Electrostatics: What power equations to use?

[hihowareu]

I was wondering what the different power equations are used for...

P=IV, P=I²R, and P=V²/R

I was solving for power loss and I noticed that using P=I²R gave me the right answer but the other two gave me different answers

 

[climber78]

They should all be equivalent. Take Ohm's Law (V=IR). If you plug in V from Ohm's Law into your first equation you get P=I^2R. If you take Ohm's Law in the form I=V/R and plug it into your first equation you get P=V^2/R. I would guess the difference was a result of the problem setup, i.e. what you define V, I, and R to be. If you show the problem I might be able to help more.

 

[elect_eng]

I was wondering what the different power equations are used for...

P=IV, P=I²R, and P=V²/R

I was solving for power loss and I noticed that using P=I²R gave me the right answer but the other two gave me different answers

I agree they should give the the same answer for resistors. Maybe you can give us details so we can understand exactly what you are doing. There is of course the question of which V you are using with the other two equations. If you don't use the right voltage, things will not check out.

The more general of these equations is P=IV because you can apply this to nonlinear devices more easily. But, for a linear resistor there is no ambiguity and all three equations are valid.

 

[hihowareu]

Thanks for the replies

for example: An electric power company decides to ship 30 000 W of power at 20 000V over 2.5 \Omega transmission lines. How much power is lost over the lines.

I got the right answer by using is P=IV to get I. Then using P=I²R to get the power lost.

But if I use P=V²/R I get a completely different number.

 

[climber78]

In your first definition V is the voltage supplied from the generator, which is the voltage across the wire AND the voltage across the load. Solving for I gives you the correct current to use in P=I^2R.

In your second definition V is still the voltage supplied to the load and wire. That would be the power dissipated from a (load+wire) of resistance R. To make this correct you would need to find the voltage across ONLY the wire, which is not the same as the voltage supplied by the generator.

Hope this helps.

 

[elect_eng]

The above answer sounds like the most likely mistake. It is a common mistake. The formula P=I^2*R is the most conenient for power line losses.