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What happened to Ohm's law?

  1. Oct 24, 2012 #1
    Earlier I asked for help about a high voltage power supply, and I was told not to worry too much about 20k volts because they have a very low current

    I was left thinking about that

    say whatever circuit that ends in an inductor and generates that voltage of 20k volts.
    and I was told that the current runs at around 200 mA

    how does this work?

    I mean I was taught that V=iR

    and AFAIK the resistance of a wire (whatever material) is dictated by its material properties, its density and whatnot, so even tho its not a "perfect wire" it exhibits certain resistance R that should be (should be?) fixed, right?

    so If R is fixed, how can V and i be fixed too?

    I found a table that says that copper wire has 0.001588 Ohms per foot

    lets say I have a 10 ft wire running those 20kV, so that means that the current
    i= V/R
    should be 20000 V / (.001588*10) Ohms= 1259445.84383 Amps

    how come the current is 200 mA ?

    is there some alteration to Ohms law or whats going on here?
  2. jcsd
  3. Oct 24, 2012 #2
    Well, first. If someone tells you not to worry about high voltage, cross them off your friends list.

    Second a 20kV supply with a 200mA capability would be lethal in the extreme. I hope your figures are wrong because if you are playing with something like that and asking these questions... dead man walking!

    But to answer your last question (no pun intended). The 200mA is the capability of the supply - in other words it supplies what it can up to that limit then the voltage drops to whatever power output it can sustain. (There are internal resistances)
  4. Oct 24, 2012 #3

    Attached Files:

  5. Oct 24, 2012 #4
    by "not worry about 20k volts" I meant its not lethal, not that it would be fun, AFAIK police tasers run over 30kVolts. . .and they are non-lethal. . . current is the cause of death AFAIK

    can you elaborate on "capability?"

    take this video for example:

    go over to 1:35, when the lifter takes off
    the guy reads from his power supply:

    "We are operating at 23 kVolts and half a miliamp"

    how do you explain that?
    so the PSU is lying and its only sending half a miliamp? what about the voltage?
    Is Ohm turning in his grave?
    Last edited by a moderator: Sep 25, 2014
  6. Oct 24, 2012 #5


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    -The source supplies a Voltage.
    -Based on that voltage, the load demands an Amperage.
    -The source supplies that Amperage or dies trying.
  7. Oct 24, 2012 #6


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    The source can only provide a maximum amount of electron flow (current). This means, if you short circuit it (e.g. by putting in a basically resistance-free wire between it's positive and negative nodes), the source can no longer supply any voltage.

    The source cannot maintain a 20,000V difference across a conductor, in that case, the voltage across the conductor is much less than 20,000V.
  8. Oct 24, 2012 #7
    Yes indeed, although if you did receive a 20kV jolt at 1 microamp you'd know about it.

    200 milliamps is enough to kill the All Blacks rugby team.

    For your information 200 mA at 20kV is

    200x10-3 x 20x103 = 4kilowatts.

    Oh, and someone should mention that high voltages in inductive circuits arise from transient effects. The DC version of Ohm's law does not apply.
  9. Oct 24, 2012 #8
    from Wiki

    in that diagram the resistor in the "load" would be the wire itself (since AFAIK diagrams assume perfect wires with no resistance)

    anyways, would that demand of amperage be based on Ohms law? (V=iR) or how is the load demanding amperage from the source?
    and if so, wouldnt that be the calculation I did earlier (i=V/R) which is not close to that half a miliamp

    I like the last step btw
    and thnkz for the replies

    that seems interesting, do you have any links where I could read about that?
  10. Oct 24, 2012 #9


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    So, as you can see in that diagram, if I replaced R_L with 0, then the maximum current is V_s/R_s. If this is a large number, it's possible the resistor R_s just blows and you've just broken the source. Hopefully, the designer of the load put in some sort of fail safe systems to prevent that from happening (by e.g. adding in a fuse that shuts off the system if some amperage is reached).
  11. Oct 24, 2012 #10
    I'm getting some mixed messages here. (Most likely my interpretation is at fault). Sometimes I hear people say "When it comes to injury, just consider the current." Here, it seems you suggest that power (P=IV) is what matters.

    My intuition is that it should indeed be current that injures you, since the only dynamic component to the scenario is the group of moving charges' speed and size, which is quantified by the current.
    Last edited: Oct 24, 2012
  12. Oct 24, 2012 #11
    GFCI circuits in the US trip at 5 ma at 120 volts....That's considered a non lethal
    power for a moment....The powers above are potentially LETHAL, no question...
  13. Oct 24, 2012 #12
    Spasm or death by electric current is an electrochemical effect, sponsored by current.

    It take somewhere between 5 and 50 milliamps to kill a human. The resistance depends upon the available path through the body and the human concerned this determines the current from an available voltage source.

    Absorbing the power generated by four electric fire bars on full blast is a thermal effect. You could also die of burns in that scenario.

    Gee Naty1 I see we have the same base figure of 5 milliamps.

    go well
  14. Oct 24, 2012 #13


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    You calculated it fine earlier. The nameplate voltage is what you apply to V=IR. If the power supply can supply the demanded amperage, everything is fine. If not, something fails. Sometimes circuits fail.
    Yep. So you've just calculated that the circuit will fail. The voltage will sag, the amperage will exceed the capacity of the power supply -- probably all that happens is the fuse blows, but there is always the possibility of damaging something.
  15. Oct 25, 2012 #14
    Death by electricity isn't a simple matter of 'X amount will kill you'.
    Any form of bodily injury might or might not kill you. You can survive a high powered rifle bullet or be killed by a sewing needle.

    A very rough guide is the amount of energy dissipated inside the body - which means power (watts) and time. To calculate the power you need voltage and current, which depends on resistance. The human body has a typical resistance in the order of a hundred thousand ohms.

    Any voltage over a hundred volts can be dangerous, depending on the circuit's ability to push power. Most dangerous of all is the seemingly innocent capacitor that can store huge amounts of energy at high voltage and supply it all in one blast. It doesn't look like anything special but it can be a real killer.
  16. Oct 25, 2012 #15


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    What do you mean by "in one blast"? Capacitors discharge at a set rate T=RC, if R is a couple hundred thousand ohms, then any large capacitor (Farad or higher range), capable of storing a lot of charge, would take quite a while to discharge into the human body. Small capacitors can't store nearly as much charge, so the total current it can provide is not as much.

    Is this somehow better than what most power sources are capable of doing?
  17. Oct 26, 2012 #16


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    There is a fair bit of literature on the web about this. Usual cause of death for low current electrocution is fibrillation in the heart muscle. AC is far more deadly in this regard than is DC. There is a paradoxical effect where a small current through the heart muscle can cause fatal fibrillation where a larger (but brief) current will only stop the heart momentarily. Hence electric defibrillators.
    Last edited: Oct 26, 2012
  18. Oct 26, 2012 #17
    A shock between two fingers on one hand can have very different effects relative
    to a shock between left and right hands......in the latter, current passes right past and perhaps through the heart.
  19. Oct 26, 2012 #18
    russ_watters, I'm sorry but I dont understand. . .
    are you then saying that in that video the HVPSU is actually failing? if the machine is reading 23kV at .5 mA . . .if its reading it while supplying it then it is obviously not failing. .

    and as mentioned, the current is running through a wire of set resistance (small gauge copper wire) . . .so how can there be 3 fixed "variables" , if this is the case then its not an equation, its not V= iR but V (porportional) iR but not necessarily equal ?

    again my question

    if a power supply's output is a FIXED voltage and a FIXED current. . .what happened to Ohms law once those values enter a circuit ?
  20. Oct 26, 2012 #19
    I'm sorry but youtube videos are of variable quality and this was not a reliable one.

    Looking at the cables carrying the output of the psu, including the open knobbly temporary connections shown about 3/4 of the way through do not incline me to believe in anything like 20kV.

    I saw no corona discharge.

    20kV is of the order of the voltage at the back of a CRT or generated by an auto ignition system.
    Have you seen the cables used here?

    Then the narrator claimed that the electric current created a 'gravitational potential well 'and talked seriously about 'electrogravity'.

    Let us discuss something more believable please.
  21. Oct 27, 2012 #20
    well Im not discussing abaout electrogravity. . .I know why that thing is lifting I already read tons about ionocrafts. . .what Im curious about is the circuitry behind them
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