How is it that transformers can step up the voltage of something and make it work something else that runs on a higher voltage?
Ok i admit, i suck at physics thats why i need help

heres what i know : voltage is the pressure pushing the charges
: current is the flow of charges

And V1/V2=I2/I1

this means that in a transformer as the voltage goes up, the current goes down…is it?

But if large transformers are used to convert the output of the generators in a power station to a high voltage and a low current to reduce power loss in cables, how does the low current power something else that needs a large current?
I know step-up but if the current in the power cable is already low how can steping-up get high current out of low current?

It would really help if someone can explain to me interms of what happens to the electrons themselves as they go through this all.

An analogy would be a lever. A lever can increase the force you can exert, at the expense of speed/distance. With your bare arms you may be able to lift a 50 pound weight 2 feet; with a lever (or pulleys) the same amount of work will lift a 200 pound weight 1/2 feet.

It’s more efficient to transfer power at higher voltage and low current. Another mechanical analogy; take a big truck, with engine in front and drive wheels in the rear. So there’s a drive shaft that transfers the power. To transfer a certain amount of power, you could either apply a torque of 100 foot-pounds and spin at 1000 rpm, or 1000 foot-pounds at 100 rpm. The amount of power transferred is the same. But the faster the driveshaft spins, the more energy is lost in the driveshat bearings, so it’d be better to choose the 100rpm setup. Similarly, losses in electrical wires are higher at high current, so you want to choose a combination of high voltage and low current.

As to what happens in a transformer - you’ve probably learned that changing magnetic fields induce current? The changing magnetic field can be made by physically moving a magnet next to a wire - that’s how generators work. Or it could be made by passing AC current in a coil, as in a transformer. One set of coils convert electrical power into changing magnetic fields; this field induces current in the other set of coils. Depending on how you set up the two coils, this field can push many electrons at a moderate force (low voltage, high current) or a few electrons with a big force (high voltage, low current).

The basic principle to remember in electrical transformers is that when two conductors are near each other, a changing current in one will induce a proportional changing voltage in the other. The effect is the result of the magnetic field resulting from the current in conductor 1 causing a voltage to be generated in conductor 2 by magnetic coupling. This is a simple, experimental and demonstrable fact of life and I don’t know of any more fundamental explanation for it. Maybe someone else does and can come forward with it.

The fundamental idea is that energy is conserved. The energy into the primary, or input, side is related to the product of the voltage times the current. For the moment let’s pretend that the power in is exactly equal to the input voltage times the input current. Since energy is conserved the output power must equal in the input power so V[sub]1[/sub]*I[sub]1[/sub] must equal V[sub]2[/sub]*I[sub]2[/sub]. When rearranged this results in the equation you cited.

Actual power computation in alternating current circuits is more involved than this. However that is a mere detail and doesn’t affect the main point of the above explanation in any way that matters for our purpose here.

The voltage is stepped up (like to 440000 volts) for transmission because this results in a lower line current than would otherwise be the case, so the line resistance causes smaller line loss. The power lost in the line resistance is I[sup]2[/sup]*R and since R, line resistance, is constant if you lower the current you cut the loss way down. Then at the receiving end the voltage is stepped down to some standard voltage, 440, 220, 110, etc. for use.

Electrical appliances don’t “require” a particular current in the sense you are using. What is needed is a certain amount of power and the transformer doesn’t change the amount of power that is available. In the ideal case, that is. Of course there is a small loss in the transformer because nothing is perfect, but that is tolerable in light of the flexibility given by being able to change voltage and current at will.

A transformer is a passive device. Power goes in the primary side and out the secondary side. In an ideal transformer, every bit of the electrical power that goes into the transformer must come out of the transformer.

Electrical power = voltage * current = V * I

V[sub]1[/sub] = voltage at transformer input (primary winding)
I[sub]1[/sub] = current at transformer input (primary winding)
V[sub]2[/sub] = voltage at transformer output (secondary winding)
I[sub]2[/sub] = current at transformer output (secondary winding)

The electrical power going into the ideal transformer = V[sub]1[/sub] * I[sub]1[/sub]
The electrical power leaving the ideal transformer = V[sub]2[/sub] * I[sub]2[/sub]

Therefore, V[sub]1[/sub] * I[sub]1[/sub] = V[sub]2[/sub] * I[sub]2[/sub]. (This can be rearranged to get your equation.) What this equation basically “says” is that:

If you have an ideal 1:1 transformer, then V[sub]2[/sub] = V[sub]1[/sub] and I[sub]2[/sub] = I[sub]1[/sub]
If you have an ideal step-up transformer, then V[sub]2[/sub] > V[sub]1[/sub] and I[sub]2[/sub] < I[sub]1[/sub]
If you have an ideal step-down transformer, then V[sub]2[/sub] < V[sub]1[/sub] and I[sub]2[/sub] > I[sub]1[/sub]

The voltage from the generator is stepped up. In other words,

V[sub]2[/sub] > V[sub]1[/sub] and I[sub]2[/sub] < I[sub]1[/sub]

The means the voltage on the power lines is high while the current is low.

But this voltage is too high to power most things. So it must be stepped down using a step-down transformer. And since we’re talking about ideal transformers, this also means the current on the secondary of this transformer must be higher than the current on the primary:

V[sub]2[/sub] < V[sub]1[/sub] and I[sub]2[/sub] > I[sub]1[/sub]

Not the one currently in the toy stores, at least – that’s actually “Super Fire Convoy,” from Japan’s Transformers: Car Robots cartoon last year. Wait for the upcoming Transformers: Armada toy line this fall for the next Optimus incarnation, complete with radio-controlled auto-transforming trailer base.

(Yes, I’m serious, and yes, I know this is out-geeking the joke. That’s what you get when you get the SDMB Resident Transformers Geek on the job. )

This being alternating current, AC, the electrons don’t travel much (the actual distance of travel is dependant on the speed of electrons through the transmission media and the frequency of the AC.)

Since transformation is dependant on changing fields you need the alternating current to make it all work. Back in my boy scout days an electricity merit badge instructor told us that you could take a dry cell and transform the voltage up or down to run whatever battery powered device you wanted. It took me a few years to realize the guy was full of it.

While doing what he suggested is technically true, you would need to change the battery’s direct current to alternating current to transform it to a new voltage and then change it back to direct current to power the device. I doubt that anyone has ever done this on a practical level.

Thx for all the replys and yes, nothing to do with optimusmonkey

Huh? :S, I=Q/t, so you mean i can run a big machine with just a few charges? Ok i’m crossed:S.

Got this from another theard : that electrons don’t really move through a wire and light up a bulb and it’s really the electric signals that travel.
If thats the case then i’m going back to basics…what is a charge? how is charge related to electrons? and what are electric signals?

The point that Crafter Man is making here is that Power is a constant. The variables here are the current and voltage. In the ideal transformer he mentions, power in equals power out. So the same amount of power fed by 6.9 amps @ 7200 volts (i.e. 49680 watts) would require 207 amps @ 240 volts.

The other thing that bears mention when speaking of the transmission of power is conductor size. Transmission lines operate at high voltages to keep conductor sizes to a minimum. 500MW (500,000,000 watts) transmitted at 500KV (500,000 volts) = 1000 amps. A single conductor of 954 ACSR (Aluminum Conductor, Steel Reinforced), measuring about 1 inch in diameter, is rated at 1010 amps and would be sufficient to carry this load. The same power (500MW) transmitted at a standard distribution voltage of 12,470 volts = 40,096 amps and would require 40 conductors of 954 ACSR.

Sounds like you’re thinking of wires as providing electricity to those machines. That’s not right. The wires are providing energy (or power, which is energy per unit time). Machines don’t feed on fresh new electrons flowing in from the wires. All it requires is the power to do its job, which can be provided by either a huge number of electrons flowing at moderate speed/pressure, or very few at a high speed/pressure. (This is an analogy, electrons don’t actually have pressure.)

The original poster wrote the following: “But if large transformers are used to convert the output of the generators in a power station to a high voltage and a low current to reduce power loss in cables, how does the low current power something else that needs a large current?” This statement is so basically confused that it can’t be “fixed up” with just a little tinkering. It needs to be thrown out and a new beginning made.

I answered: “Electrical appliances don’t “require” a particular current in the sense you are using.”

My answer was probably poorly phrased. What I should have done was referred the poster back to my statement that energy is conserved in a transormer. So when the voltage is stepped down from the line voltage to the particular voltage used by the appliance in question, a larger current (than the line current) is automatically available at that lower voltage.

To step down the voltage again, just add a backwards transformer!

Transformers are just like paired gears, a large gear driving a small one. Or vice versa.

Utility companies use transformers because power lines are like long drive belts, but drive belts with horrible friction. Imagine a leather drive belt that drags upon the ground. Now imagine one that’s hundreds of miles long. If you used it to drive distant machines, you’d waste all your work in dragging the leather and heating up the ground.

The solution: put some step-down gears at one end, and some step-up gears at the other. Now, when you spin the gear at on end, the leather belt moves very slowly, but with extremely high tension. At the far end, the tense and slow-moving belt makes the step-up gears spin just as fast as the “input” gear at the near end of the loop.

In an electric transmission system, the gears are replaced by transformers, and the belt is replaced by the movable electron-sea within long metal rods. The motion of the leather belt becomes electric current, and the tension of the belt becomes the voltage. In the 1800s, high voltage was even called “High Tension”, presumably because they understood this leather-belt analogy.

Incidentally, the idea to use AC generators, high voltage transformers, and extremely thin power lines was Nikola Tesla’s. He patented it and sold to Westinghouse, who defeated Thomas Edison’s DC proposal for bringing energy to Buffalo NW from the generators at Niagra Falls 30 miles away. I remember seeing the statue of Tesla on Goat Island at Niagra. I always wondered who the heck he was, since no textbook ever mentioned him. (Wasn’t electricity invented by giant corporations?)

After skimming the replies, I don’t see the magic word anywhere: Impedance. It sounds like one of the main questions in the OP is where does the additional [secondary] current come from during a voltage step-down operation. It’s all very well to invoke the law of conservation of energy and then simply proceed to deduce that there must be a resulting increase in current. But where does it come from? Remember that the secondary winding of (for example) a 2:1 step down transformer will have half the number of coil windings as the primary, and consequently a much lower impedance. The secondary winding of the transformer appears as a power source to a load, and a power source with half the impedance can deliver twice the current.

Remember that, although loads can be described as “current driven”, it is really the product of the current & voltage that powers them. In order to achieve maximum power transfer from a source to a load, the impedance of the source should match that of the load. A low impedance load (like a motor) that is connected to a high impedance source (like a photovoltaic cell) is a serious impedance mismatch, even if the voltage is correct. The source (cell) doesn’t have a low enough impedance to deliver the current that the load (motor) needs to run. A transformer can help this by matching the impedance between source & load to provide the load with a voltage source that has a low enough impedance that will allow sufficient current to flow.

(Note- I realize that a PV cell wouldn’t work at all since we are talking about transformers, I just used it for illustration 'cuz I couldn’t think of another well-known high impedance power source.)