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Transformer resistance question

  1. May 8, 2008 #1
    Does the resistance in the secondary circuit connected to a transformer provide for a certain resistance in the primary circuit?
  2. jcsd
  3. May 8, 2008 #2
    The question is a little unclear and I would reframe it as this: Can one measure the secondary resistance from the primary side? The answer to that is clearly yes, but it requires an involved testing procedure.
  4. May 8, 2008 #3


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    Hi Gear300,

    Yes, that's right. If you start with just a resistor connected to an AC voltage source (and no transformer), you know of course that you can find the current by V=IR. (V and I are max or rms values)

    Now when we connect the voltage source to the primary side of a transformer, and the resistor to the secondary side of the transformer, what is the current from the AC voltage source? It will be given by:

    V = I R \left(\frac{N_p}{N_s} \right)^2

    So the idea is that the voltage source sees an effective resistance of

    R_{\rm eff} = R \left(\frac{N_p}{N_s} \right)^2

    You can derive this by combining Ohm's law, the transformer voltage relation, and the power voltage relation.

    (Of course this is for an ideal transformer, with a 'good' driving frequency, etc. Dealing with real tranformers to a high precision will complicate things, but the above will give you the general idea.)
  5. May 9, 2008 #4
    Alright...I'm following up with you on this, although, I'm stuck at one more thing. For a step up transformer, in which the secondary circuit is a length of transmission lines, there is a current I running through the secondary and a resistance R along the wires, that all occur by a voltage V induced in the secondary. I am supposed to measure the power loss across transmission lines, in which, when I use I^2*R the value differs (considerably) from when I use V^2/R...I'm assuming this might mean that using the secondary voltage may not be right...but it sort of has me confused (since the process doesn't seem to obey Ohm's relation).
  6. May 9, 2008 #5


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    I hope I'm understanding the situation, so let me know if it sound like I'm saying something wrong!

    The correct expression for the power loss would be I^2 R, because the R is the resistance of the transmission wire and I is the current in the transmission wire.

    The expression V^2/R is not the correct expression, because the V is not the voltage from one end of the transmission wire to the other. The wire is going somewhere, and so V is the voltage from the beginning of the transmission wire to the ground.

    So if the transmission wire is connected to a building, then the V you found is the potential difference across the combination of (transmission wire + building). If you knew the resistance of the building, you could find the total power used (power loss in wire + power used by building) by doing V^2/(R_wire + R_building) .
  7. May 9, 2008 #6
    But what if there was no building and that the secondary was simply a loop consisting of the inductor and resistance? If the power delivered to the circuit is 20E6W and the stepped up voltage is 230E3V, then the current in the secondary would be 87A. If the net resistance across the secondary was 2 ohms, then when using P = I^2*R, it differs largely from P = V^2/R.
  8. May 9, 2008 #7


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    Hi Gear300,

    From your post it sounds like you are just picking 2 ohms out as an example. This is not the right thing to do, because the power delivered to the circuit depends on the value of R.

    To see why, let's pick a simple case: one AC voltage source with V1 attached to the primary side that has turns N1, and one resistor R attached to the secondary side that has turns N2. Let's say that's all we knew; here is one way you could go about finding the other values:

    1. Find secondary voltage V2 by using normal transformer equation (V1/N1=V2/N2).
    2. Find secondary current by using V2=I2*R
    3. Find primary current by using power relation I1*V1 = I2 * V2

    However, an alternate procedure that is more in the sprit of your example would be this:

    1. Find effective resistance of primary coil by R_eff = R (N1/N2)^2
    2. Find primary current by V1 = I1*R_eff
    3. Find power on primary side: V1*I1
    4. Find secondary current I1*V1 = I2*V2

    So step #3 in this last procedure I think shows where the problem in your example is. Your power of 20E6W depends on the resistance. (I think we could find the resistance in your example by just doing V2/I2.)

    Of course if your circuit has other resistances on the primary or secondary side you'll have to take those into account also.
  9. May 9, 2008 #8
    I actually took those values from a book...but from your last sentence, it came to thought that if there was resistance in the primary, it would also affect the secondary...and I would have to take that into account. But I was just wondering: If I1*V1 = I2*V2 in an ideal step up transformer and since V2 is larger than V1, then I2 would have to decrease to compensate. In Ohm's relation, if V increases, so should I, which is sort of opposite of what happens here. This is pretty much where the root of my confusion holds...because of this, I'm not sure how effective Ohm's relations are, even though they are used to derive several of the transformer equations.
  10. May 9, 2008 #9


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    Sorry about asking again, but I want to be clear that we are talking about the same thing: did you get the value of 2 ohms as the net resistance from the book? I could tell that the power and stepped-up voltage were probably given values, but I was wondering about the resistance.

    Let's break up the predictions from Ohm's law into two cases:

    1. Ohm's law predicts that for constant resistance, the current increases with increasing voltage. The transformer circuit obeys this behavior. If we increase V1, then V2 increases (it just depends on the turns ratio). R_eff on the primary side is the same, so I1 will get larger. Then using the power equation (I1 V1= I2 V2) the current on the secondary will also get larger. So all currents get larger with larger voltage V1.

    2. Ohm's law predicts that for constant voltage, the current decreases with increasing resistance. Here V1 and V2 are unchanged, but we increase R. Now R_eff gets larger, so I1 will get smaller, so you will have a smaller power. This means the power equation predicts a smaller current for I2, and so the transformer obeys this also. All currents get smaller with larger resistance R.

    About the seeming contradiction that you mention between the power law equation and Ohm's law: Ohm's law relates voltage, current, and resistance between two specific points. So it predicts what will happen to the current between two points if you change the potential difference across those same two points. But the power expression I1 V1 = I2 V2 relates two different places in the circuit. So there is no contradiction with Ohm's law if you say since V2 > V1, then I2 < I1. Both sets of values (I1, V1) and (I2, V2) are obeying Ohm's law on their different places on the circuit, and their respective resistance values (R_eff and R) are different.
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