# Transformer Question

1. Sep 22, 2013

### Staff: Mentor

When stepping up/down voltage in a transformer, what limits the current flow through the secondary? Is it the impedance of the windings/circuit?

2. Sep 22, 2013

### meBigGuy

Copied from Wikipedia:

Core losses collectively called magnetizing current losses consisting of:
Hysteresis losses due to nonlinear application of the voltage applied in the transformer core
Eddy current losses due to joule heating in core proportional to the square of the transformer's applied voltage.
Whereas the ideal windings have no impedance, the windings in a real transformer have finite non-zero impedances in the form of:
Joule losses due to resistance in the primary and secondary windings
Leakage flux that escapes from the core and passes through one winding only resulting in primary and secondary reactive impedance.

So I'd say primary and secondary resistance up to saturation.

3. Sep 23, 2013

### hisham.i

The current is related to the connected load on secondary side, I2 = U2/RL (suppose a resistive load RL is connected across the secondary windings of the transformer).

4. Sep 23, 2013

### Crazymechanic

in simple words , some key factors limit the current in the secondary , wire diameter would be first , wire resistance and the coils impedance would be second the core size and material would be third ,
quite honestly drakkith I think you kinda know this already:D
well the frequency of course is of matter and the source power also.

from my personal experience I can say that even using a say old russian tv transformer which has a large core and the transformer rated at 300w just because there is large space for the windings I rewound the primaries with a thicker wire and the secondary also with a thicker wire respectively , say the primary was something like 230VAC 50hz wire diameter something like 1.05 now the secondary was for audio purposes with a 2mm diameter wire , now trust me as I say I didn't measure directly but the secondary output was way above 300w so technically the core is not that much of a limit if it is good metal with reasonable size.
You can " overclock" a transformer significantly just be increasing the wire diameter.

5. Sep 23, 2013

### Staff: Mentor

I just suck at asking questions. I was trying to figure out why pumping the voltage up to like 1,000 volts doesn't cause a huge current flow in the secondary. I assume it's because of impedance and back EMF generated from the secondary coils?

6. Sep 23, 2013

### sophiecentaur

hisham gave you the answer in his post. It depends upon the secondary load (in nearly all cases). It is very rare to short circuit the secondary of a transformer - just as it is rare to short circuit a battery.
If your transformer is designed right, it will 'easily' give you 1000V secondary volts with loads of current. (Could be a bit pricey though!
Note- if you short the secondary, the secondary volts will be zero.

7. Sep 23, 2013

### Staff: Mentor

Oh. Well that changes things.

8. Sep 23, 2013

### sophiecentaur

Just a bit. Lol

9. Sep 23, 2013

### hisham.i

If you have an idea on your load power and the voltage applied we can calculate the current that will be drawn from secondary.

10. Sep 23, 2013

### Staff: Mentor

I don't. I was just trying to understand how the step up/step down thing works in a transformer. If you step voltage up you reduce the current, keeping the power the same, right?

11. Sep 23, 2013

### AlephZero

This isn't quite so simple as it seems.

Suppose you connect no load to the secondary. Is there any current in the primary? Yes. To a first approximation, it is limited by the inductance of the primary (and the AC frequency), and of course it is 90 degrees out of phase with the primary voltage, so your home electricity meter doesn't respond to it. But if the inductance of the transformer is low, this current will be high, and it in still wastes $I^2R$ of power in heating the transformer winding. So the inductance needs to be high enough to keep that under control.

Therefore, when you connect a resistive load to the secondary, you actually have an LR circuit to analyze (actually two LR circults, coupled by the mutual inductance of the transformer).

But "simple" explanations of what a transformer does to the current and voltage usually ignore the above, and just talk about the component of the current that is in phase with the voltage.

12. Sep 24, 2013

### Staff: Mentor

Ah, okay. Thanks Aleph.

13. Sep 24, 2013

### sophiecentaur

It would be easier if you get your 'cause and effect' the right way round. You say you have a step up transformer, so the volts are higher. If you want the 'same' power, then you would need to have a higher load resistance - which would result in less current (so you need different loads, with and without the transformer, to keep VI the same). The transformer can also be looked upon as transforming the load resistance (load Impedance, actually) to a lower value, as seen by the source. In fact it makes the load, with the transformer, 'look' the same as a load, without the transformer, that dissipates the same power. (Hence the concept of a 'matching Transformer', in audio and RF work.)
This all refers to an ideal transformer, of course, but there is no essential difference as long as the transformer has enough copper and iron in it and is wound 'properly'. Having finite inductance in the primary, iron and copper losses plus incomplete linkage of the magnetic flux can be thought of as adding other components into any 'equivalent circuit' you could represent a real transformer with.

14. Sep 24, 2013

### Crazymechanic

Ok if we have gone so far I might add a question myself I'm sure drakkith approves this.
What happens if we have limited current from the source but this step up transformer has both primary and secondary superconducting ? The impedance is still there but the resistance is zero so ususally the current should be less stepping up voltage but in this case?
My mind tells me it should also be less because the total energy must be conserved.

15. Sep 24, 2013

### sophiecentaur

By "limited current from source", you are implying the existence of the source resistance of the supply to the primary. For an ideal transformer (forget the "superconducting" bit; it's taken care of in the word 'ideal'), this will transform the source resistance to a higher value (as seen by the load) and will result in a lower value of output current 'capability' before the output volts start to sag.
A full equivalent circuit for the whole set up needs to include source emf, source ('internal') resistance, transformer equivalent components (of which you can ignore most, with an ideal transformer), load resistance.
As you say, you will always have energy conservation and you will lose power in the source as you try to take more power into the load. The transformer doesn't change that basic fact any more than an ideal gear or lever changes the work / energy situation in a machine.

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