How does increasing voltage lower current in a transformer?

[fawk3s]

I know this question is prolly stupid and with a simple answer but I just can't figure it out.

Well, the electromagnet, which is basically what changes the voltage and current. It is said that you can lower the current by raising the voltage and vice versa. I know WHY they do it, but what I don't understand is how can highering the voltage lower current?
I=U/R
R stays the same
I changes according to U
U is raised

"I" should raise too.

I would completely understand when lowering the voltage would lower the current and vice versa.

This prolly a very stupid question so I understand all the "lolling" but I need to know hte answer.

Thanks in advance,
fawk3s

 

[diazona]

I can't say for sure without knowing what physical system you're talking about, but it probably has something to do with energy conservation. The power (energy transferred per unit time) in a circuit is $P = IV$, and if energy/power is to be conserved, P must be a constant. So if you raise I, V gets lower, and vice-versa.

 

[fawk3s]

Ah, sorry about my equations. Its not the same as in English or where ever. I live in Estonia so, I=current, U=V=Voltage, R=obstacle (or I don't know how you call it).

P must be a constant. So if you raise I, V gets lower, and vice-versa.

But that's what I want to know, how.
In this case, the Ohm's law doesn't comply.

 

[uart]

Hi fawk, think about this problem.

When I try to spin something (like a bicycle wheel or a fan for example) I've noticed that the faster I want to spin it then the more torque (turning force) I need to apply. So I've learned that more speed equates to more torque and visa versa.

Now recently someone told me about this thing called a "gear box". I have no idea about how it works but I've been told that with the magic of this "gear box" I can get more torque in the output shaft by lowering its speed (via using a lower gear ratio). So what's going on here? I'd always thought that more torque means more speed but now with this thing called gear box the opposite is happening.

Can you explain this to me? If you can then you've answered your own question as in a way the transformer acts very much in an analogous way to a gear box. The equivalent of the gear ratio in a transformer is the turns ratio.

 

[fawk3s]

In a gearbox you win with force, but loose with distance and vice versa. (If I am right).
I know the same thing is with voltage and current, but I just can figure out how is it done. I give up.
So could you please explain?

 

[Bob S]

When you make an electromagnet, you need a certain number of amp turns, NI. This coil has a resistance R that depends on the diameter of the wire and its length. For example you could choose 100 turns of wire that can carry one amp, or 10 turns of wire that can carry 10 amps. The 10 amp wire is about 3.3 times the diameter of the 1 amp wire, and its resistance per unit length is 10 times less. Because the 10 amp wire is 1/10th the length of the 1 amp wire, its end-to end resistance is 1/100 the end-to-end resistance of the 1 amp wire. So the voltage drop in the 10 amp wire is about 1/10 of the 1 amp wire. So I times V (current times voltage) is the same for both coils.

 

[fawk3s]

Aah. Now I get it. The thing I hadnt noticed was the thickness of the wires. So I basically did the same thing, only leting R be 10 times less in one winding that the other. That caused the voltage to be equal in both, and that confused me since it is clearly said that when the current is raised, the voltage is LOWERED.
Damn, how didnt I see that? Dran I am stupid.

Big thanks you guys! This thing was giving me headaches for times.