## Transformer ideal if wires have resistance?

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a.s.h.
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### Transformer ideal if wires have resistance?

Can a transformer be considered ideal if we don't assume the wires to have no resistance? Can the formula 'Power in = Power out' be used if for example the output coil has some given resistance?
joe
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### Re: Transformer ideal if wires have resistance?

Can the formula 'Power in = Power out' be used if for example the output coil has some given resistance?
Provided that the 'Power out' term includes heat production (and radiation), then yes, it's just a statement of conservation of energy:

(Electrical power in) = (Electrical power out) + (rate of heat production in primary coil) + (rate of heat production in secondary coil) + (rate of heat production in core) + (rate of energy radiated as low f EM radiation)

The second last term is due to hysteresis in the magnetisation of the core: the H,B curve makes loops of finite width. Of course, makers take some effort to keep the last four terms small. However, if you touch the casing of a transformer in use, you'll usually notice that it is warm.

Joe
a.s.h.
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### Re: Transformer ideal if wires have resistance?

Thanks Joe.

Is this solution to this problem (24 (b), attached) correct then?

That's a question from this year's HSC Physics paper. How can we say that "V_p I_p = V_s I_s" if the secondary coil has resistance (so there should be heat loss)? And if we can't say that, is that problem possible to solve with the given information?
Attachments 2014 HSC Physics Exam Q.24.jpg (81.65 KiB) Viewed 27110 times
joe
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Location: Sydney

### Re: Transformer ideal if wires have resistance?

You are correct: the answer is inconsistent. There are a couple of problems in the question. The first is a practical one: because it is a step up transformer, it is likely that the primary winding will have the greater resistive losses, not the secondary.

Then part (a) should say something like 'neglecting ohmic losses in either coil' because, unless you neglect (or calculate) these, you can't work out the turns ratio. Then part (b) should say something like 'use the turns ratio that you calculated in part (a)'.

Now it turns out that the loss is only 1%. In practice, one would assume no ohmic losses to get the approximate turns ratio, use that to get the losses, then see how good the initial approximation is. (If you do this, you can't give 3 sig figs.) But for that, the question should be rewritten.

(to be continued)

Joe
joe
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### Re: Transformer ideal if wires have resistance?

Continuing because of word limit:

Combining the two parts, can one solve it anyway? We have
elec power in = elec power out + rate of ohmic losses
given current and resistance of the secondary gives the ohmic losses
can we now calculate the turns ratio?

This is possible but not trivial, because of phase differences in
the applied emf to the primary,
the back emf in the primary,
the reflected load from the secondary load (which also depends on the angle of its impedance) and ohmic loss.
a.s.h.
Posts: 110
Joined: Tue Jan 28, 2014 3:50 pm

### Re: Transformer ideal if wires have resistance?

Would the problems with this question be taken into account during the marking (like they will be for the question about photoelectrons' kinetic energy)?

Or do the Board of Studies not even realise there are problems with this question?

Also, the 2014 paper has been put online by the Board of Studies: http://www.boardofstudies.nsw.edu.au/hs ... hysics.pdf
joe
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### Re: Transformer ideal if wires have resistance?

I don't know what the BoS is doing. You could write to them: maybe no-one has mentioned the problem yet.

Joe
joe
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### Re: Transformer ideal if wires have resistance?

Another teacher has told me that he has raised the issue with the Board of Studies and recommended that it be brought to the attention of examiners and markers.

Joe
joe
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Location: Sydney

### Re: Transformer ideal if wires have resistance?

Can a transformer be considered ideal if we don't assume the wires to have no resistance?
No. Further, even if the wires are superconducting, there are usually very significant losses due to hysteresis in the core—see the earlier discussion on that.
Can the formula 'Power in = Power out' be used if for example the output coil has some given resistance?
Well, if you included resistive heat out in the 'power out', and if there were no core losses, then it's sort of true in a very peculiar way. But that would be a very unusual meaning for 'power out'.