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I have referred to many books and all of them apply loss in potential across a resistor only in case of circuits. However, the electric field is present in the whole conductor. Hence the electrons should lose potential all along the wire and not only across the resistor?

P.S. There could be some assumption in this topic and I'm apparently unaware of it.

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  • $\begingroup$ Nobody makes that assumption outside of simplified student exercises. Neither engineers nor working experimental physicists treat wires as ideal, lossless conductors. Conduction in materials (electrical and heat) is one of the most important topics in solid state physics, with a wide range of theoretical work to understand it. $\endgroup$ – CuriousOne Jan 12 '16 at 15:14
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Ok, that's a bit tricky: These books propably use the assumption that the wire is a perfect conductor. What follows is that any difference in the electric potential $\Phi$ allong the potential will create a field (as you said). BUT, and here is the clue: Since it is a perfect conductor, and electrons have little mass compared to the force that the electric field generates, these books (and pretty mutch everyone) makes the assumption that the electrons are being distributed along the conductor in a way that there is no electric Field anymore.

Any electric Field would create movement of electrons, until they are distributed in a way that the electric field vanishes. This process may take a finite time, and - here is the assumption - one assumes that this time can be neglected, because it is small compared to every other thing one wants to calculate.

Result: There is no electric field in the conductor, but only in the resistance, and hence electrons don't lose potential when they are travelling in the conductor. Note that the part of the conductor before the resistance now has another potential than the part of the conductor after the resistance, and if you would connect this parts, then the different charge densities in this wires would create a field along the connection.

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  • $\begingroup$ @RaghavKhaterpal: It might be more accurate to say it is an approximation rather than an assumption. It is an approximation that is sufficiently accurate for many useful purposes. $\endgroup$ – RedGrittyBrick Jan 12 '16 at 14:45

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