# Transformer Help

1. Apr 27, 2015

### DWarrington

Hello everyone,
I need some help with transformers as I am very confused....
I understand the role of a step up transformer - the secondary coil has more turns thus has a higher voltage induced. This in turn reduces the current and the overall heating effect is reduced. This is why transmission lines have such high voltages i.e so that they do not melt.
So, when I connect a lamp to a step-up secondary coil the lamp gets brighter. I understand it so that if voltage increases then current DECREASES as with transmission lines. So why does it get brighter - what am I missing?

Thanks for any help!

2. Apr 27, 2015

### CWatters

You are assuming the load draws constant power which is not correct.

It gets brighter because the voltage is higher.

The current in the secondary (Is) is determined by the load (the light bulb) not the transformer. It works like this...

The step up transformer creates a voltage Vs. The current in the secondary is determined by the resistance of the filament and is given by Is = Vs/R.

The transformer equation (Vs/Vp = Ip/Is = turns ratio) allows you to calculate the primary current Ip

3. Apr 27, 2015

### DWarrington

Hi - thanks for this.

So I shouldn't think that a transmission line behaves like a filament lamp then? Is that because it is non-ohmic or is it because a transmission line has negligible resistance compared with a filament? Therefore because the transmission line has zero resistance then current goes down when voltage goes up. And because a filament's resistance increases when voltage goes up then current goes up too?
I this right?
Thanks again

4. Apr 28, 2015

### CWatters

No it's nothing to do with the difference between a transmission line and a filament...

When they build a transmission line they put a step up transformer at the power station end and a step down transformer at the town end. So the town always gets 110V (USA), 230V (UK/Europe).

Your experiment only mentioned adding a step up transformer. In that case the filament gets a higher voltage than it did originally. It's the higher voltage which causes the filament to draw more current and burn brighter. Had you used a step up and a step down the brightness would be unchanged.

5. Apr 28, 2015

### lonely_nucleus

I like old military documentaries

6. Apr 28, 2015

### DWarrington

Thanks for taking the time to help me.

I am still confused. I guess I don't get this as much as I thought!
I thought that by having a step up transformer this meant that voltage would be increased and that current would be decreased. I thought that was the point of the step up transformer. So if current is decreased how can a filament become brighter?
I get that more voltage means a greater 'push' of electrons but surely resistance would increase and current drop using Ohm's law?

7. Apr 28, 2015

### rumborak

No, the lowered current only happens when the power consumption remains the same as with the original voltage. Mind you, to keep the power consumption steady with a different voltage you need to adjust the resistance (or impedance, in general). They do this with transmission lines, but your lightbulb filament doesn't.

I think your confusion stems from the fact that indeed in transmission lines the point of the step-up transformer is to lower the current. But, again, the current is lower because they draw the same power as if they did with the original voltage.
The reason for lowering the current in transmission lines is because the losses along the line are proportional to the square of the current (I^2*R).

8. Apr 28, 2015

### CWatters

It is (in a power transmission system) but you are misunderstanding what determines the current. It's not the transformer but the load and the voltage it sees that determines the current.

In a transmission line system the voltage is stepped up at the power station and down again at the load. This means the load sees the correct voltage and draws it's rated current. The current on the transmission line between the two transformers is reduced.

In your circuit you omitted the step down transformer so the load saw the higher voltage and drew more than it's rated current.

A

9. Apr 28, 2015

### lonely_nucleus

I think you answered the OP's question good response.

10. May 1, 2015

### DWarrington

Thank you once again. This is starting to make sense mathematically.

I did an experiment so I was able to see this so it would make even more sense (I am a kinaesthetic learner!).
I got a 1200 coil primary coil and attached a 120 secondary coil to it and measured the current and voltage using a multimeter. I then changed the secondary coil from 120 turns with one of 600, then one of 900, then one of 1200 and then one of 2400. Each time the current went lower and lower. How is this explained - surely the current should go up when the number of turns on the secondary coil increases??

11. May 1, 2015

### sophiecentaur

That's good!!
The essence of understanding what goes on here is to get a clear view of what actually effects what (cause and effect). The glib statement that "increasing the volts reduces the current" can be very confusing (it was for you) if you take it at face value but you have to be careful about the actual message in the statement. Your experiment would show you very well what you get when you tweak different things.
The load resistor is the pivotal factor, governing what Power goes through the transformer (zero power if it's an open circuit, irrespective of the supply volts). Thereafter, it's the same power (VI) at each stage. You have to look in both directions at once to get what happens. Maths does this vey well, of course - as you have seen.

PS I don't understand what your experiment actually shows. What load resistor were you using?

12. May 1, 2015

### DWarrington

Thanks!
I connected the secondary coil was connected to a filament lamp so that provided the resistance.

13. May 1, 2015

### sophiecentaur

Problem with a filament lamp is that its resistance is very much dependent upon its temperature so that can't be considered as a constant load R. As the supply volts to the lamp increase, the temperature rises and so will the resistance (10:1 ratio at least!!!!)
Use a heater element (low power soldering iron, perhaps?) and you experiment may behave much better because the heater element doesn't get white hot!

14. May 1, 2015

### DWarrington

Ok - I will have a go at that and get back to you!

he original issue I had was that I thought that a step up transformer increased the voltage and reduced the current. As I increased the number of coils on the secondary coil I would have expected the current to drop. But it didn't - the filament lamp got brighter suggesting current increased which was not what I expected.
I now think that as the resistance drops, due to higher voltage, the current increases. A second transformer is needed to see voltage increase and current fall. Is this sounding right?

15. May 1, 2015

### Staff: Mentor

The original issue in the OP is a common one in physics problems (also seen a lot in fluid flow/venturi questions). The issue is not recognizing that the equation applies to a uniqe/static situation, not to what happens when you change a circuit. When a transformer exists in a circuit, the power is equal on both sides of the transformer. But when you add a transformer to a circuit, the power is not equal in the circuit before and after the transformer is added.

16. May 1, 2015

### sophiecentaur

Hang on. As the supply volts increase, the filament gets hotter and its resistance Increases. A hot filament has a (10X) higher resistance than a cold filament so you really can't say anything straightforward about what's going on in your experiment.
As I said before, you have to be careful to get your causes and effects in the right order if you want to get it right in 'yer head'. Use the simplest model to learn on - which means you need a load with an unchanging resistance.

17. May 1, 2015

### CWatters

Please stop using a light bulb and use an ordinary resistor instead! Otherwise you will get really confused.

Which current? The current in the primary or the secondary?

18. May 1, 2015

### sophiecentaur

I was trying to find a link that would explain fully what the OP needs to know, without getting too deep.
I have just looked at a load of google hits on simple transformer theory. There is a problem. The secondary school level pages all concentrate on the turns ratio equalling the volts ratio and the fact that VI is the same for primary and secondary. They do not seem to deal with the 'real situation' of a loaded transformer, like the OP is using. Secondary teachers also seem to trip out at that level and will often struggle to explain the practicalities. (Using high voltage transmission lines is just something they repeat parrot fashion.)
There is a big jump in the complexity of the higher level of google hits, which includes the other elements in a real transformer system (losses in the transformer, due to its own resistance etc.)
If someone could find a suitable link, at the right level, it could help the OP a lot.

19. May 1, 2015

### sophiecentaur

I liked the film but have the same problem with the fact that the current in the secondary can only be known when the load resistance is known. and that is not made at all clear. No wonder people have problems when there is a history of education being delivered like that.

20. May 1, 2015

### rumborak

To add insult to injury, at some point you can't neglect the resistance of the secondary coil anymore either once it has enough turns.

That's why, while it's understandable to want to do things kinaesthetically, doing the circuit analysis at some point becomes inevitable.