Current through body as a result of touching a neutral

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gar,

I must be thick-headed tonight. I have read your posts several times but still do not see 1/2 voltage from feet to hands. Feel like making a quick sketch or something?

add: I can see 1/2 voltage from heart to hands or heart to feet but don't see the relevance.

add: nevermind...the housing is grounded?
 
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When I originally set this up, I took a reference from a study on resistance of the human body, under varying conditions of contact. Dry skin with contact of minimal pressure and surface area resulted in apparent resistance of around 10 K Ohms. However, the higher the pressure and the larger the contact surface area and the more electrolytes and moisture was present, the further the resistance dropped. Low readings of approximately 100 Ohms were observed.

For my hypothetical guy, working in soaked conditions on a job site at 3:30 on a Friday afternoon, I arbitrarily chose 500 Ohms for his through body resistance. I added another 500 Ohms for the footwear.

The wire resistance is simply from NEC Chapter Nine.

I left the open circuit source voltage at a low value.

This is the circuit diagram that I got.

img37.gif


Reducing this to its equivalent, I got:

img39.gif
 
090730-2009 EST

don:

I basically use this type of analysis to show why there can be a major problem when connecting a computer via a direct RS232 connection to a CNC machine.

Usually totally different branch circuits are used to supply the equipment at both ends of this serial connection. The EGC reference point for the computer and CNC is at the main panel. If a HOT to EGC short occurs at the CNC (CNC cabinet and machine frame are connected to the CNC EGC), then there is a voltage drop on the CNC EGC, but not on the computer EGC. Thus, there is a potential difference between the CNC chassis and the computer chassis equal to the voltage drop on the CNC EGC.

.
 
al hildenbrand said:
...Reducing this to its equivalent, I got...
Click to expand...
Double-checking your work for fun:

I get 56.7047926665 volts from the low-side to the hand. The current is 55.5410963166 mA through the apprentice.

The equivalent impedance from "c" to the low-side is 0.301910933444 ohms. Inserting the apprentice in series with the parallel grounds then this in parallel with an impedance called Rneutral yields Rneutral = 0.302001029702 ohms.

I'm going to go out on a limb here and call your numbers close enough. :grin:
 
gar said:
Thus, there is a potential difference between the CNC chassis and the computer chassis equal to the voltage drop on the CNC EGC.
Click to expand...

Sans the frame ground you have a path via the signal ground and it lets the smoke out... ... ... ask me how I know. :roll:


Al, nice drawings. They were enjoyed.
This is a great example, and I have been the little resistor guy in the middle before.
It started to rain, I started to tingle! I didn't need any further object lesson.
Needless to say I quickly destroyed my supposed cord-end GFI.

AND I now test my field use GFI's periodically! (With a tester, not my body)

My 2?
Doug S.
 
philly said:
Even if you cut the neutral and then somehow put your body in between the neutral wire from the load and the other side of the cut neutral wire returning to the source, you would still have 120V across your body, because your body would now be in series with the load and current would return in series on the neutral with the voltage across you body being determined again by a voltaeg divider?
Click to expand...

Thanks to the retarded electrician that wired my house, I got to try this out first hand a little while back.

I was changing a wall light fitting, and had the breaker feeding the circuit off, and was working away changing the light and got shocked. On investigation the electrician had run out of cores from the garage to the panel, so used the wall light neutral for the garage and outdoor light return. I got between ground and open neutral as I undid the wiring for the old wall lights...

I s'pose I'm glad that the only thing on was the outdoor CFL and thus it was a somewhat current limited shock... and this is on 240V.
 
al hildenbrand said:
When I originally set this up, I took a reference from a study on resistance of the human body, under varying conditions of contact. Dry skin with contact of minimal pressure and surface area resulted in apparent resistance of around 10 K Ohms. However, the higher the pressure and the larger the contact surface area and the more electrolytes and moisture was present, the further the resistance dropped. Low readings of approximately 100 Ohms were observed.

For my hypothetical guy, working in soaked conditions on a job site at 3:30 on a Friday afternoon, I arbitrarily chose 500 Ohms for his through body resistance. I added another 500 Ohms for the footwear.

The wire resistance is simply from NEC Chapter Nine.

I left the open circuit source voltage at a low value.

This is the circuit diagram that I got.

img37.gif


Reducing this to its equivalent, I got:

img39.gif
Click to expand...


Al from your diagram above it looks like it shows the fault to ground happening on the neutral side of the drill load which I'm assuming is
"c". Since we are not breaking the neutral is this an example of my question from the origonal post, showing what happens when someone touches the neutral of a circuit that is still grounded?

This diagram doesn't look like the other case you talked about where the hot conductor made contact with the drill casing. Or am I missing something?
 
steve66 said:
That's pretty much the case as long as the neutral wire has a good connection back to complete the circuit.

It's also important to remember as soon at that connection is gone, the neutral wire becomes a "hot" wire.

I've seen a journeymen splicing neutral wires on live circuits. He didn't realize (and neither did I at the time) that as soon as he cut the neutral wire (to install a wire nut) that one of those ends suddenly had 120 Volts on it.

Steve
Click to expand...

Ok, we've discussed how there can be 120V on the end of a neutral that has been cut for a single branch circuit with only one load. I'm assuming that the same applies for a branch circuit with mutiple loads tied into a single neutral where you cut the neutral anywhere on the neutral wire.

Does the same principle apply for the neutral connected to multiple single phase circuits on a three phase system?
 
gar said:
090730-1519 EST

philly:

Al provided a question earlier assuming 150 ft of #12 wire and 10 A thru the neutral. Effectively he asked what is the voltage drop on that neutral? Here is the answer --- 1000 ft of #12 copper is about 1.6 ohms, 150 ft is 0.24 ohms, 10 A and 0.24 ohms is 2.4 V. Under certain conditions 2.4 V can kill you.

To add to Al's last question. If your bare feet are on wet earth, your bare hands are holding the metal case of an electric drill, the extension cord is of most any length, all three wires in the cord are the same material and diameter, and there is a direct dead short between the hot wire and the drill housing, then the approximate peak voltage between your feet and hands is about 0.5*120*1.414 = 85 V. You will probably be dead.
Click to expand...

Gar, One small point, the standard practice is to use RMS values when calculating stray voltage/current, not peak values. There are no standards to speak of addressing this, but it is the practice used by the public service of wisconsin and also addressed in a paper by the Resource, Agriculture, and Engineering Service, which discusses stray voltage measurement techniques.

I know we are talking about contact voltage and not stray voltage, but electrically it is the same principle, only the voltage/current magnitudes are different.
 
philly said:
Ok, we've discussed how there can be 120V on the end of a neutral that has been cut for a single branch circuit with only one load. I'm assuming that the same applies for a branch circuit with mutiple loads tied into a single neutral where you cut the neutral anywhere on the neutral wire.
Click to expand...

Yes, the same thing would apply with most 120V loads. It's basically just using the voltage divider formula. All the applied voltage is dropped across the open circuit. So if you have 20 lights in parallel, and you cut the neutral to the last one, you will still see about 120V across the two ends of the cut wire.

philly said:
Does the same principle apply for the neutral connected to multiple single phase circuits on a three phase system?
Click to expand...

That gets a little more complex. The voltage that would be present depends on how well the loads are balanced. With a balanced load, you wouldn't get any voltage at the neutral when you cut it.

The same thing can also apply for single phase circuits if two different phases share a neutral. If the loads are balanced, then you don't have the same situation.
 
090731-0930 EST

philly:

You can use RMS for a comparison if you are working with a sine wave, and I simply calculated the peak assuming a sine wave.

However, peak voltage is really the important criteria, especially if you are looking at electronic circuits. The typical failure mode of an integrated circuit or most other semiconductor devices is not related to the RMS or average value of the applied voltage, but rather to the maximum voltage (peak).

See
http://www.datasheetarchive.com/dow...m/pdf-datasheets/Datasheets-21/DSA-408744.pdf
Click on PDF Datasheet.
Look at Maximum Ratings 25 deg C.
The maximum input is +/-30 VDC. It does not say RMS. Since this is the maximum value it means that a sine wave applied to this input with an average value of 0 should not exceed an RMS value of 30*0.707 = 21 V. If I have a DC input of 30 V then the RMS value is 30 V. If I have a repetitive pulse of 0 to 30 V with a duty cycle of 1% (30 V for 0.01 seconds and 0 V for 0.99 seconds), then the RMS value is 3 V and the average is 0.3 V.

Most breakdown problems result from peak values and not RMS values. Gaseous discharge breakdown is related to peak voltage.

.
 
090731-1253 EST

philly:

Based on the various questions you have asked on this thread I suggest that you draw the schematics of the circuits in question, break the wires where you want, and then analyze what happens.

Al has provided enough schematics and rough values that if you use those values in your schematics you should be able to answer your own questions.

.
 
philly said:
Al from your diagram above it looks like it shows the fault to ground happening on the neutral side of the drill load which I'm assuming is
"c". Since we are not breaking the neutral is this an example of my question from the origonal post, showing what happens when someone touches the neutral of a circuit that is still grounded?

This diagram doesn't look like the other case you talked about where the hot conductor made contact with the drill casing. Or am I missing something?
Click to expand...

I was thinking that the diagram was showing the neutral wire but it appears that it is showing the EGC wire, and "c" is where the Hot to EGC fault takes place. I see not that this is an EGC and how current can still go through the body even with an EGC.
 
philly said:
I had a discussion recently about what would happen if someone were to touch a neutral conductor (grounded neutral) that had current flowing through it.

I was thinking that because there is no voltage potential on the neutral in relationship to ground, there would be no new ground currnent introduced through the body. However because there is current flowing through the neutral your body would set up a current divider for this current with the neutral wire connected back to the source, and a small amount of current would go through your body. The amount of current would be dictated strictly by the amount of neutral current determined by upstream loads, and by the impedance ratio between your body and the neutral path returning back to the source.

Am I looking at this correctly?
Click to expand...

Yes, you are looking at this fairly accurately (you neglected to mention the neutral wire resistance/body resistance in the determination of the total current in the circuit).

Consider the following:

Picture1-5.jpg


First, the amount of current flowing in the entire circuit would be determined by the total resistance of the circuit. The total resistance of the circuit is the hot wire resistance of .25 ohms plus the load resistance of 12 ohms plus the resistance of the parallel path of the neutral wire with the resistance of the body/earth path.

The neutral/body earth parallel path is as follows by the product/sum method:

R(neutral wire and body earth) = (.25 x 5005)/(.25 + 5005) = .2499875 ohms

The total resistance of the entire path is:

R(t) = .25 + 12 + .2499875 = 12.4999875 ohms

The total current through the load is I = E/R = 120/12.4999875 = 9.6000092 amps

Using the ratio of neutral wire impedance to body and earth resistance to determine current flow through body:

9.6000092 amps x .25/5005 = .0004795 amps

The current flowing through the body in the diagram is .0004795 amps, less than 1/2 milliamp.


The same conclusion can be reached by determining the voltage drop across the neutral/body parallel resistance, then using that voltage applied across only the body/earth resistance as follows:

R(parallel path = .2499875
Amps = 9.6000092
Voltage across parallel path = I x R = 9.6000092 x .2499875 = 2.3998823 volts

Use that voltage across body/earth path:

I(through body) = 2.3998823/5005 = .0004795 amps, the same result as with the ratio of impedances.

The actual voltage across the body would be .0004795 x 5000 = 2.3974848 volts.


To sum this up, under normal circuit conditions, touching a neutral wire that has current flowing through it is not going to hurt anyone. On the other hand, touching a broken neutral, or a neutral with high resistance connections at the panel may have serious consequences.
 
philly said:
I was thinking that the diagram was showing the neutral wire but it appears that it is showing the EGC wire, and "c" is where the Hot to EGC fault takes place. I see not that this is an EGC and how current can still go through the body even with an EGC.
Click to expand...
Hi, Philly.

Just, finally, returning my attention to cyberspace.

You are correct, that "C" is a "point", a point of contact within the drill body, where the hot conductor has gotten loose and banged up against the conductive metal body (EGC).

Now, I could've gone down a rabbit hole, chasing endless little resistances at the plug contacts and terminals, the panel and receptacle contacts and terminals, panel bus impedance, etc, etc. I chose to assume that the workmanship was very good, and materials were fresh enough to not be affected by any degradation, and the associated impedances were negligible. That left the impedances of transformer, conductors, contact with earth, and the human body.

My scenario is a simple setup, that I hope you can "feel".

Now, having considered my scenario, let's return to your question about the "touch voltage" of a neutral, and imagine that it is the neutral of a subpanel that is some length of wire away from the service disconnect (with its Main Bonding Jumper and Grounding Electrode System connection).

If a hot to neutral short occurs at the subpanel, the voltage at the subpanel neutral will rise to roughly 1/2 of the hot open circuit voltage, assuming the wires constituting the hot and neutral are of essentially equal gauges and lengths.

Perhaps you can suggest some possible through-human "shock" scenarios that might happen, for a human that is in contact with a neutral within the vicinity of the subpanel.

Lastly, I'm not sure what your phrasing is saying here:
philly said:
I see not that this is an EGC and. . . .
Click to expand...
 
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