Why does touching an un-grounded circuit theoretically not shock you?

Thread Starter

Merrick_097

Joined May 14, 2017
6
Hello everyone!

I am new to electronics and am working on an associates program as a means to expand my knowledge. I come from a background in Chemistry so I'm all stuck in my head about reactions going to an "equilibrium".

In the case of my question in the subject line, I was reviewing Chapter 3 in the All About Circuits textbook regarding electrical safety and shock path. In the example of a person touching an un-grounded circuit, the author details that ideally a grounded person can touch any portion of the circuit without getting shocked (ideally). I understand that in order for current to flow, there needs to be a complete path through a circuit, which the person is not a part of.

That being said, as I understand from the book, voltage is the potential energy caused by the separation of charge. If ground is our reference for "0V", then wouldn't there be a difference in voltage should you be touching the positive terminal of a voltage source? In that case wouldn't current flow through you to ground in an attempt to balance the charge?

CHPT3_DIAGRAM.png
 

MrChips

Joined Oct 2, 2009
21,042
That drawing is not completely correct. Do not attempt this as an experiment.

What is not shown is the total capacitance of the circuit.
Every object has electrical capacity, that is, the ability to hold a charge.
If the capacitance is sufficiently large, a significant current could flow through your body in order to remove the charge on the object that is touched.
 

crutschow

Joined Mar 14, 2008
24,911
voltage is the potential energy caused by the separation of charge. If ground is our reference for "0V", then wouldn't there be a difference in voltage should you be touching the positive terminal of a voltage source? In that case wouldn't current flow through you to ground in an attempt to balance the charge?
Ground is a 0V reference for a voltage positive terminal only if there as a connection between the voltage minus terminal and the ground.
If they voltage is floating then the voltage between the source and "ground" is undefined.
Repeat after me as many times as it takes:
"Current cannot flow without a complete path for the charge to follow."
 

Thread Starter

Merrick_097

Joined May 14, 2017
6
Ground is a 0V reference for a voltage positive terminal only if there as a connection between the voltage minus terminal and the ground.
If they voltage is floating then the voltage between the source and "ground" is undefined.
Repeat after me as many times as it takes:
"Current cannot flow without a complete path for the charge to follow."
Could you clarify the term "floating"? I assume that means the absolute voltage between terminals on a source are not known only that the difference is what is stated on the supply. While I do understand we need a complete circuit for charge to flow, does charge not balance between two regions? Unless is ground not always 0V? Does ground not always guarantee a difference in voltage in the case of the diagram with the person touching the positive terminal?
 

SLK001

Joined Nov 29, 2011
1,543
Could you clarify the term "floating"?
A "floating" ground could be -21V, +74V, or even +115kV different from "earth" ground (what you are standing on now). It doesn't matter to the circuit, as the voltage are normal (+5V, +12V, etc).

If you think that you cannot get shocked by an ungrounded circuit, you are destined to die from electrocution. If you touch a "floating" ground and you are at another ground potential, your body will conduct current until the two grounds are the same. The current may be high enough to cause burns, or to even seize your heart.

If you want verification of this, watch the repairing of +115kV transmission lines with a helicopter. Before a worker is lowered to the line, a grounding rod is moved toward the line, where as much as 6 feet of arc jumps from the line to the helicopter. The current is low, but the potential difference between the two is huge, which results in the arc as the two potentials are equalized. Once the helicopter's ground is floating at +115kV, the repair can proceed.
 

Thread Starter

Merrick_097

Joined May 14, 2017
6
A "floating" ground could be -21V, +74V, or even +115kV different from "earth" ground (what you are standing on now). It doesn't matter to the circuit, as the voltage are normal (+5V, +12V, etc).

If you think that you cannot get shocked by an ungrounded circuit, you are destined to die from electrocution. If you touch a "floating" ground and you are at another ground potential, your body will conduct current until the two grounds are the same. The current may be high enough to cause burns, or to even seize your heart.

If you want verification of this, watch the repairing of +115kV transmission lines with a helicopter. Before a worker is lowered to the line, a grounding rod is moved toward the line, where as much as 6 feet of arc jumps from the line to the helicopter. The current is low, but the potential difference between the two is huge, which results in the arc as the two potentials are equalized. Once the helicopter's ground is floating at +115kV, the repair can proceed.

Okay I think I'm starting to get it! So in this case the textbook on this web-site is incorrect then, a current will pass through you to ground to equalize the charge gradient between the line and the un-grounded energized circuit.

I understand that different grounds can have different voltages, but with the mention of these floating voltages I have another question, how can you be sure the ground you choose is at a lower voltage than your circuit? While I know this would be highly unlikely, could you unknowingly connect a +3V battery to a +74V "ground"? In this case would you have current moving into your circuit from ground?
 

MrChips

Joined Oct 2, 2009
21,042
Okay I think I'm starting to get it! So in this case the textbook on this web-site is incorrect then, a current will pass through you to ground to equalize the charge gradient between the line and the un-grounded energized circuit.

I understand that different grounds can have different voltages, but with the mention of these floating voltages I have another question, how can you be sure the ground you choose is at a lower voltage than your circuit? While I know this would be highly unlikely, could you unknowingly connect a +3V battery to a +74V "ground"? In this case would you have current moving into your circuit from ground?
No. Different grounds cannot have different voltages.
By definition, GROUND must be at 0V if we are referring to EARTH GROUND.

If a circuit ground is not at 0V then it should be labelled as COMMON, or what we would classify as FLOATING GROUND.
It is a circuit REFERENCE and not true GROUND. Sometimes we call this SIGNAL GROUND.
 

MrChips

Joined Oct 2, 2009
21,042
CIRCUIT REFERENCE or COMMON does not have to be a lower voltage that the circuit. You can create a split power supply using two 9V batteries as shown:

upload_2017-5-14_20-27-34.png

You can also create a SIGNAL REFERENCE or virtual-GROUND from a single supply by splitting the supply voltage in half:

upload_2017-5-14_20-31-14.png
 

Thread Starter

Merrick_097

Joined May 14, 2017
6
CIRCUIT REFERENCE or COMMON does not have to be a lower voltage that the circuit. You can created a split power supply using two 9V batteries as shown:

View attachment 126788

You can also create a SIGNAL REFERENCE or virtual-GROUND from a single supply by splitting the supply voltage in half:

View attachment 126789
I think I see it now, especially from the first picture. So your floating/common ground voltage "value" is insignificant, since it is tied to right terminals on the battery you can split the voltage and make a relative +9V/-9V supply. While I don't have all the know-how for the second circuit I do recognize the voltage-divider, which I take it creates a similar case to the above circuit which once again splits the supply into a relative +V/-V, in which we can classify the voltage divider output as the virtual ground relative to the circuit.
 

MrChips

Joined Oct 2, 2009
21,042
I think I see it now, especially from the first picture. So your floating/common ground voltage "value" is insignificant, since it is tied to right terminals on the battery you can split the voltage and make a relative +9V/-9V supply. While I don't have all the know-how for the second circuit I do recognize the voltage-divider, which I take it creates a similar case to the above circuit which once again splits the supply into a relative +V/-V, in which we can classify the voltage divider output as the virtual ground relative to the circuit.
You've got it!
 

crutschow

Joined Mar 14, 2008
24,911
Note that the arc between the wire and the helicopter is largely due to AC current through capacitance between the chopper and ground as the arc does not stop until the equalizing probe actually touches the HV wire.
So it's the capacitance that provides the return path.
 

Thread Starter

Merrick_097

Joined May 14, 2017
6
Note that the arc between the wire and the helicopter is largely due to AC current through capacitance between the chopper and ground as the arc does not stop until the equalizing probe actually touches the HV wire.
So it's the capacitance that provides the return path.
This might be above my current level but I do recall learning that a capacitor blocks DC current but allows AC current to pass through. A quick googling suggests the phenomena you are describing is capacitive coupling. So in the case of the original diagram I posted, if there was AC current traveling through the circuit, this would functionally create a "capacitor" between us, which would allow AC to travel between us and provides the return path to the circuit as you said. This is great information to have!
 
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