Circuit completion : is it necessary?

[westom]

If you have 6V/m then you would measure 6V half way across. This assumes the instrument is ideal.

You have also assumed transformers are ideal. Parasitic components inside transformers are why a primary and secondary are never fully isolated. A floating secondary could be anywhere between 0 and 33,000 volts. Another reason why the secondary is earthed.

 

[sophiecentaur]

This might be a dumb question, but if there is 60 volts between you and the ground, but the ground doesn't complete a circuit with the source, how do you know how much current will flow through you in that brief instant you touch the wire and the ground? Is it enough to be dangerous?

Not a dumb question at all. What counts here is the Capacity involved, which wouldn't be much for just a human body. If you were joined to a long transmission line or a large steel building then there may be enough charge to harm you when it's passed through you.
If you touch a Van der Graaf ball, you just feel a small prickle - a man on a high voltage cable, doing live maintenance, needs to wear a Faraday suit to make sure that the charge doesn't flow through him when he connects up, even though he's on an insulated platform. Similar volts but different capacities - so different charges.
The volts in your example are very low so the charge would be miniscule and harmless. But you can die from a 60V shock under the right conditions so, perhaps with a supercapacitor but no complete DC circuit. . . . . .

 

[sophiecentaur]

You have also assumed transformers are ideal. Parasitic components inside transformers are why a primary and secondary are never fully isolated. A floating secondary could be anywhere between 0 and 33,000 volts. Another reason why the secondary is earthed.

You are just asserting something on the grounds of "parasitic components". If you want parasitics inside then you have to have parasitics outside too. Leakage between windings and Earth are involved on both sides - particularly if a screen is involved. Until you can quote some actual figures relating to a suggested mechanism then your 33kV is just a made up number. Any statements you make on this sort of forum need, really, to be backed up with a good reference or some convincing sums if you want to be believed. I am always open to being convinced - honest I am.

 

[Evil Bunny]

Where, for example, does it discuss wire impedance?

It doesn't. Is it important to the grounding discussion? I would think the lower impedance the better... that's the whole idea behind a conductor in the first place, right?

Where does it discuss "why" a grounding wire must be shorter rather than thicker?

It doesn't. Can you explain the relevance?

Also not discussed is "why" both longitudinal and transverse currents are relevant.

Right again... Not discussed. Can you expand on this for us?

You brought them up, so I'm assuming they must be important points... Would love to hear more.

 

[RedX]

What counts here is the Capacity involved, which wouldn't be much for just a human body. If you were joined to a long transmission line or a large steel building then there may be enough charge to harm you when it's passed through you.
If you touch a Van der Graaf ball, you just feel a small prickle - a man on a high voltage cable, doing live maintenance, needs to wear a Faraday suit to make sure that the charge doesn't flow through him when he connects up, even though he's on an insulated platform. Similar volts but different capacities - so different charges.
The volts in your example are very low so the charge would be miniscule and harmless. But you can die from a 60V shock under the right conditions so, perhaps with a supercapacitor but no complete DC circuit. . . . . .

The formula for charging a capacitor is:

I(t)=\frac{\epsilon}{R} \left(e^{-\frac{t}{RC}} \right)

120 V can be dangerous to the human body if you touch the hot and the neutral at the same time (assume the neutral is not grounded for simplicity). But now say you only touch just the hot wire. Then you form one plate of a capacitor, and the other plate would be the outer surface of the neutral wire itself. I assume that the capacitance between you and the outer surface of the neutral wire is really small. So using the formula for the current, the emf (120 V) divided by your resistance is enough to kill you, but the only thing that could possibly save you is the exponential decay of the current. A low capacitance would drop the current really fast, but the resistance R of the human body is high and this counteracts that (the time constant is RC, so high resistance can make up for low capacity). Moreover, for small time intervals, the exponential doesn't drop off by much (at time=0, the current flowing through your body is the full current, which is enough to kill you).

My question is:

1) Am I correct to model a human who touches just the hot wire as a resistor and capacitor in series with the power source, with resistance being the resistance value of a human, and the capacitance being between the human and the outer surface of the neutral wire? So would you assume a human is a spherical capacitor of radius equal to the waist size, and the other "plate" of the capacitor is a long cylindrical tube (the outer surface of the neutral wire), and the dialetric between them is air?

2) Using the RC formula, is the low capacitance C enough to overcome the high resistance of the human R in the exponential, and even if it is capable of doing so, then at least for some small time interval doesn't the human experience the full, capable-of-killing current: emf/(human resistance) ?

My understanding of the Van der Graff ball is that the ball will run out of charge so that it can't maintain it's voltage, and also by having a fixed amount of charge the current is limited. But in the example I have above, the emf will always maintain its voltage of 120 V. So I'm not sure the Van der Graff ball applies.

 

[sophiecentaur]

It is the charge that does the damage (ionising vital bits inside us). A brief high current is not particularly harmful. The Van DDR Graaf is an example of a harmless jolt because the Capacity is small. Leyden Jars on a Whimshurst are far more dangerous because their capacity will hold a more "dangerous" quantity of charge. Not used in Schools any more!
Btw it's not the capacity between body and neutral wire that counts. It't total capacity to ground, which is more. Also, the Capacity does not "overcome" the resistance; it just prolongs the time for which current flows.

 

[Evil Bunny]

1) Am I correct to model a human who touches just the hot wire as a resistor and capacitor in series with the power source...

I don't think so. I think the capacitance here would be so small that we can essentially consider it an open circuit. You would not be capacitively coupled as you would with a "real" capacitor circuit that you describe above.

I don't have any fancy equations or engineering models to back this up... only personal experience. I have on several different occasions, been in contact with a 120V "hot" wire without being in contact with the return and have not felt so much as a tingle.

Usual disclaimer... 120V can kill you so do not play games with it!

 

[sophiecentaur]

I don't think so. I think the capacitance here would be so small that we can essentially consider it an open circuit. You would not be capacitively coupled as you would with a "real" capacitor circuit that you describe above.

I don't have any fancy equations or engineering models to back this up... only personal experience. I have on several different occasions, been in contact with a 120V "hot" wire without being in contact with the return and have not felt so much as a tingle.

Usual disclaimer... 120V can kill you so do not play games with it!

Equations aren't "fancy" my boy. They are just a way of stating things in a way that is least likely to be misinterpreted. If you want an arm waving forum, then this is not one of them.

You are right that the capacitance between your body and the Earth is not great but, under many circumstances, you PLUS a lot of other conducting stuff may be joined together and then the capacity is more significant. The same principle applies whatever Volts and Capacity are involved - it's just that the numbers make a difference to your experience.
Q = CV, the initial current is V/R and the energy stored is CV²/2 whatever the circumatance. Those formulaearen't "fancy" are they?

 

[Evil Bunny]

Seems to me that if you want to figure out the "real" capacitance between yourself and a conductor, especially with "a lot of other conducting stuff" around... the equations involved would probably be pretty fancy

I'd like to see someone figure out the "actual" capacitance in such a situation.

Maybe it's simple... But to a lowly arm-waver like me, it would look pretty fancy!

 

[Dmytry]

Calculating the AC current into capacitor, directly from basics.
lets say capacitance is 100pF = 1E-10 F, let's say amplitude is 120*sqrt(2) volts (the 120v is rms), then the voltage is
v=120*sqrt(2)*sin(t*60*2*pi)
and it's time derivative is
dv/dt=120*sqrt(2)*60*2*pi*cos(t*60*2*pi)
Current is charge per time. Charge on capacitor is voltage*capacitance.
So the current is
I=c*dv/dt
or
I=120*sqrt(2)*60*2*pi*1E-10*cos(t*60*2*pi) = about 6.4 micro amper peak.

Or you can use the formula explained here:
http://www.allaboutcircuits.com/vol_2/chpt_4/2.html
reactance = 1/(2*pi*f*c)
and to find the AC current for sine wave you take AC voltage and divide by reactance.
You can even calculate things for arbitrary circuits consisting of capacitors, resistors, and inductances, the same way you would for resistor network, but using complex numbers for 'resistances'.

Safety notice: the current in mains is not quite sine wave, and there may be high frequency components leaked by some circuitry (such as cheap PC power supply, or perhaps circuit trip and all the connected inductances shooting the voltage up). Meaning it is not safe to touch mains even if you are well insulated.

Note for guessing capacitance: the capacitance of 1 m^2 plates at distance of 10cm is about 88 pF.

 

[Evil Bunny]

What about the calculation for capacitance between a human and a wire... and possibly some other metal stuff that might be around?

 

[RedX]

I understand there is more capacitance between the ground and a person just touching the hot wire (and insulated from the ground) than there is between the person and the neutral wire, but I assumed for simplicity that the neutral wire was not grounded. So if you wanted to connect the person, the power source, and the ground in one big circuit, you'd have to include the resistance between the neutral wire and the ground R_{n-g}, which decreases the overall current:

<br /> I(t)=\frac{\epsilon}{R_{\mbox{human}}+R_{\mbox{n-g}}} \left(e^{-\frac{t}{R_{\mbox{total}}C}} \right) <br />

since the first factor would be decreased with the increased resistance R between the neutral wire and the ground.

In reality, the neutral wire is grounded, so there would be no resistance between the neutral wire and the ground, so you would consider the capacity between the Earth and the person touching the hot wire rather than the person and the neutral wire.

In other words, if there is no grounding, the least resistance is offered if you consider the capacitor as being between the human and the neutral wire.

I think what confused me is I've seen charts that state the amount of current it takes to kill you, but it's really charge and not current, so you don't have to worry about being a capacitor if you touch just one wire.

You might need to worry when the voltage is really high before it has been stepped down, since the voltage is 1000 times more. In fact someone just did a calculation in post #160 (using AC) and got 6.4 microamps. 1000 times more voltage would be 6.4 milliamps which looks to be on the threshold of dangerousness:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/shock.html

So I guess it's not dangerous to just stand on the ground beneath a power line (on a real power line the neutral is grounded), since the capacitor would be between you and the cylindrical hot wire. But if you touch the hot wire on an insulated platform, it would be between you and the earth.

Anyways, the formula I used for current is actually only good for DC power, so I guess it doesn't apply for AC. Although if the current dies off faster than 1/60 seconds, then maybe it would be a good approximation.