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What exact wire diameter allows approximately $6.25\cdot 10^{18}$ electrons to pass in 1 second? Will a thinner diameter or a wider diameter wire allow the same approximate number of electrons in a second? If the same approximate number of electrons enter and exit the resistor, what do they lose to cause that voltage drop? And if one resistor drops a voltage to zero, how come one other resistor placed after the first will still find voltage?

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closed as off-topic by John Rennie, user36790, mpv, user10851, CuriousOne Mar 4 '16 at 22:33

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    $\begingroup$ Have you attempted to calculate how much charge 6250000000000000000 electrons add up to? Hint: one electron has a charge of about $1.602 \times 10^{-19}$ coulombs, and one coulomb per second is a current of one amp. $\endgroup$ – John Rennie Mar 4 '16 at 6:59
  • $\begingroup$ Like John Rennie said, 6.25e18 is a current--about 0.39 A. There is no single "exact wire diameter" that can carry that current. For a given set of environmental conditions, there will be some smallest diameter of copper wire that can carry it without melting. and basically, any wire larger that that would also be capable of carrying the same current. $\endgroup$ – Solomon Slow Mar 4 '16 at 21:50
  • $\begingroup$ Re, if one resistor drops a voltage to zero... That's not what resistors do. To learn more, read about voltage dividers and Kirchoff's Voltage Law. $\endgroup$ – Solomon Slow Mar 4 '16 at 21:58
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Just think of it as a hose pipe carrying water. The higher the pressure (voltage) the more water passes through it (charge). However, the pressure drops along the pipe. The current is the amount of water per second flowing. The pipe offers a resistance to flow. The thinner the pipe, the higher pressure is needed to get the same flow.

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