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>> When people ask that question the answer is generally given in
>> simplistic terms.
>
> The most common one being "the voltage is high, but the current is low",
> which don't make sense.
Presumably they mean that the maximum current is low. If you neutralise
the difference in charge faster than the system can replace it, the
voltage *changes*. So U=RI is not violated, yet the system *does* have a
maximum currentl.
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Warp <war### [at] tag povray org> wrote:
>
> The most common one being "the voltage is high, but the current is low",
> which don't make sense.
>
> --
> - Warp
It does make sense if you know that a voltage source always has a series
resistance. The series resistance of the high voltage generator is in the kilo
ohms, whereas the series resistance of a mains supply is in the single digit
ohms or below.
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Invisible <voi### [at] dev null> wrote:
> >> When people ask that question the answer is generally given in
> >> simplistic terms.
> >
> > The most common one being "the voltage is high, but the current is low",
> > which don't make sense.
> Presumably they mean that the maximum current is low. If you neutralise
> the difference in charge faster than the system can replace it, the
> voltage *changes*. So U=RI is not violated, yet the system *does* have a
> maximum currentl.
This is exactly what I'm talking about. It's impossible to get a clear,
straight, understandable answer from anybody.
I = U/R has no "maximum". If you increase 'U', 'I' will increase likewise,
when 'R' stays the same. The relationship is linear. There's no "maximum".
There's no point after which when you keep increasing 'U' but 'I' doesn't
increase because it has reached some magical "maximum".
If 'U' is 1 volt, then 'I' will be 1V/R. If 'U' is one billion volts, then
'I' will be 1 billion volts / R. There is not magical "maximum".
--
- Warp
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Invisible <voi### [at] dev null> wrote:
> It's current that kills you, not potential difference or anything else.
No, it's not the current that kills you. It's the power transfer.
You yourself said it: Even if the current is tens of kilovolts, it may
still not kill you.
--
- Warp
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Aydan <hes### [at] hendrik-sachse net> wrote:
> > The most common one being "the voltage is high, but the current is low",
> > which don't make sense.
> It does make sense if you know that a voltage source always has a series
> resistance. The series resistance of the high voltage generator is in the kilo
> ohms, whereas the series resistance of a mains supply is in the single digit
> ohms or below.
But the voltage difference is between the wire and the ground, and
there's nothing else between them than the body of the person touching
the wire.
The current *is* very high. But only for a really small amount of time.
It's not the current that kills you (as the saying goes), but the power
transfer (the amount of current over time). I think the unit is Watt.
--
- Warp
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> I = U/R has no "maximum".
As an abstract equation, no.
In a real system, if you decrease R far enough, eventually U starts to
decrease too, in each a way that I remains fixed. Thus I has a "maximum
value".
Put simply, there is a limit to how fast the system can replace the
charge you're neutralising. It's no different than the fact that a water
pump has a maximum flow rate, no matter how slight the resistence of the
water pipes becomes.
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Invisible <voi### [at] dev null> wrote:
> Put simply, there is a limit to how fast the system can replace the
> charge you're neutralising.
You are not talking about U=RI here (not solely). You are talking about
how the voltage/current changes over time. There's a different unit for
that. And that's the key for the answer.
--
- Warp
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Warp <war### [at] tag povray org> wrote:
> Aydan <hes### [at] hendrik-sachse net> wrote:
> > > The most common one being "the voltage is high, but the current is low",
> > > which don't make sense.
>
> > It does make sense if you know that a voltage source always has a series
> > resistance. The series resistance of the high voltage generator is in the kilo
> > ohms, whereas the series resistance of a mains supply is in the single digit
> > ohms or below.
>
> But the voltage difference is between the wire and the ground, and
> there's nothing else between them than the body of the person touching
> the wire.
>
> The current *is* very high. But only for a really small amount of time.
> It's not the current that kills you (as the saying goes), but the power
> transfer (the amount of current over time). I think the unit is Watt.
>
> --
> - Warp
The thing with the series resistance still holds true. The series resistance is
inside the device. If you measure the voltage with a voltmeter, the voltmeter's
resistance is usually several gigaohms, so there will not be a noticeable
voltage drop.
In the special case of the cattle fence there's the question of how the high
voltage is generated. Usually it's done with sort of a transformer. The ratio of
the transformer then is something like 1 : 5000 and the primary is driven by a
12V battery. There's a timer in the device that switches the primary on and off.
The current through the primary will increase "slowly" (kinda) because of the
inductance of the coil. But when you switch the supply off, the current wants to
keep going (magnetic energy stored in the coil) and be discharged over a sparc
gap. The rapid drop in current in the primary will induce a much higher current
in the secondary. The energy is very limited as well as the source impedance
(read resistance) because the secondary coil will have to have a much thinner
and longer wire (higher resistance) because of the many more windings it needs.
Disassemble an ignition coil of a car or motorcycle if you want to know how this
looks like.
As for simplicity of describing something: For some things you just need a basic
understanding of the matter discussed.
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Invisible <voi### [at] dev null> wrote:
> > I = U/R has no "maximum".
>
> As an abstract equation, no.
>
> In a real system, if you decrease R far enough, eventually U starts to
> decrease too, in each a way that I remains fixed. Thus I has a "maximum
> value".
>
> Put simply, there is a limit to how fast the system can replace the
> charge you're neutralising. It's no different than the fact that a water
> pump has a maximum flow rate, no matter how slight the resistence of the
> water pipes becomes.
Your explanation is abstract as well. ;o)
The reason for the drop in voltage is a series resistance inside the supply.
This series resistance is not necessarily constant. It might increase with
current drawn.
There's no such thing as an ideal voltage source (read unlimited current supply)
in reality.
If you short a battery for exampe it will source a maximum current. Measure that
and measure the voltage without drawing current and you get the source
impedance. If you connect an load with this resistance to the battery, you'll
get half the voltage out of it.
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> However, the relevant thing is what happens immediately after the
> connection is made. What happens is that the voltage of the wire decreases
> very rapidly until it approaches the voltage of the ground. Thus the
> voltage difference between the two will be 40 kilovolts only for a really,
> really small amount of time.
Exactly - and if you were to plot the resistance (between you and the wire)
over time as you made the connection, you'll probably find the
transistor-based current limiting circuit can reduce the voltage faster than
you can touch the wire...
> Thus I think I wouldn't be really wrong when I say "it's not the current
> that kills you, it's the electrical power transfer (over time)".
All the material I've read states a level of current that is dangerous, not
power. The "power" would vary depending on how good your connection to
ground was.
As for very short pulses of dangerous current, I don't know what the outcome
would be, and don't want to try it!
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