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I am having a bit of trouble understanding what an ideal wire is.

Let's assume there is a positive charge on the positive terminal of the battery and a negative on the other side which will give us the same results. Now electric potential is given as integral of E.dr. Therefore the potential will decrease with the distance .How is it possible that there is same potential across a wire. Electrons would not move if this was the case.

Also it is said that potential is dropped across a resistor and all the energy lost is made in heat energy,and that the battery provides this energy,now when an electron moves closer to a proton because of Coulomb attraction potential energy is also lost but no one has to provide it in this case? Also when it moves closer energy is lost and similar happens in a resistance although resistance provides an obstruction the energy lost in that should be different how can we equate these two and make them heat dissipation?Can someone answer these questions

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An ideal wire is really a limit case. Consider the limit of a resistor as the resistance goes to zero. The electric field approaches zero, but the voltage across it also approaches zero. As long as the current is limited by other components (such as other resistors), the result will approach that of an ideal wire.

If you have a short circuit, you don't have any other component limiting the current, and you find that the equations diverge. There is no valid solution for a short circuit involving ideal wires.

In the end, every wire has a little resistance. It can typically be ignored by using this limit case of an ideal wire. The only thing which doesn't have a resistance is a superconductor. They behave completely differently from conductors, and require a very different set of theories to model them.

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  • $\begingroup$ But Voltage also depends on distance and voltage will drop or increase with distance and that is not dependent upon the resistance $\endgroup$ – BlackSusanoo Jun 18 '20 at 7:11
  • $\begingroup$ It depends on how you fix things. In most circuits, the voltage is not "chosen" by the component. Its chosen by things like the battery and larger resistors nearby. But one way of thinking about it: you have a circuit with a 3V battery and a 1kohm resistor. In one case you have short wires with a resistence of 0.00001ohm. In another case you have long wires with a resistence of 0.0001ohm. In both cases, we find the voltage drop across the wire is negligable $\endgroup$ – Cort Ammon Jun 18 '20 at 7:14
  • $\begingroup$ Ohk if the wire is of 0.00000Ω and it's length is 1m then and the electric field produced by the battery is say 5N/C let's take the potential at a point A in the wire at distance 0.5m from the source of electric field therefore potential at that point will be 2.5V now at the end of the wire which is of length 1m potential will be 5V.Now when an electron will move from 2.5V to 5V the electron will lose potential energy $\endgroup$ – BlackSusanoo Jun 18 '20 at 7:45
  • $\begingroup$ You have not answered the second part of the question $\endgroup$ – BlackSusanoo Jun 18 '20 at 7:46
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If you short an ideal voltage source (V) with an ideal wire (R=0) then an infinite current I will flow such that V=IR . In practice a fire or explosion will result to remind you that idealisations are approximations to reality.

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