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Assume we have a source @100 DC volts in series with a 200 ohm resistor. According to Ohm's Law, current flow in 0.5 amps. Voltage drop across this resistor is 100 volts and equal to the source voltage. If voltage drop is 100 volts(meaning none is left on the "outflow" or neg. side of the resistor), then how can current continue to flow to the neg. term ? There is no electrical pressure(voltage) left to push the electrons.

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  • $\begingroup$ I think in your description occurs a common misconception. The voltage-difference is between the two terminals of the battery. Since the terminals are connected via the wires this voltage difference is applied across the whole circuit. This voltage difference does not simply vanish after the resistance. Therefore, I also think that it is very misleading of teachers and textbooks to talk about a voltage drop. $\endgroup$ – F. Ha Mar 3 '16 at 9:23
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This problem is an idealization; real batteries have an internal resistance, and so do the wires.

In this idealized circuit the wires have no resistance, nor do the connections. You can logically collapse them to points -- and this brings the battery terminal into contact with the resistor; the current has obeyed Kirchoff's rules, and has arrived at the battery.

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You say 'no pressure (voltage) to push the electrons', but doesn't that mean there's no acceleration of the electrons? They're already in motion, and they stay in motion. Were the current to drop, there would no longer be an equal I*R to the applied V. That would break Kirchoff's rule for voltage.

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  • $\begingroup$ Are you talking about a perpetual motion machine? Just saying... $\endgroup$ – george rocks Mar 5 '16 at 22:43

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