If we ignore circuits for a moment and just think about a fixed charge and a test charge in space, we know that if we release the test charge, its PE changes and KE changes such that PE + KE is constant. However, in a circuit, we never talk about voltage dropping because a charge is converting PE to KE; we only talk about voltage dropping because a charge goes through a load. Why is this? Does voltage drop when a charge moves at all?
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asked Mar 16, 2018 at 7:52
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Because we are using the approximation of an ideal conductor. Within an ideal conductor, the electric field is zero; particles rearrange in a perfect manner to cancel out the electric field.
When there is no field, there is no change in the kinetic energy of particles - thus, there is no change in the potential energy either.
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$\begingroup$ “Within an ideal conductor, the electric field is zero”. Then what causes the current? If there is no electric field shouldn’t the charges just stay put? $\endgroup$ Commented Mar 16, 2018 at 14:14
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$\begingroup$ The electric field accelerates the charges. It is not needed for a constant kinetic energy; in the classical model, the current sees the whole circuit before deciding on how much it moves. There's a great question on that on the electric engineering stack exchange, something along the line of "how does the current know how much to move". $\endgroup$ Commented Mar 17, 2018 at 11:17