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From what I understand, voltage is required to motivate electrons to flow through a circuit. With KVL we have that the voltage will essentially be depleted at the return point (ground). So, if this is the case then why can we have current flow (at the return point)?

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The ground point/connection and KVL are completely unrelated. Your circuit has no ground point (you did not draw one) so we cannot know at which potential to ground your circuit is, that is not a problem. KVL still applies. You could place a ground in your circuit, you can place it anywhere, it would not influence KVL in any way. A ground point is just like saying: at this node we define that the voltage is 0 V and that's it, no more no less. It is just a reference point. No current flows into it, if it did you would be doing something wrong. \$\endgroup\$ – Bimpelrekkie Sep 6 '17 at 11:50
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    \$\begingroup\$ The short answer is that in real systems the wires have a slight resistance, so the voltage is not exactly equal at the ends and there is a small potential causing the flow. \$\endgroup\$ – pjc50 Sep 6 '17 at 12:11
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    \$\begingroup\$ Either you consider the GND wire to have non-zero resistance, then you can not treat it as ONE node but TWO nodes connected by a non-zero resistor; or you consider it to have zero-resistance, then the WHOLE wire is the node and it doesn't make sense to talk about current INSIDE a node. \$\endgroup\$ – Curd Sep 6 '17 at 12:16
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Voltage cannot be depleted, it is simply a difference in potential between two points. Conventional current flows from higher voltage potential (the positive of the battery) to lower voltage potential (ground, or 0 volts).

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    \$\begingroup\$ Also good to note, that while conventional current flows from more positive to less, in actuality electrons move from less positive to more. I know you know this, just adding for completeness. \$\endgroup\$ – Kyle Hunter Sep 6 '17 at 12:40
  • \$\begingroup\$ Correct, it is always very important to note whether we are describing conventional current flow or actual current flow. \$\endgroup\$ – DerStrom8 Sep 6 '17 at 13:08
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If you study physics voltage actually derives itself from the strength of the electric field of a charge distribution. The electric field from a setup charge distribution causes current to flow with energy proportional to the the strength of the electric field. Voltage actually represents energy per unit charge. So basically voltage determines the amount of energy 1 coulomb of charge delivers. You see the electric field generated imparts energy to the flowing electrons in a wire that we call current. So when it is zero voltage it just means the electrons have zero energy relative to when it first started because it was used up after flowing through devices that used the energy.

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  • \$\begingroup\$ This is a good way of thinking about it. Saying that "electrons have zero energy relative to when it first started" is good because we can't measure energy at a single point. \$\endgroup\$ – Snoop Apr 11 '18 at 11:01
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If the voltage at Ground is 0V, why is there a current flow?

The ground potential is one node in a circuit where we define the potential to be 0V. This gives us the basis to express voltage potentials for every other node in our circuit since they are always expressed relative to the ground node. When we can choose any node in our circuit for that purpose at own will, we surely do not influence any current with that decision. And indeed: choosing the ground potential does not say anything about any current in the circuit.

With KVL we have that the voltage will essentially be depleted at the return point (ground).

KVL tells you that in a closed loop the sum of all voltages are 0. Depending on how you draw the loop, you will have a positive value for your voltage source(s) and negative values for your consumers -- or the other way around. The result would be the same: the sum of all sources and all consumers will equal 0V. The word "depletion" should not be used here. But feel free to comment what you mean with it.

So, if this is the case then why can we have current flow (at the return point)?

As said above: the voltage potential of a node has nothing to do with the current flow at that point. In theory you could strip the cord of the power cable to your running computer and touch the wire which is connected to neutral (0V). You would not get electrocuted since there is ideally no voltage difference between you and the neutral wire. Still you know that there is a current running through that wire to power your computer.

(PLEASE DON'T TRY THIS AT HOME. ONLY A PROPER MEASUREMENT REVEALS WHICH CONNECTION IS CARRYING GROUND POTENTIAL. DON'T EXPERIMENT WITH LINE VOLTAGE!)

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