# Proof for magnetic formulae

1. Jun 11, 2004

### Cheman

Hi,
Can anyone please supply me with an algebraic proof for the transformer equation? ie - Vp/Vs = Np/Ns.
Also, can anyone prove the other transformer equation VpIp = VsIs WITHOUT using the theory of conservation of energy as a basis? Please do it the algebra using other formulae, etc.

2. Jun 12, 2004

### Staff: Mentor

This "transformer equation" follows from Faraday's law and the properties of inductance. It should be available in any textbook. Or the web: http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/transf.html
What do you have against conservation of energy?

3. Jun 19, 2004

### Cheman

But is there ANOTHER to prove that formula not through the conservation of energy? Cause if you think about it, the power in= power out does not really make sense if yuou think of it in terms of conserving energy, cause one coil isntlosing energy and one gain it - after all, what you are actually doing is not channelling the magnetic field from one to the other but the field of the primary is magnetising the core, which itself then induces the secondary coil. So why should the energy of 1 be rerlated to the other?
Therefore, I just wondered if anyone had another proof, or could prove that what I have just said is not true. :-)
Thanks.

4. Jun 19, 2004

### zoobyshoe

As a matter of fact, it is. If you draw current from the secondary, (add a load) the primary must draw more current from its source.
Looking at it interms of watts, you should be able to see they are intimately tied to each other. The secondary cannot draw more watts than the primary puts out. The ratio of voltage to amperage will be different in the secondary's output, but if you convert them to watts, they will be the same as the primary's watts, minus losses.

The power that is available from the secondary is directly dependent on the strength of the magnetic field induced by the primary.

5. Jun 20, 2004

### Cheman

So, if you were to put a voltmeter across the primary coil, would there be a potential difference because energy has been lost? And where has it been lost to?

6. Jun 20, 2004

### zoobyshoe

Yes, there will be a potential difference. There is resistance in the coil because it is, really, a very long piece of wire. The power lost here goes to heat.

I think that voltage is the wrong place to be looking for the answer to why the secondary is not a completely separate system from the primary. If you look at the more important contribution of amperage to the strength of electromagnets and also the contribution of the strength of a magnetic field when it is being used to generate electricity, you'll understand how the two (primary and secondary) are inextricably linked.

7. Jun 21, 2004

### Cheman

8. Jun 21, 2004

### zoobyshoe

Just examine a generator situation. Take for example the magneto on a gas powered lawn mower. Here a permanent magnet is passed by a coil, generating electricity in that coil.

Do some research and find out how each part of the set up contributes to the output of the coil. You will find that the strength of the field of the permanent magnet is important and can't be neglected.

Go back to your transformer. The strength of the field created by the primary is equally important to what you get out of the secondary, and can't be neglected.

Then do a little more, very easy, research and see how the strength of the field created by the primary is proportional to the amount of current flowing in the primary, and how that effect is further dependent on the number of turns of wire in the primary.

9. Jun 21, 2004

### zoobyshoe

I'm having another look at this to try to isolate the problem in your logic.

It seems to me that your reasoning is following these lines: you put current through a wire, you are using x amount of power. You wind the wire around an iron core drawing the same voltage and current, and you get a magnetic field still only drawing x amount of power. Therefore, it costs nothing more to produce a magnetic field than it does to simply put current through the wire in the first place.

Since you already have this magnetic field, you seem to be reasoning, you can wind a secondary around it, take power off, and there is no additional strain on the primary. Therefore, you conclude, conservation of energy is beside the point.

If this is your logic, then what you are forgetting is that you are expending energy and creating a magnetic field just by putting current through a wire in the first place.

So, power in, power out works perfectly if you bear in mind that electricity is being used whether it's just going through an uncoiled length of wire, or if it's going through a wire coiled around a core. That is the energy that is being conserved.

10. Jun 26, 2004

### Cheman

No, thats not what I mean. A certain current is folwing through that coil, at a certain voltage. This causes a magnetic field to develop around the coil. This magnetises the iron which in turn induces the current in the second. Where in that is any kind of energy conservation!?! Does it require energy to create a magnetic field? etc? Thanks. :)

11. Jun 26, 2004

### Cheman

Is there any other way to prove the equation without conservation of energy?

12. Jun 26, 2004

### zoobyshoe

Yes, indeed, it requires energy.
Magnetic fields go hand in hand with current flow: a magnetic field doesn't exist by itself. The magnetic field is a phenomenon that arises along with current flow. If the magnetic field is your goal, you must first make current flow, and, of course, it requires energy to create a current.

Whenever there is current flowing, there is also a magnetic field. In many cases that field is ignored. There is a magnetic field around the wires that lead the power into your computer, for example. In this situation, that magnetic field is ignored because the current is what is desired.

In the transformer the magnetic aspect of the current flow suddenly becomes very important. It can be used to change the voltage and amperage by induction for use in different auxiliary circuits.
-------
I don't know what you mean when you say "prove the equation". Those equations are descriptions of what happens.
They are, in a sence, instructions for how to transform voltage and current. If you need 15 volts to operate a given circuit, and you are starting with 110 volts from the wall socket they tell you the ratio of the number of turns to make in the primary and secondary.

We can do this as a ratio. Let us say your primary coil has 400 turns.

110 is to 400 as 15 is to ?

or 110:400 :: 15:?

The product of the means equals the product of the extremes:

We have the means: 400 and 15. Multiply. You get 6000. Divide this by 110. You get 54.54.

To get 15 volts out of you secondary you will put 54.54 turns of wire on it. (round it off to 54 or 55).

The only "proof" of this I can think of, is for you to actually make a transformer according to these instructions.

13. Jun 27, 2004

### zoobyshoe

And scroll down a bit to the illustrations that show the magnetic field around a current-carrying conductor.

The magnetic field arises simply by virtue of current flowing in a circuit. In fact, if you cause current to flow, you can't prevent a magnetic field from arising around the conductor.

So, in your transformer, the energy to create the magnetic field is coming from the current flowing in the primary.

If you weren't putting energy into your primary, there would be no changing magnetic field to induce current in the secondary. This is the energy that is being conserved such that thw power out equals the power in, minus losses.

14. Jun 27, 2004

### Cheman

By "prove" I mean prove algebraically where the equation is derived from. eg - you can algebraically prove the formula for the sin rule or cosine rule or phythagoras or achamedes principle.