electrons velocity and Potential energy

I'm confused about when we say the power is equal to the product of UxI but I is for a cross sectional area and U is the electric potential between 2 terminals and by coulomb of charge what i mean is those charge dq = Ixdt how they will pass both terminal?

 Does it mean that dq as quantity that enters the resistor the same amount leave? And not as an " object"?
 Recognitions: Gold Member Science Advisor Yes. The path taken by the charge doesn't matter in this analysis. In fact an individual electron goes every which way as it makes an average movement from end to end and sometimes at great speed. All you need to consider is dQ in and dQ out of each section of a circuit. This is not a cop-out but an appropriate level of treatment. Mechanical or pictorial models are pointless and even misleading.
 so it's only like an abstraction it's like we are saying that the same object travelled through the resistor but actually it's the same amount.. thank you again
 Recognitions: Gold Member Science Advisor "object"? I get the impression there's something about this that you're hanging onto which is giving you difficulty. "Abstraction" is a good way to go.
 what is making me a difficulty is when in electrostatics we say that the work done by the electtric force(qE) on a charge q to make it pass between point A and B is equal to loss or gain of electric potential energy than it's like a mechanical object..
 Recognitions: Gold Member Science Advisor Do you not see the similarity - the direct correspondence, in fact? I thought we'd already agreed that, after a long time, nearly every charge can be thought of as getting all the way through the resistor. The ones that do not have a complete journey (after switch on or just before switch off) will only have a portion of the PD to 'drop through'. But, the same number of charges make incomplete journeys at the start and the end of the experiment and the total Potential energy of those charges converted will be the same as for one 'resistor-full' of charges. Why not do a series of sketches showing charges at various points along the resistor and at different potentials? Imagine I have a 1kg mass and I let it fall distance s, working some machine. I take a different 1kg mass and let it fall from the end point of the first mass's journey a further distance s. The two masses will have converted 2sgJ of energy - just the same as one mass, falling through 2s. If this were water in a hydroelectric dam and the valve was opened, the water at the bottom wouldn't supply much energy before exiting the system and neither would the water that only gets a few metres down from the top before the valve is closed. But, altogether, we are always dealing with a pipe full of water when we want to work out the energy obtained. Likewise, we are always dealing with a Resistor Full of charge when calculating the Power dissipated when the circuit is connected. The speed that the charges move is totally irrelevant in that respect. I don't think I can help you any more with this. I've put it in as many ways as I can think of. It's up to you to convince yourself now. The theory is well established as correct and the models are valid.
 what im saying is the power given per seconde = U*I is it equal also in the same second to the power given by the charges inside the resistor? because in a second the dqin didnt pass through out the whole resistor but we are making an abstraction on it because dqin=dqout
 Recognitions: Gold Member Science Advisor Is there a problem with that? Why does the same bit of charge have to go the whole way?
 no it's not a problem:P just because i was confused about the work done on the charge and the path taken because of its low velocity but it's ok now i think FINALLY understand it and really thanks a lot Sophie i believe i annoyed you with my questions:P
 Recognitions: Gold Member Science Advisor I've had much worse and we did get somewhere, which was why I kept going.
 hi again:P just i want to make sure of a thing:P in a wire the electrons do not lose of their potential energy and we dont know what is their potential right?
 Recognitions: Gold Member Science Advisor No / low resistance means no / low drop in potential - correct. We don't need to know their potential - just how much they lose. But please call them 'charges'. No one in electronics uses 'electrons' except when discussing the internals of a device, and that's only when absolutely necessary.
 Okay thank you again