If an electrical circuit only contains 1 resistor, and electron's potential difference (voltage) before and after passing through a resistor is equal to the e.m.f, how does the electron have potential energy after passing through the resistor to flow to the positive terminal of the battery?
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1\$\begingroup\$ First of all an electron doesn't have a potential difference. Apart from that you are answering your own question, without a potential difference, there will be no current. Did you consider the possibility that there is no such thing as exactly 0V or +V(bat) in your circuit? In other words the conductors, the wires, are non ideal and do have an however small resistance. With your logic, not a single electron would exit the battery in the first place for the same reason. ;o) \$\endgroup\$– jippieCommented Feb 13, 2016 at 10:00
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\$\begingroup\$ Your entire premise is flawed. 1- An electron doesn't have a voltage, it has a charge. 2- This charge is not reduced in any way by having the electron pass through a resistor. \$\endgroup\$– brhansCommented Feb 13, 2016 at 14:35
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\$\begingroup\$ Does charge or potential difference cause the flow of current? \$\endgroup\$– Raphael LowCommented Feb 14, 2016 at 11:26
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\$\begingroup\$ If a superconductive loop has fifty amperes at exactly zero potential, why do the electrons keep moving in a circle? If you stir your coffee, and all the coffee in each circle is at the same height, how can the coffee molecules keep turning in closed loops? \$\endgroup\$– wbeatyCommented Jan 27, 2019 at 8:58
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2 Answers
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Firstly, remember that the current is a migration of electrons jostling and pushing each other in the general direction of the positive terminal rather than each electron shooting along the wire from negative to positive. Pushing an electron in at one end of the wire pushes one out the other end. This happens close to the speed of light even though the individual electrons are moving much, much slower than that.
Secondly, remember that the potential difference causes current to flow. Let's use another dodgy water analogy.
simulate this circuit – Schematic created using CircuitLab
Figure 1. Yet another water analogy.
Figure 1 shows a water tank with a down-pipe of resistance R1, a horizontal pipe which is relatively wide bore and has no resistance and another down-pipe of resistance R2. Water flows in the circuit due to the pressure difference between the top of the tank and the outlet (at the bottom of R2).
Question: There is no height difference (potential difference) across the horizontal pipe so how can the water have the energy to flow from left to right?
By now the answer should be fairly obvious. It's getting pushed by the potential difference in other parts of the circuit.
In the electrical circuit of your question the electrons can't all just pile up at the end of the resistor they have to continue to return to the battery. If they didn't the battery would get out of balance and refuse to supply any more out the other end.
I hope this helps.
Comment:
However if a circuit only has 1 resistor, electrons which pass through the resistor is not in between 2 points of different resistance. The electrons and the battery have no potential difference between them any more after passing through the resistor. So why will the electrons flow?
Figure 2. A single resistor water analogy.
Again, pushing the dodgy water analogy a bit further, we have two horizontal big pipes and one vertical small pipe, R1. We all know that the tank will drain despite the horizontal sections. What's driving the current is the potential difference between the top of the tank (battery +) and the open end of the pipe (battery -).
Keep asking questions if it's not clear. You are close to understanding but judging by your other question, Electric potential in a cicuit, you've got static and current slightly jumbled in your thinking.
answered Feb 13, 2016 at 10:12
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\$\begingroup\$ Thanks for the help. However in your analogy above the position without potential difference is in between 2 resistors, hence the potential difference between the 2 resistors push the water in the middle area. However if a circuit only has 1 resistor, electrons which pass through the resistor is not in between 2 points of different resistance. The electrons and the battery have no potential difference between them any more after passing through the resistor. So why will the electrons flow? Thank you for your help \$\endgroup\$ Commented Feb 14, 2016 at 11:33
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\$\begingroup\$ Thanks a lot! I think I've cleared up most of my misconceptions thanks to you :) \$\endgroup\$ Commented Feb 16, 2016 at 12:39
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In the theoretical world, wires have zero resistance - in other words they are perfect superconductors. It requires zero potential to push an electron along a conductor of zero resistance.
In the real world, all wires have a small resistance. That means there isn't just one resistor in the circuit, but several in series - one big one and several small ones.
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1\$\begingroup\$ ... and consequently, there is a potential difference between the "bottom" of the resistor and the negative battery terminal, equal to the voltage drop across the small resistance of that conductor. \$\endgroup\$– scannyCommented Feb 13, 2016 at 10:50
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2\$\begingroup\$ And what if that portion of the circuit was superconducting? \$\endgroup\$ Commented Feb 13, 2016 at 11:13
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1\$\begingroup\$ There's no voltage drop across a superconductor. V = IR, so if R is 0, then so is V. \$\endgroup\$– Simon BCommented Feb 13, 2016 at 16:29
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\$\begingroup\$ So then (my point was) it's not the potential difference along that conductor that's moving the electrons. They're getting pushed along by a potential difference elsewhere in the circuit. \$\endgroup\$ Commented Feb 13, 2016 at 17:09