# Voltage and Current in transformers

In transformers, the ratio of the voltages equals the ratio of the turns - so double the output coil's turns and the output voltage doubles. Then, in order to conserve energy, current halves.

This makes perfect sense in terms of $\mathrm{P=VI}$, but what happened to $\mathrm{V=IR}$? Doubling voltage and halving the current seems to completely contradict this basic law. That is, of course, unless the resistance in the output circuit changes, with R proportional to $\mathrm{V^2}$ - but I don't see how this is possible.

So how can a transformer obey both laws? Can resistance change or am I missing something else?

• You miss that both laws are independent. The first (constancy of power/energy) has to be obeyed across the transformer, Ohms law has to be obeyed in both cicuits, primary and secondary independently. And simple transformers in textbooks are thought to be "ideal", no ohmic resistance in windings, no losses, no stray inductance. Nov 1, 2011 at 20:58
• I think you've missed my point slightly. The theory may not be exact experimentally, but it must be close or it would be worthless. Theory does say that, in the output circuit, I α 1/V, which seems to contradict I=V/R. This is what I was getting at, and what is answered below. Nov 2, 2011 at 16:02
• I wrote a blog post about this a few years ago: blueraja.com/blog/194/do-transformers-obey-ohms-law Jun 17, 2014 at 0:31
• @BlueRaja-DannyPflughoeft , I read your blog, and it was amazing. I want to make sure I got it: In a step up transformer, since the number of coils in secondary is more than primary, the inductive reactance is more, and hence, as voltage increases, current decreases. Is that right? Also, if you have the time, can you answer this: physics.stackexchange.com/q/235551/93868 , a similar question, but slightly more intuitive. Thanks a lot! Feb 14, 2016 at 15:04
• Jun 29, 2016 at 19:19

There is a well known transformation law for the effective load seen through a transformer.

Let $R_o$ be the load in the output circuit.

$V_o = I_o R_o$

Assuming all power is transferred into the output circuit,

$V_o I_o = V_i I_i$

It then follows simply that

$V_i / I_i = (V_i / V_o)^2 R_o$

This is the effective load seen by the input circuit.

• It took me a bit, but I think this answers the question. I don't know why it was downvoted. The questioner sees how $VI$ is kept constant for both sides of the transformer, but doesn't see how $V^2/R$ is. The answer is that $R$ is different for the two sides, as detailed (correctly I think) here. Nov 2, 2011 at 3:56
• Yes, this is a correct answer.+1 Nov 2, 2011 at 6:59
• I asked how it was possible for resistance to change, and if my understanding is correct, user1631 is saying that the transformer itself acts as a source of resistance. Plus, this confirms that the load is proportional to the voltage squared as I suspected. Accepted and +1'd, thanks Nov 2, 2011 at 15:32

as pointed out by @Georg, $V=IR$ is a different thing and conservation of energy accross the transformers a different!!

you can validate the Ohm's Law $V=IR$ in case where you have current flowing through a resistive medium, which as a matter of fact applies everywhere expect the less abundant superconductors...

If you want to apply Ohm's law in transformer then you can do that provided you do that in primary and secondary independently since primary and secondary in a transformer is not electrically coupled rather they are magnetically connected. So applying Ohm's law between primary and secondary is not correct. In fact when you actually see the conservation of energy though primary and secondary you will notice that they are not balanced i.e. when current is halved, theoretically voltage should double; but practically this doesn't happen since resistance takes its share and reduces the voltage slightly below its expected value.

Since ohms law is applicable only to pure conductors and at constant temp. Therefore transformer can't obey this rule.