# I don't get "electrically common-ness"

Discussion in 'General Electronics Chat' started by Amaterasuu, Oct 10, 2010.

1. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Hi, I'm fairly green to this field, but I'm having a great time reading the awesome ebook series to get myself into this.
I am at the third chapter and I am still being confused by the term "electrically common".
I understand what it means, but I have no grasp of how to determine such a property.
How do things become electrically common to each other?
Grounding a giant wire makes the ground "electrically common" to the wire, does that mean any physical contact makes it possible?
If so, let's take a look at the following:

The toaster is now electrically common with the ground, right?
What happens if both the hot and neutral wire touches the metal case (of the toaster)?
Also, how is Ohm's Law applied to electrically common points?
I = 0/R...=0?
Thanks

Last edited: Oct 11, 2010
2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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It refers to equilibrium. You have two reservoirs of charges, once you connect them, the charges flow until the two types of charges are equally distributed, equilibrium is reached.

Another way to think of it is hot and cold. Put the two objects together and the final result is something in between. The whole will be warmer than cold object, but cooler than the hot object, they achieved a common temperature.

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3. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Thank you for the swift reply, I understand what you are saying, but another question arises:
The earth is in fact a giant resistor, because it has poor conductivity.
But still, you can achieve equilibrium with physical contact.
How come "normal resistors" (the normal ones) require voltage drop cross them instead of also being electrically common with the wires touching them?

4. ### shteii01 AAC Fanatic!

Feb 19, 2010
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You forget what resistors do. They impede the flow of the electrons. The voltage drop that you see across the resistor is the energy needed to force electrons through the resistor.

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5. ### Wendy Moderator

Mar 24, 2008
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The breaker blows, since this is a short. In older houses the wiring catches on fire.

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6. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Exactly, I did not forget that.
In order to illustrate my question:

How come the the ground points have zero voltage? (I know, it's because they are ele. common)
But isn't the earth effectively a resistor between those two points?
Indeed because it is, there is a high voltage drop between such a circuit:

In this case, the earth acts like a resistor, and 2000+ voltage is needed to push electrons pass the earth (right?).
Now the question again: Why does the eletrcially commonness of the earth with the neutral wire in the second case not cause a 0 voltage drop between the two points?
By the way, thanks Bill_Marsden too!

Last edited: Oct 11, 2010
7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In the first figure the circuit current is returning to the source via the low resistance common return conductor - rather than via the more uncertain path through the earth itself. This circuit doesn't appear to be suggesting a case of fault condition.

In the second circuit the current at the downed line fault point returns to the source earth connection via the bulk resistance of the earth itself. There is a voltage gradient from the source earth point to the fault point. If this occurs over a distance of say 2.39km there will be a mean voltage gradient in the earth between the fault and the source earth point of 1 volt per meter displacement.

In reality the voltage gradient near the fault and the source earth points will be appreciably higher owing to the higher current density at both the fault and the earth return connection at the supply. As the current 'returns' to the source it will distribute somewhat arbitrarily throughout the earth before re-concentrating at the source common point. I've ignored the nominated current flow convention shown in the diagram.

Last edited: Oct 11, 2010
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8. ### GetDeviceInfo Senior Member

Jun 7, 2009
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'Common' by definition does not really explain the general term as applied to electricity. In electricity, the term has a more absolute value, that of zero resistance.

I disagree with the 'equilibrium' analogy, although it could be applied to the dictionary definition of 'common'.

In regards to earth grounding, you are correct that the earth is not a homogenous common, due to varying resistance. The terms 'touch' and 'step' voltages can be found in power station jargin. In other words, take baby steps when high voltage lines are down.

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9. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Thanks, but is there a reason why the current "chooses" to not use the earth as a return conductor, but instead the low resistance common return conductor?
Is it because there already is a lower resistance path?
Allow me to further illustrate my question, and thanks again for the support!

Contact with the resistor means contact with that wire.(which is the same, almost)
What would happen in that scenario?

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10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yes that's the reason. The lower resistance path being via the conductor.

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11. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Thank you so much, so that means that in the last uploaded picture. the current would still flow through the wire instead of passing the resistor at all, right?

12. ### GetDeviceInfo Senior Member

Jun 7, 2009
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As you've presented the circuit, there is no force that would drive a current through the human.

BE AWARE that the circuit as you've presented it, does not reflect real world transmission. DO NOT play with live transmission wire.

13. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Sorry, I'm back for once more!
Is there even a voltage drop between 1 and 2 in this?:

I assume not, I also assume that touching the entire wire both hands is not dangerous, right?

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14. ### shteii01 AAC Fanatic!

Feb 19, 2010
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There is. Your voltage source is in parallel with the two branches. Each branch will have a voltage across it that is equal to the voltage of the voltage source.

15. ### Amaterasuu Thread Starter New Member

Oct 10, 2010
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Sorry for more question, but imo, the follow 2 circuits are analogous to each other:

I have added a resistor in the first picture to create the analogy.
Why is there a parallel voltage drop in the first one, but none in the second one? (The ground being the resistor)

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16. ### JasonL Active Member

Jul 1, 2011
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Hi sorry for bringing this old thread up. But I had the same questions within this thread and many of them were answered, thanks. Instead of making a new thread on the same subject, I would like to bump this because I am too curious about why there wouldn't be a parallel voltage drop in the 2nd picture.

I'm just guessing but I think that there actually is a voltage because the diagram looks like it is a parallel circuit. But because Earth has such a high resistance, the current across the man would be small and not lethal because the current will take the path of least resistance away from the man. So the man is not shocked. Am I wrong?

17. ### MrChips Moderator

Oct 2, 2009
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This discussion is off on the wrong path. You cannot model the earth as a high resistance. The original telegraph and telephone systems used only one wire.
What you are ignoring is that the earth has huge capacitance. Hence regardless of how many amps you pump into it, it still remains at zero potential (relatively speaking).

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18. ### Adjuster Well-Known Member

Dec 26, 2010
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To say that the current "chooses" the low resistance path is a simplification. What actually happens is that, since the two paths are in parallel, the current shares between them according to the ratio of their conductances (that is, the inverse of their resistances).

For this reason it is important that the earth connection is high conductance (low resistance). If the earth connection has too much resistance, an unacceptable proportion of the current may flow in the other path. You may prefer to think of this in terms of the voltage dropped along the earth line being too large, but the effect is the same: a shock hazard.

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19. ### Adjuster Well-Known Member

Dec 26, 2010
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You may debate the nature of earth resistance (it certainly is not purely ohmic), but any practical earth connection will have some, and quite enough to cause important practical effects, including lethal ones.

The ground has distributed resistance behaviour, and can be particularly dangerous to quadrupeds like cows and horses which can be killed by events such as nearby lightning strikes and underground cable faults.

The Earth may be modelled as an equipotential sphere at a (very) macro level, but this is not the whole story.

20. ### MrChips Moderator

Oct 2, 2009
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True. But I believe what we are discussing here is what are are potential hazards to personnel when we use improperly grounded equipment. This discussion is ongoing in three other threads on floating power supplies, use of isolation transformers and why oscilloscopes are grounded. These are all very important issues that everyone on this forum need to know about.

(On another thread an OP was measuring EEG that requires placing electrodes on the scalp. I asked the question how were the amplifiers being powered. He never answered the question.)

Last edited: Jul 3, 2011