated values i had, the current source was placed in at 4.71 which gave me the values of the current i needed.

    The No Load is Current is 5% less than the full load current. From what Hersch was explaining earlier i need to have an Ie and Xm but from the data i have given above this is all i have.


    Attached Files:

  17. May 3, 2015 #16


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    There are 75w of iron losses. These don't change [much] from no load to full load. Have you used this to calculate Re to account for these iron losses?

    How did you come up with a primary current of 4.71A for your simulation?

    Is this stated in the question, or is it something you interjected?
  18. May 3, 2015 #17
    Hi Nascent, Thanks for the reply,

    This was stated earlier on in the question since the Load Current is 5% higher than the non load current.

  19. May 3, 2015 #18
    From my earlier calculations,
    I can confirm the 110W copper loss with I1 = 3.33A R1 = 0.35ohms I2 = 20A R2 = 0.265 Ohms but this is at full load. Which to me says that those currents are for full load.
    I have a 75W loss at open circuit on the primary side so can i not rework this to find the current on the primary side?

    As for the Full Load Copper Load, i can assume that this is the 110W since it has been calculated at the full load values.
  20. May 3, 2015 #19


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    But as stated it doesn't make sense. So we have to try to work out what was intended, and it's not proving easy.

    First, have you translated this from another language?
  21. May 3, 2015 #20


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    Yes, those all seem right.

    That's what I asked you! Have you done this calculation?
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