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Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Dr. Slack wrote ...
7200V with an amp behind it will kill you.
1MV with an amp behind it will kill you, and destroy the evidence.
This is a rather non-scientific answer, intended to produce a smile. See also Chuck Norris jokes.
Speculating what sort of damage would occur when an unrealisable power supply with some of its parameters accurately specified will do when connected to a broadly described load with some assumed characteristics is, of course, impossible.
Perhaps the most practical (least impractical) supply that meets constant current with limited output voltage is a short-circuited inductor with 1A flowing through it. This also meets the 'voltage applied after contact' if you grab the short circuit wire with both hands and break it. It would store only so much energy, so doesn't meet the indefinite energy output of an ideal constant current source, as modelled in SPICE for instance.
Perhaps the main things are a) 1A is more than enough to cause death through fibrillation b) 7200V is more than enough voltage to jump through the thickness of dry clothes, and to drive more than a lethal current through a typical body c) 7200V will only sustain a very short arc, so once our hero has collapsed, the arc stops (assuming he was standing) d) 1MV will sustain an arc comparable with the length of a human, so the arc continues
Model the load accurately in SPICE, including changes in resistance due to temperature, carbonisation, posture, and we could do a simulation with fewer assumptions.
Registered Member #57401
Joined: Sat Sept 19 2015, 08:06PM
Location: Huntsville, AL
Posts: 10
ScottH wrote ...
That makes it a lot clearer. So the excess voltage/power will just "cruise" through you at 1MV after the body absorbs set amount of energy? Due to the body's resistance, the energy dissipation at set amps will be the same, regardless?
I hope I worded this right.
You still haven't quite got it - there will be no excess voltage or power - a constant current source will only produce the voltage required by the load. For a purely resistive load, this means that either source will provide exactly the same voltage difference when presented with the same load.
Now, back to the specifics you provided - assume a 50kOhm load. Pushing 1 amp through 50k of resistance requires a pressure of 50kV, so your 7200V supply fails to maintain constant current, and the current drops to 144mA. Your 1MV capable supply provides the required 50kV and maintains 1A of current for a power draw of 50kW - the assumptions you started with are not valid in this case since they would not "both easily push 1 amp of current through the body"
However, as others have stated, in the case of a human body things get more complex than a constant resistance load. Think of the body as being a string of resistors in series, with say a 24.5k resistor for the skin, a 1k resistor for internal tissues, and another 24.5k resistor for the skin at the second contact point. For a certain current through this string, almost all of the voltage change will occur across the high resistance of the skin (3528V drop at each contact point, and 144v across the "wet" tissues for your 7200V source) This means that most of the power must be dissipated by the skin - this equated to over 500W at each contact point for the 7200V case. Unfortunately the skin on your finger tips doesn't retain it's properties for very long when heated with 500W. Let's say as the skin chars it's resistance drops to become negligible - Now your 7200V capable source can easily provide the 1000V necessary to push 1A through the remaining 1kOhm resistance. The same thing happens with your 1MV capable source with the end result being that both supplies drop to 1000V to maintain 1A though a 1kOhm body, which dissipates the 1000W in the internal tissues depending on their respective resistance. This resistive heating would cause irreversible internal damage at even relatively short durations even if the heart's electrical system were not overwhelmed.
Again, I think you are being confused by your intuition, which figures a 1 MV capable source must surely do more damage. And you are generally correct in that a real life supply will not immediately drop voltage to limit current. When you are dealing with transient conditions such as resistance changing over time and voltage and current responding though some feedback control system, you have a much harder time finding the "ideal" source that your hypothetical problem requires.
I think the real answer to your question goes back to Ohm's law - for a fixed resistance the voltage is directly related to the current by a CONSTANT. Providing a power supply which is CAPABLE of higher voltage does nothing to break this law. The voltage it provides is identical to that provided by a less capable supply, assuming that both have at least enough capacity for the specified current. Try setting your benchtop power supply to constant current mode and hook it up to a small lamp. Crank up the voltage and once the current hits the specified level, you will find that that voltage holds constant and the bulb stays at the same brightness even if you continue to crank up the voltage knob. You cannot choose both the voltage and the current - you choose one, and the other adjusts to meet the demand of the load.
Registered Member #61373
Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87
KrowBar wrote ...
ScottH wrote ...
That makes it a lot clearer. So the excess voltage/power will just "cruise" through you at 1MV after the body absorbs set amount of energy? Due to the body's resistance, the energy dissipation at set amps will be the same, regardless?
I hope I worded this right.
You still haven't quite got it - there will be no excess voltage or power - a constant current source will only produce the voltage required by the load. For a purely resistive load, this means that either source will provide exactly the same voltage difference when presented with the same load.
Now, back to the specifics you provided - assume a 50kOhm load. Pushing 1 amp through 50k of resistance requires a pressure of 50kV, so your 7200V supply fails to maintain constant current, and the current drops to 144mA. Your 1MV capable supply provides the required 50kV and maintains 1A of current for a power draw of 50kW - the assumptions you started with are not valid in this case since they would not "both easily push 1 amp of current through the body"
However, as others have stated, in the case of a human body things get more complex than a constant resistance load. Think of the body as being a string of resistors in series, with say a 24.5k resistor for the skin, a 1k resistor for internal tissues, and another 24.5k resistor for the skin at the second contact point. For a certain current through this string, almost all of the voltage change will occur across the high resistance of the skin (3528V drop at each contact point, and 144v across the "wet" tissues for your 7200V source) This means that most of the power must be dissipated by the skin - this equated to over 500W at each contact point for the 7200V case. Unfortunately the skin on your finger tips doesn't retain it's properties for very long when heated with 500W. Let's say as the skin chars it's resistance drops to become negligible - Now your 7200V capable source can easily provide the 1000V necessary to push 1A through the remaining 1kOhm resistance. The same thing happens with your 1MV capable source with the end result being that both supplies drop to 1000V to maintain 1A though a 1kOhm body, which dissipates the 1000W in the internal tissues depending on their respective resistance. This resistive heating would cause irreversible internal damage at even relatively short durations even if the heart's electrical system were not overwhelmed.
Again, I think you are being confused by your intuition, which figures a 1 MV capable source must surely do more damage. And you are generally correct in that a real life supply will not immediately drop voltage to limit current. When you are dealing with transient conditions such as resistance changing over time and voltage and current responding though some feedback control system, you have a much harder time finding the "ideal" source that your hypothetical problem requires.
I think the real answer to your question goes back to Ohm's law - for a fixed resistance the voltage is directly related to the current by a CONSTANT. Providing a power supply which is CAPABLE of higher voltage does nothing to break this law. The voltage it provides is identical to that provided by a less capable supply, assuming that both have at least enough capacity for the specified current. Try setting your benchtop power supply to constant current mode and hook it up to a small lamp. Crank up the voltage and once the current hits the specified level, you will find that that voltage holds constant and the bulb stays at the same brightness even if you continue to crank up the voltage knob. You cannot choose both the voltage and the current - you choose one, and the other adjusts to meet the demand of the load.
That makes sense.
I'm still baffled about this: I know this was very dumb and dangerous to try, but when I was 14 I made a HV supply with a basic automotive ignition coil and a 12v source. I held one HV wire with one hand, and a nail in the other.
With the nail in hand, I made a spark jump 1cm to the positive terminal on the coil. The spark was capable of jumping 1 inch between positive and negative, which is 25,000v. If my internal resistance is 1,000 Ohms, and the spark jumped from me 1cm, then there was 10,000v between me and the positive. Why would a 10kv spark jump when it only takes 5v to push 5ma through 1kohm?
I estimate 5ma of current. It felt like a continuous static shock. Not much juice.
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
Let me describe what is happening in your dangerous experiment over time: 1) The voltage between the nail and where the spark jumped to was the same voltage as the output of the ignition coil. 2) You brought you hand closer to where your sparked will jump to. 3) When your hand was 1 cm away, the air could not sustain the high voltage across it. The air broke down, current flowed and the resistance of that air gap went from megaohms to ohms, the current flowing caused the voltage from the ignition coil to drop as the ignition coil has its own internal resistance. 4) Now the voltage across the gap is quite low, much less than the voltage from the ignition coil. The rest of the voltage is across your body, and you are feeling its effects.
An ignition coil isn't a very good example because it is a pulsed device. The peak current was probably much higher than 5mA, but that current was only flowing 1% of the time, so it felt much less.
Registered Member #61373
Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87
hen918 wrote ...
Let me describe what is happening in your dangerous experiment over time: 1) The voltage between the nail and where the spark jumped to was the same voltage as the output of the ignition coil. 2) You brought you hand closer to where your sparked will jump to. 3) When your hand was 1 cm away, the air could not sustain the high voltage across it. The air broke down, current flowed and the resistance of that air gap went from megaohms to ohms, the current flowing caused the voltage from the ignition coil to drop as the ignition coil has its own internal resistance. 4) Now the voltage across the gap is quite low, much less than the voltage from the ignition coil. The rest of the voltage is across your body, and you are feeling its effects.
An ignition coil isn't a very good example because it is a pulsed device. The peak current was probably much higher than 5mA, but that current was only flowing 1% of the time, so it felt much less.
According to my calculations, 10A flowed through me for a tiny fraction of a second in short pulses in order to make 10kv go through me? Correct me if I'm wrong.
Registered Member #11591
Joined: Wed Mar 20 2013, 08:20PM
Location: UK
Posts: 556
ScottH wrote ...
hen918 wrote ...
Let me describe what is happening in your dangerous experiment over time: 1) The voltage between the nail and where the spark jumped to was the same voltage as the output of the ignition coil. 2) You brought you hand closer to where your sparked will jump to. 3) When your hand was 1 cm away, the air could not sustain the high voltage across it. The air broke down, current flowed and the resistance of that air gap went from megaohms to ohms, the current flowing caused the voltage from the ignition coil to drop as the ignition coil has its own internal resistance. 4) Now the voltage across the gap is quite low, much less than the voltage from the ignition coil. The rest of the voltage is across your body, and you are feeling its effects.
An ignition coil isn't a very good example because it is a pulsed device. The peak current was probably much higher than 5mA, but that current was only flowing 1% of the time, so it felt much less.
According to my calculations, 10A flowed through me for a tiny fraction of a second in short pulses in order to make 10kv go through me? Correct me if I'm wrong.
no, that is too high, mostly because the extra resistance (about 15 kOhms, due to the very thin and long the secondary is wound from) built into the transformer. This is in series with the circuit and will have to be added to your body's resistance to find the current.
Oh, and voltage doesn't go through you. Current goes through you, voltage goes across you. (Or anything else)
Registered Member #57401
Joined: Sat Sept 19 2015, 08:06PM
Location: Huntsville, AL
Posts: 10
ScottH wrote ...
According to my calculations, 10A flowed through me for a tiny fraction of a second in short pulses in order to make 10kv go through me? Correct me if I'm wrong.
Also, remember that your skin wasn't charred in this case, so it's a gross underestimation to assume your resistance was 1kOhm - it was probably closer to 100k (grab your multimeter to see).
Registered Member #61373
Joined: Sat Dec 17 2016, 01:45PM
Location: San Antonio, TX
Posts: 87
KrowBar wrote ...
ScottH wrote ...
According to my calculations, 10A flowed through me for a tiny fraction of a second in short pulses in order to make 10kv go through me? Correct me if I'm wrong.
Also, remember that your skin wasn't charred in this case, so it's a gross underestimation to assume your resistance was 1kOhm - it was probably closer to 100k (grab your multimeter to see).
Isn't skin resistance negligible when a high voltage spark is hitting you? Just like a stun gun spark will go through many sheets of paper without burning it? Remember, the skin has pores too, so it can bypass outer skin resistance.
In a system where you calculate resistance and Ohms Law, do you have to account for the resistance of the coil/circuit and wires too, or just the body?
Registered Member #60240
Joined: Mon May 16 2016, 07:01PM
Location:
Posts: 304
Hii ScottH
"Isn't skin resistance negligible when a high voltage spark is hitting you? Just like a stun gun spark will go through many sheets of paper without burning it? Remember, the skin has pores too, so it can bypass outer skin resistance.
In a system where you calculate resistance and Ohms Law, do you have to account for the resistance of the coil/circuit and wires too, or just the body?"
According to Kirchhoff`s law you have to consider the complete loop:
1. resistance of the coil/circuit 2. wires 3. body
AND ALSO the current or voltage source respectively.
Registered Member #3343
Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311
Some comments :
1- You wrote "The amperage of both outputs are the same, but the 1MV supply has 1MW of energy, vs 7,200W. " >> 1 MW is power, not energy.
2 - you wrote "Assuming that both supply's have a limited current output " If the current is limited to one ampere, does not make difference if the SOURCE voltage is 10kV or 100000000V. The voltage applied in your body always will be the one ampere times your body resistance.
3 - If the voltage is applied for very short time the effect could be just a discomfort. No burns, no shock.
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