# A circuit with no voltage difference, but current flowing

From Michael on Skeptics Stackexchange:

How about a wire that's grounded? Safe to touch, right? WRONG.

   ________________ 30 amps -> ________________
|                                            |
+                                            |
-                                            |
|______(YOU ARE HERE)______<- 30 amps________|
|
Ground


The wire you touched was not only at 0 volts, but also grounded, and yet, you are feeling pretty shitty in this diagram. You have ceased to be as a human, and you are now a part of a circuit, functioning as part of a return leg (pictured above) or as a "parallel path to ground" (not pictured above.)

I don't get how this can work. If the wire is at 0 volts and you are at 0 volts, then there is no potential difference and hence I'd expect no current. Is this physics correct?

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As a practical matter, touching a grounded wire will not shock you. However, I've been shocked on touching a ground wire. The reason is that it wasn't grounded. Their advice is accurate; avoid touching electrical wires. If the circuit is hooked up correctly you won't be shocked. But plenty of circuits aren't hooked up correctly. –  Carl Brannen Apr 3 '11 at 7:28
In a forum like sceptic... this question is "legal" maybe. What I do not understand, is that "current without voltage" is not opposed immediatly in a physics forum. This is answered now, but only in an indirect form. In another thread there was the statement that electrons "follow the current". :=( –  Georg Apr 3 '11 at 12:05
@Georg: just because a post include a phrase that describes something unphysical, it doesn't mean it's unwelcome here. Nowhere in the question does Casebash claim that a current flows without a voltage difference. IMO questions of the form "this doesn't seem to follow established physics, so what's going on?" (such as this one) are perfect for this site. –  David Z Apr 4 '11 at 3:51
I do not object this question in toto (I think I wrote what I miss) –  Georg Apr 4 '11 at 9:50
That's why electricians keep one hand in their pocket when working in panels. –  joebanana May 26 '11 at 21:50

The issue is that by touching a wire, you're augmenting the circuit with yourself as a resistor. (At first I wrote "inserting yourself" but as mmc's comment pointed out, that is a misleading phrase to use.) And whenever you change the layout of an electrical circuit, all the potentials and currents are subject to change. So the wire that is at ground potential before you touch it won't necessarily still be at the same potential after you touch it.

In this specific case, I'd guess that the $30\ \mathrm{A}$ value is the current that flows with only the battery and the load.

You can calculate that the resistance of the load has to be $7.3\ \Omega$.

If you could actually insert yourself into the circuit as shown in the diagram, you would be adding to the resistance of the circuit, which reduces the current. The resistance of the human body varies greatly depending on several factors, but a "typical" value might be on the order of $10\ \mathrm{k\Omega}$ which reduces the current to a small fraction of an Ampere.

Even with that small current, though, you're going to experience a large voltage drop because your resistance is so large compared to the rest of the circuit. In the diagram above, there would be a voltage drop of $219.8\ \mathrm{V}$ across your body. By inserting yourself in the circuit, you've prevented the wire coming out of the load from being grounded.

Note that if you're trying to determine whether this is a dangerous situation to be in, the value you should be looking at is the current of $0.022\ \mathrm{A}$, not the $219.8\ \mathrm{V}$ voltage drop. That's what it means to say that it's the current that kills you, not the voltage.

If you instead grab on to the wire while standing on the ground (which seems like a more realistic situation), you're not actually inserting yourself in the circuit. Instead you wind up with a setup more like this,

In this case the wire coming out of the load resistor is still at potential zero because it's connected to ground via a zero-resistance path. (Keep in mind that this is an ideal model; real wires do have some small nonzero resistance so in reality the wire wouldn't quite be at zero volts.) So the voltage difference across your body is going to be basically zero.

Besides, any current that flows between the circuit and the ground can do so by one of two paths, either through you or through the open wire. Since your bodily resistance is much higher than that of the wire, essentially all the current will go through the wire, not through your body.

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@Casebash It's important to note that the original answer in the Skeptics SE is wrong. One cannot insert oneself in the return loop by touching a grounded wire. –  mmc Apr 3 '11 at 4:18
@mmc Your assumption is unwarranted. In practice, you can easily insert yourself into a circuit as depicted in my diagram or David's second diagram if the circuit is open, and your body completes the circuit while you are, for example, tinkering with the connection at the load, at its double-throw switch, at its service panel, etc. If you merely grab onto the grounded return leg, and if you are grounded, then you become a parallel path to ground, as seen in David's final picture. You will have current flow through your body, but as David points out, it will usually be insignificant. –  Michael Apr 3 '11 at 6:33
@ David Zasalavsky an off topic question...what did you use to draw these circuits? –  Sandeep Datta Apr 3 '11 at 6:46
This comment-discussion is typical. Maybe by writing some realistic values for the resistance of the leads (eg some dozen milliohms) would have enhanced understanding. Another problem is not reading carefully. The question was about "touching a grounded wire". Is there something ambiguous in that words? –  Georg Apr 3 '11 at 11:59
Blind leading the blind. –  Janne808 Apr 5 '11 at 18:10

If you are in the circuit at that point then there is a 240V potential different across you! One side of you is at 0v and there is no potential difference between you and the 240V line (except for the wire resistance) and so that side of you is at 140V

If you mean you are standing on the ground and touch the circuit at that point then, no the line is at approximately 0 and there is no potential difference.

Obviously in the real world don't do this anyway - just in case!

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I am stunned -- STUNNED! -- that no one mentioned the difference in voltage between "neutral" and "safety ground". Anyone with a multimeter can jam the probes into the "neutral" and "ground" slots of a standard 3-pin electrical outlet and measure for himself the difference in voltage between those two points. That difference is not zero -- even when both lines are properly and solidly connected to "earth ground" at a single point.

This situation involves small resistances that normally can be neglected, but not in this case.

This is not a situation with "no voltage difference".

   ________________ 30 amps -> ________________
|                                            |
+                                            |
-                                            |
|__________________________<- 30 amps________|
|                       |
|______(YOU ARE HERE)___|
|
|
Ground


The wire you touched was not only at 0 volts, but also grounded

No. One end of the wire connected to the ground post is grounded at 0 volts, but the other end that you are holding is not exactly 0 volts.

... ...

I don't get how this can work. If the wire is at 0 volts and you are at 0 volts, then there is no potential difference and hence I'd expect no current. Is this physics correct?

The wire is not at 0 volts, and therefore there is current through your body.

Let's redraw this diagram, including a few resistances that are normally negligible:

   ________________ 30 amps -> ________________
|                                            |
+                                            |
-                                            |
|_r2__________r6_______r4____<- 30 amps______|
|                         |
|r3_____(YOU ARE HERE)_r5_|
|
|
Ground


When 2 physical wires touch each other, there is a contact resistance. (Rule of thumb: it's typically around 0.1 Ohm). That's r2, r3, r4, r5.

When you have current flowing from one end of a long wire to the other end, the resistance (r6) is often very negligible, but not in this case. A copper-conductor 16 AWG extension cord has a resistance of about 4 milli-Ohm per foot. So the "neutral" conductor of a 20 foot long cord has about 0.08 Ohm of resistance from the copper (not including the contact resistance).

Even though most of the 30 A bypasses the human and flows in the copper wires, a small amount of that 30 A goes through the human.

The voltage across the human is approximately 30 A * ( 0.1 Ohm + 0.08 Ohm + 0.1 Ohm) =~= 8 V. Out of 220 VAC, a mere 8 VAC is often negligible -- but not in this case. If some human has a resistance of 100 Ohms (it's typically an order of magnitude more), The current through that human would be approximately 8 V / 100 Ohm =~= 80 mA.

This is more than the 60 mA AC which can cause death from ventricular fibrillation.

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There was no mention of neutral vs. safety ground in the original post. It's a question about an ideal circuit. Although it never hurts to emphasize the point that you shouldn't go around touching wires in real life, you're pulling in some extra information that isn't necessary to this question. (Also, 100 ohms for human resistance? That's an exceptionally low figure) –  David Z Apr 26 '11 at 5:02
You're right -- human skin resistance is normally a few orders of magnitude higher. –  David Cary Jun 14 '11 at 1:25

There will be voltage across you since you are part of the circuit. How much? Well, apply Kirchoff's law,

$220 = IR + v$

where $I$ is the current in the circuit and also through your body which is $30 amp$ in this case, $R$ is the resistance of the load. $v$ is the voltage across you.

$v = 220 - IR$ or $v = 220 - 30R$

This is the voltage at a point just right of (you are here). You see it is not at 0 volt. The voltage at a point just left of (you are here) is 0 however.

Hence 30 amps current will flow through your body which is dangerous.

Don't try this experiment.

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Please explain the down vote. I am an electrical engineer with years of experience. What is wrong in this answer? –  user1355 Apr 3 '11 at 4:22
I wasn't the downvoter, but perhaps whoever did was objecting to the assertion that the current flowing through you is actually 30 amps. In order to make that amount of current flow through the body, you'd need a lot more than 220V (because the body's resistance, although it does vary greatly, is always more than 220V/30A = 7.3 Ohm AFAIK). –  David Z Apr 3 '11 at 4:38
@david : it is given in the problem that 30 amps is passing and there is no extra branch and the current has to pass through the body. –  user1355 Apr 3 '11 at 5:03
@sb1: OK, but I think the 30A as given in the original question is just not realistic. I looked at some websites to research human resistance values and although your resistance certainly does drop by a large factor when you're wet, the 7-ohm value required for the given current and voltage is exceptionally low. –  David Z Apr 3 '11 at 5:26
This answer is wrong. If 30 A were flowing through the human body, it would require at least 16 kV. If 220 V were applied across your body, it would produce less than 400 mA. Both can't be true at the same time. Ask an electrical engineer. –  endolith May 25 '11 at 18:56

## protected by Qmechanic♦Jun 29 at 21:30

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