1. Jan 28, 2009

I know that P = VI where P is power in watts (joule/second), V is voltage in volts (joule/coulomb), I is electric current in amps (coulomb/second).

My idea of work is like this. I use force (in newton) to pick up a stone and take it from point A to point B with a distance d in meters. I think that's work in joule (Work = Force x distance). If I would apply this to electricity, the stone pertains to an electron.

My idea about power is, say I did what I did above in 1 second. Thus Power is equal to joule/second.

Now my question is this. Say that the pressure (Voltage) is 120 V and the flow 10 Amps is less (electric current). That means I have 1200 Watts of power. What if I switch values as in 10 V, 120 Amps. This clearly states that the pressure (Voltage) now is less while the flow (electric current) is much. Now my question is this. In the situation where the pressure is high and current is low, shouldn't the power be less? And in the situation where the pressure is low and the current is high, shouldn't the power be much?

2. Jan 28, 2009

TVP45

The concept is clearer if you don't compare power in the electric instance to work in the mechanical one. If you think about the work done in moving a charge from one potential to another, it's W = qV. So, just as you get the same work if you lift one stone two meters or two stones one meter, you get the same work if you move one Coulomb two Volts or two Coulombs one Volt. Very similar to the mechanical case.

Power has to do with how fast you do the total work. You can use 10 W of power to do a billion Joules of work, just not very quickly. But, to compare work and power is something akin to comparing distance and speed - they can only be compared under very specific conditions.

3. Jan 28, 2009

gunslingor

The answer to your question is no, power would be equal. I think you are leaving out the consept of resistance in your reasoning.

4. Jan 29, 2009

gunslingor, thanks for the help. And you TVP45. I really appreciate your help.

After reading gunslingor's comment on this, the way I understand it is like this. In the first situation where there are more voltage, low current, and high resistance, the work needed to move 1 coulomb of electron is so much greater because of high resistance than moving 1 coulomb of electron in the second situation where there is low resistance. So basically the work done in the two situations in a second is equal although there are few electrons which are moved in the first situation than in the second situation because of their difference in the level of resistance.

5. Jan 29, 2009

gunslingor

Pretty much. let me modify as follows:

After reading gunslingor's comment on this, the way I understand it is like this. In the first situation where there is a stronger voltage, low current, and high resistance, the work needed to move 1 coulomb of electron is so much greater because of high resistance than moving 1 coulomb of electron in the second situation where there is low resistance; but since the voltage is higher in this case, the same work is performed. So basically the work done in the two situations in a second is equal although there are few electrons which are moved in the first situation than in the second situation because of their difference in the level of resistance and voltage levels.

One more example. Imagin I am building the great pyramids. starting on level one, I need a lot of blocks. On level two I need less blocks, but have to move them higher. And so on, until you only have to move one block at the top level, but it is very very high. Therefore, in conceptual theory, the total power to produce the pyrimad remains constant throughout the process even though the number of blocks changes (electrons, i.e. current) since the hieght (resistance) changes inversely to the number of blocks and the work potential (height per block or resistance times current) changes proportionally to the numeber of blocks (or inversely to to amount of current.