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This seems like a question which has been asked multiple times on forums. I researched the problem but I still was unable to figure out the answer which I can at least intuitively understand. So in this simple circuit why is there no current leakage to earth? Assume that it is a true earth ground connection (a rod in the earth for example).

Update: I am giving researched answers below, but not as the only options. I am not satisfied with any of them, and give my explanation as to why. So I am hoping for either an elaboration or an alternative answer.

schematic

simulate this circuit – Schematic created using CircuitLab

The answers I have found:-

  1. Here With only one connection to ground there is no circuit for the current to flow through. It can't flow "to" ground, because there is nowhere for it to flow to. There's no difference between ground and a wire dangling in the breeze. However for current to flow there must be potential difference. And because the ground and the battery return has the same potential - why would current prefer to flow specifically to the return point. Especially taken into account that earth can absorb all electrons generated by the battery (on contrary for example to just piece of wire).

  2. Here The ground isn't a great conductor and while it is in parallel with the service, the amount of current returning through the ground is so small it is effectively zero - Firstly this answer implies that there's some leakage occurs which just depends on earth resistance. Also I do not understand which resistance is taken into account. Where it is measured? I think also the answer relates to the several connection of the circuit to the ground.

So anyone can give real physical explanation of this?

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  • \$\begingroup\$ 1) If there is a potential difference, it is equalized instantly after the connection. So yeah, there would be an instantaneous current. 2) I think it is not an answer. \$\endgroup\$ – Eugene Sh. Oct 12 '17 at 18:36
  • \$\begingroup\$ @EugeneSh. The question why would it stop then. The current flow can take either path to the ground or to the return.Both with the same potential. Why the second one is preferred? \$\endgroup\$ – Boris Oct 12 '17 at 18:38
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    \$\begingroup\$ 2) is an is more applicable to the AC mains situation where there can be some leakage and/or capacitive coupling to ground causing small (AC!) currents to flow. But your example is DC so 2) simply does not apply. \$\endgroup\$ – Bimpelrekkie Oct 12 '17 at 18:38
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    \$\begingroup\$ Let's turn that around, why would the current flow into the ground? What would happen if it did? How would you make that current flow into ground? Would there be a stream of electrons going from the circuit into the ground? Is that sustainable? \$\endgroup\$ – Bimpelrekkie Oct 12 '17 at 18:41
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    \$\begingroup\$ @Boris, no they are not at the same potential, and even if they were, the bottom of the battery is negative with respect to the top and the electrons move around the circuit from the bottom to the top. No extra electrons are added or removed from the circuit, as would be the case if something crossed the ground connection. \$\endgroup\$ – Trevor_G Oct 12 '17 at 20:38
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This is explained by Kirchhoff. He has two laws: the voltage law and the current law.

I won't explain the voltage law, as that is not what your question is asking, but the current law simply states:

The sum of all currents into a node is zero

alternatively:

The sum of all currents out of a node is zero

What this means in the case of your simple circuit is that the current \$I_1\$ flowing out of the top of your battery is equal to to the current \$I_2\$ flowing into the battery.

Since the battery and resistor are in series with the resistor, then the current flowing into the bottom node via the resistor is \$I_1\$.

Let's call the current flowing to earth \$I_3\$.

\$I_3 = I_1 - I_2 = 0 \$ since \$I_1\$ = \$I_2\$.

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  • \$\begingroup\$ Not sure Kirchhoff is applicable here as adding earth connection breaks the assumption of "Lumped element model" Kirchhoff is applicable to. Especially the second assumption about the constant charge as earth effectively makes the charge "dissapear" from the circuit. \$\endgroup\$ – Boris Oct 12 '17 at 20:14
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In reality there is current in the ground connection, but not enough to measure above ambient noise level.

Below is your equivalent circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

ADDITION

As such we normally ignore that leakage current in DC circuits. So let us look at your question another way.

For the moment consider what would happen if an electron fell down that "drain" to ground. What happens to the rest of the circuit?

Suddenly it will have a missing electron and will have a net positive charge. The ground now has an extra electron and has net negative charge. That means there is a reverse voltage on the ground wire which will immediately cause the electron to return to the circuit.

The opposite would happen if somehow an extra electron were to arrive from the ground. The circuit would be at a negative potential and ground would be positive. The visiting electron would be immediately repelled.

In actuality, that returning force is what prevents the electrons from crossing the connection in the first place. It is a self stabilizing state.

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  • \$\begingroup\$ Hmmm - I'm not sure about the precision of your 1e16 resistor ... \$\endgroup\$ – brhans Oct 12 '17 at 18:46
  • \$\begingroup\$ @Trevor Air resistance? And if the circuit is enclosed in vacuum? And only ground wire is "piercing" the vacuum chamber. \$\endgroup\$ – Boris Oct 12 '17 at 18:48
  • \$\begingroup\$ @brhans ya that was a very rough approximate and will vary widely... it's the exponent that is important though .. huge # = b.all current ;) Things change a lot when its a high frequency voltage source though. \$\endgroup\$ – Trevor_G Oct 12 '17 at 18:52
  • \$\begingroup\$ @Boris indeed, in a vacuum you still have an impedance.. but it gets complicated. \$\endgroup\$ – Trevor_G Oct 12 '17 at 18:53
  • \$\begingroup\$ @Boris, the point is you need a loop back to a higher potential for current to flow through that wire. Now you will get a loop, be that through leakage in air or the impedance of a vaccum, but the current that will flow round that loop is so small you can take it as zero for a DC circuit. A high frequency power source though would be different. \$\endgroup\$ – Trevor_G Oct 12 '17 at 18:58
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In the circuit from your question, the earth ground connection function is of a reference point, i.e. a zero potential point. From this point, you can specify all voltages in circuit nodes. It's a convenience for the analysis, say, nodal method.

There isn't relation with current. Is a convention to assign a node as a zero-potential node.

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  • \$\begingroup\$ i specifically mentioned that it is not reference point. But real ground connection. \$\endgroup\$ – Boris Oct 12 '17 at 18:51
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So in this simple circuit why is there no current leakage to earth? Assume that it is a true earth ground connection (a rod in the earth for example).

The circuit diagram shows no elements that would cause a current to flow to ground. If current DOES flow to the 'earth ground' node, is it positive? Negative? AC? DC?

Simply put, this circuit doesn't create any current output, unless other circuit elements exist, which are not shown. If there's a storm cloud, and lightning strikes, there WILL be current to ground. That, however, isn't part of the schematic, so isn't part of the equation. Thus, we're left with 'zero' current predicted.

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