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Joined: Oct 2000
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Got a question. 120v system supplying an outside Light Fixture. It is on. Assuming worst case you are standing in a puddle with your Ground rod or by your pool's bonding points (which are connected back to panel) If you touch the screwshell (grounded) how much current will flow through you back to source? (This gives us a seriesparallel circuit)  120v
 Lets use a bulb resistance (hot) of 120 ohms
 body resistance of 1000 ohms
(can it go lower? I don't know)
What would it be when the light is first turned on and that resistance is say 20 ohms? Please include work so We can follow calculations. Bill [This message has been edited by Bill Addiss (edited 05272001).]
Bill




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Bill, Assuming that the grounded conductor to the fixture is intact, there will be almost no current through the body. The only voltage to ground from this point on the circuit will be the voltage drop on the grounded conductor. Assuming the fixture is fed with 100' of #14, the voltage drop will be about 1.8 (1/2 of the voltage drop for the circuit as we have only sent the current through 1/2 of the wire at this point) volts for the cold lamp and 0.3 volts for the hot lamp. I=E/R so the current through the 1000 ohm person would be 0.0018 amps for the cold lamp and 0.0003 amps for the hot filiment. A GFCI is designed to trip at 0.004 to 0.006 amps. This current flow is well below the GRCI trip point. Don(resqcapt19)
[This message has been edited by resqcapt19 (edited 05282001).]
Don(resqcapt19)




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Bill,
Great Qs!!!
One of the major factors here is what the load current is past the lamp's screwshell. This will give the complete voltage drop [total current in circuit] at that point.
This Q kinda' looks like it's the spawn of the Mike Holt Forum thread about standing on the ground rod and grabbing hot wires  which turned into a Hot File [>15 posts].
You have a valid thought on the Inrush current the Incandescent Lamp will have  it might have an effect on the results. That inrush, for the most part, is looking at maybe a R of 17 OHMs for a coldfillament on a 120 VAC 60 Watt A17 lamp. That would last maybe 1/30 second [2 cycles], upto 1/8 second [8 cycles], then taper down to a running value [hotfillament] within 2 seconds [120 cycles].
Don has given better numbers than I can, so I'll not complicate the thread with jibberish.
Scott SET
Scott " 35 " Thompson Just Say NO To Green Eggs And Ham!




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Don & Scott, Thanks for the quick reply. Does anybody know how low the resistance of the body can go? If it wouldn't be too much trouble could someone help out with the math? (show work)?? It would be a good refresher. Bill
Bill




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Phil,
Thanks, good Link!
Scott,
Yes it's a spawn of that thread. I've conceded the point that voltages and current will be low under normal conditions or even these 'test' conditions mentioned.
I would like to do some permutations sometime to see what scenarios could be possible aside from a broken neutral. I figure that there could be 20 to 30 connections in the circuit who's integrity would have to be good to match our test Question. I'm wondering what variations could be expected in the field and still be a working circuit, and how those numbers affect the outcome here.
Bill
Bill




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Originally posted by Bill Addiss: Phil,
I would like to do some permutations sometime to see what scenarios could be possible aside from a broken neutral. I figure that there could be 20 to 30 connections in the circuit who's integrity would have to be good to match our test Question. I'm wondering what variations could be expected in the field and still be a working circuit, and how those numbers affect the outcome here.
Bill How about the farmer who had a light on a pole between the house and the barn. The farmer wanted to switch the light on/off from the house or the pole but there were only two overhead wires existing from house to pole. If the farmer added the three way switches and made the circuit work without changing the wiring between house and pole, what would this do to your calcs?




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If the voltage drop is 1.8 for the return path, then the resistance is 0.015 ohm (1.8 V / 120 V). For a nominal 120 W bulb the current flow is ~1 A (~120 VA / ~120 V). (Don's number may have been figured on a larger current).
The parallel resistance for a standing human (nominal 1000 ohms) plus the copper is 1/((1/0.015)+(1/1000)), which is 0.0149998 ohm.
The current that flows is inversely proportional to the resistance of each parallel path.
The human to the copper ratio is 1000 ohms / 0.015 ohms = 66667, meaning that 66667 times as much current flows through the copper as through the human. 1 A / 66667 = 0.000015 A through the human.
I have done this in rather round numbers. More exact calculations would require the resistance of both halves of the circuit with one including the bulb, the resistance of the human pathway back to the circuit source, and the exact voltage at the circuit source.
The problem I with Don's calculation using a 1.8 V drop is that it does not invoke the parallel resistance formula. If I provide a parallel path with a resistance of 0.015 ohms, his formula yields an amperage of 120... and you know it's not a 14500 W bulb.




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The problem I with Don's calculation using a 1.8 V drop is that it does not invoke the parallel resistance formula. If I provide a parallel path with a resistance of 0.015 ohms, his formula yields an amperage of 120... and you know it's not a 14500 W bulb. DS, I don't understand what you are telling me here. Don
Don(resqcapt19)




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Okay, Don, forgive me for copying your words above and changing a few...
Assume the fixture is fed with wire having a voltage drop of 1.8 volts for the cold lamp and 0.3 volts for the hot lamp. I=E/R so the current through the 0.015 ohm metallic path would be 120 amps for the cold lamp and 20 amps for the hot filament.
What I did was instead the human I substituted a metallic path with a 0.015 ohm resistance. I kept your math the same.



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