# Voltage vs Current

1. Apr 26, 2012

### Finris

I understand that power(watts) is equal to voltage(potential energy) X current(flow of electrons).

I understand that is is more efficient to increase the voltage using a tansformer to send less current but come out with the same power.

i.e 12 volts X 10 amps = 120 watts

24 volts x 5 amps = 120 watts

but how does less amps/current give the same power output? Would half the current travel twice as fast? I understand current to be electrons moving along a copper wire in a magnetic field so if I doubled the voltage I would end up with half as many electrons but moving twice as fast?

Perhaps similar to bandwidth in a computer where a 1 Ghz processor can send data down a 16 bit bus at the same speed a 500Mhz processor can send data on a 32 bit bus.

2. Apr 26, 2012

### DragonPetter

Volts itself is not potential energy, but rather potential energy per a coulomb of charge. Current is the flow of charge, and each individual electron isn't necessarily moving much, but rather the net direction of charge adds up to a net current.

The reason it is more efficient to use high voltage and low current is because with low current, less voltage is dropped across the wires, and so less power is dissipated by them while still delivering the same intended amount of power to the load.

It has nothing to do with current traveling at any speed over a distance. That is actually not a true picture you have and you will understand electricity better if you don't think of current as having a speed dependening on how much current or voltage is present. Current travel speed does not change simply because more voltage is driving it. Think of it more as how much energy each coulomb of the current has; if current is halved but voltage is doubled, then the current will have half as many coulombs/sec, but each of those coulombs will have twice as much energy.

The bandwidth and data rate is pretty much completely unrelated to the current/voltage power relationship, although there might be some algebraic similarities (bw x bus width = #bits/sec, current x volts = #joules/sec). I would try not to draw any relationship between the two if I were you.

Last edited: Apr 26, 2012
3. Apr 26, 2012

### Fuxue Jin

This is interesting but confusing.

For the same distance of power delivering, we are talking about the high voltage, how can it be "less voltage" across the wire (distance)?

4. Apr 26, 2012

### phinds

You need to study ohms law and realize that transmission lines have resistance.

5. Apr 26, 2012

### Fuxue Jin

No offensive, but Yes, I am major in Physics, and currently work as an electrical engineer.

6. Apr 26, 2012

### Antiphon

It's more voltage across the wires.

But it's less voltage dropped along the wires. It's known as I-squared-R loss of power in the wires before you get to the load.

7. Apr 27, 2012

### Fuxue Jin

You may be confused the power lost on transmission line vs the power delivered after the current passing through the transmission line.

If the transmission line is super conductor, zero resistance, then it doesn't matter high or low voltage because there will be no loss during the delivering process. The power will be the same at the starting point and end point.

If the transmission line is normal conductor with resistance, then there will be I^2 R power loss on the line, and you don't get 100% of the power delivered.

In order to minimize the loss during the power transfer, it is better to have low current. Low current requires high voltage in order to have high power.

Thinking of carrying a bucket of water from point A to point B, but there is a hole on the bucket. So you will get more water at point B if the hole is small (current is low).

Last edited: Apr 27, 2012