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scott wrote:
>> And current will flow until that charge has been exhausted. There's
>> nothing mysterious about that. But claiming such a system has "no
>> voltage" and yet there's current flowing through it is just silly.
>
> As you know, a sine wave is only at zero for an infinitely short period
> of time, so actually the voltage is never exactly zero for any finite
> period of time. But at the instant it is zero, it is normal to have a
> non-zero current when dealing with capacitors and resistors.
Well OK, but you can make the voltage arbitrarily small while making the
current arbitrarily large. Just by asking nicely.
In my experience - and hell, what would I know about reality? - things
don't just move of their own accord. There must be a *force* driving
them. Otherwise we'd all by driving perpetual motion machines to work by
now...
> Someone posted this here a while ago:
>
> http://www.falstad.com/circuit/
>
> The default circuit that opens is exactly the point here. See how the
> green and yellow lines on the scopes at the bottom are out of phase?
> Green = voltage, yellow = current.
I've played with this thing before. Never been able to make much sense
out of it though - too much happening at once to really take it all
in... That guy does a lot of really cool physics stuff though. (I'm
currently addicted to the wave tank...)
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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scott wrote:
> I really don't see the problem. Do you understand what happens if you
> short a battery with a wire? You will get a huge current flowing even
> though the voltage across the wire is virtually zero.
How the hell do you figure that?
Surely the potential at one end of the wire is +4.5 V, in the middle
it's 0 V, and at the other end it's -4.5 V, and therefore the potential
*difference* across it is 9 V. How is that zero?
> If you put a
> super-conductor across the terminals of a battery, would you expect a
> current to still flow? After all, the voltage across the superconductor
> would be zero...
Where would you measure the difference? You need two points.
Anyway, presumably a superconducting magnet is bizare enough that Ohm's
law doesn't apply. (Hmm, I = V/R where R = 0. Yeah, that looks pretty
undefined to... oh, wait, you're that guy who things that division by
zero is defined, aren't you?)
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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> Well OK, but you can make the voltage arbitrarily small while making the
> current arbitrarily large. Just by asking nicely.
Of course, but then in another part of the cycle you will have to have a
large voltage with a small current.
> In my experience - and hell, what would I know about reality? - things
> don't just move of their own accord. There must be a *force* driving them.
A capacitor stores energy, this can generate force.
> Otherwise we'd all by driving perpetual motion machines to work by now...
I know you like maths :-)
http://en.wikipedia.org/wiki/Deriving_capacitor_impedance
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>> I really don't see the problem. Do you understand what happens if you
>> short a battery with a wire? You will get a huge current flowing even
>> though the voltage across the wire is virtually zero.
>
> How the hell do you figure that?
Because the resistance of the wire is way lower than the internal resistance
of the battery. DId you never notice that the more current you draw from a
battery the lower the voltage across its terminals? If you *short* a
battery, the voltage is pretty much zero.
>> If you put a super-conductor across the terminals of a battery, would you
>> expect a current to still flow? After all, the voltage across the
>> superconductor would be zero...
>
> Where would you measure the difference? You need two points.
The two terminals of the battery. Connect a super-conductor between the two
terminals of a 9V battery and put your volt-meter across the battery
terminals. It *will* be zero volts.
> Anyway, presumably a superconducting magnet is bizare enough that Ohm's
> law doesn't apply. (Hmm, I = V/R where R = 0. Yeah, that looks pretty
> undefined to... oh, wait, you're that guy who things that division by zero
> is defined, aren't you?)
You don't need to do that, batteries have an internal resistance, Ri, so the
current flowing will simply be Vbattery / Rinternal. With zero volts across
the super-conductor, but a current flowing.
Ohms law works perfectly well, the fact that R is zero simply means you can
have whatever current flowing you want with no potential difference. Which
is exactly how superconductors work.
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On Wed, 02 Apr 2008 10:45:08 +0100, Invisible <voi### [at] dev null> wrote:
>Right. So the potential difference between the terminals of a battery is
>9 V, unless there happens to be a capacitor connected to them, in which
>case the potential difference is magically 0 V despite the fact that a
>vast current is being generated?
You are simplifying it to the extreme and using rigorous logic and
that does not work. A battery can be considered as a current source
with an internal resistance. N.B. we are talking dc here and not ac.
When you put a high impedance voltmeter across an open battery you
will get a reading of about 9.5 volts. This is the open circuit PD.
When you put a resistor across the battery it forms a resistor -
resistor circuit. When you put a cap across the terminals it forms a
R-C circuit. You would not normally do this as you should have a
limiting resistor in series with the capacitor. (And in real life a
shunt resistor across the capacitor to slowly discharge it when there
is no power applied) Also remember Michael was talking about
instantaneous voltages which are different from steady state ones
which are different again from ac.
Or are you just having fun winding us up? :)
--
Regards
Stephen
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scott wrote:
>>> I really don't see the problem. Do you understand what happens if
>>> you short a battery with a wire? You will get a huge current flowing
>>> even though the voltage across the wire is virtually zero.
>>
>> How the hell do you figure that?
>
> Because the resistance of the wire is way lower than the internal
> resistance of the battery. DId you never notice that the more current
> you draw from a battery the lower the voltage across its terminals? If
> you *short* a battery, the voltage is pretty much zero.
I thought the voltage of a battery only decreases if you try to draw
current faster than the chemical processes inside the battery can
restore it? (I.e., it's a flaw parculiar to chemical batteries.)
>> Anyway, presumably a superconducting magnet is bizare enough that
>> Ohm's law doesn't apply. (Hmm, I = V/R where R = 0. Yeah, that looks
>> pretty undefined to... oh, wait, you're that guy who things that
>> division by zero is defined, aren't you?)
>
> You don't need to do that, batteries have an internal resistance, Ri, so
> the current flowing will simply be Vbattery / Rinternal. With zero
> volts across the super-conductor, but a current flowing.
>
> Ohms law works perfectly well, the fact that R is zero simply means you
> can have whatever current flowing you want with no potential
> difference. Which is exactly how superconductors work.
Heh. Superconductors truly *are* weird...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Stephen wrote:
> You are simplifying it to the extreme and using rigorous logic and
> that does not work.
> Or are you just having fun winding us up? :)
Well, let me put it this way.
As I understand it, moving electrons round a wire is like moving water
round a pipe. You need a *force* to push it. However, you confidently
claim that you can make electricity move with no force at all. If
somebody claimed that they could push water round a network of pipes
just by writing an equation, they'd be laughed at. And yet you claim the
exact same thing is possible with electricity.
Personally, I don't believe it. I believe that electrons won't go
anywhere unless you *drive* them there. And that the amount they move
must be directly proportional to the amount of force driving them and
inverstly proportional to the force resisting them. But you
[collectively] claim that actually the amount of movement is unrelated
to the force applied or the resistance to be overcome. If that really is
true then as far as I can see, all of electronics is pure black magic
and every electronic circuit I've ever built shouldn't have worked...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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> I thought the voltage of a battery only decreases if you try to draw
> current faster than the chemical processes inside the battery can restore
> it? (I.e., it's a flaw parculiar to chemical batteries.)
No, it decreases the instant start to draw any current. Maybe trying to
teach you about complex AC circuit theory and power electronics was a bad
starting point :-)
When you connect a resistance, R, to a battery, the circuit you have made is
identical to a true, fixed voltage source (say 12V), with two resistances
connected in series. One of the resistances is R, the one you connected,
the other, called the internal resistance, Ri, is representing the resistive
parts of the battery that you have no control over.
You can analyse this circuit easily. You have a voltage, V of say 12 V, and
a combined resistance of R+Ri. So the current is found by Ohm's Law, and is
equal to 12 / (R + Ri).
Now that you know the current in the circuit, you can work out the voltage
across each resistor, again using Ohm's law.
For the resistor you just connected, R, the voltage across it will be given
by I * R. You know I because we just calculated it, so the voltage is 12*R
/ (R + Ri).
If Ri is very small compared to R, then this is approximately just 12 volts.
But when R gets small enough the voltage can no longer be approximated by 12
volts. And when R gets very much smaller than Ri, the voltage essentially
becomes zero.
If the voltage R *is* zero, ie you have connected a thick wire or
super-conductor between the terminals, then the voltage will be given by
12*0/(0+Ri), ie zero volts. *But* the current is still defined by 12 /
(R+Ri), so that simply becomes 12/Ri. This is the maximum current a battery
can deliver.
This explains why on old cars the radio switches off momentarily as you
start your car - due to the huge current drawn from the starter motor the
battery voltage drops, often to as low as 6 or 8 V. Also when charging the
battery the opposite happens, and the voltage often goes up to 13 or 14
volts.
I'll leave it as an exercise to work out the maximum power the battery can
output...
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On Wed, 02 Apr 2008 11:50:39 +0100, Invisible <voi### [at] devnull> wrote:
>Well, let me put it this way.
A quick reply with out thinking too much.
I think that you are getting mixed up between Potential Difference and
voltage at a point. Consider opening a drain pipe in a tank, the
pressure at the tap is ambient (or a little above) this equates to
zero volts but the head of water is still high this is the PD of the
battery.
The fault is with your thinking you are analysing a childish analogy
with an educated adults reasoning power.
You are wrong we are right if you find out why you are wrong life will
not be any easier :)
--
Regards
Stephen
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> As I understand it, moving electrons round a wire is like moving water
> round a pipe. You need a *force* to push it. However, you confidently
> claim that you can make electricity move with no force at all.
Capacitors and inductors can STORE ENERGY!
> If somebody claimed that they could push water round a network of pipes
> just by writing an equation, they'd be laughed at.
What if I connected a vertical pipe several metres high to your pipe network
in the middle somewhere? As you applied some external force through your
system, I opened up a valve for a second to let water flow into my vertical
pipe, then closed it. You would notice a slight increase in force needed.
Then, once you switched off your external force later, I open my valve and
let my water flow back into the system and it would force more water out the
other end.
Would you be puzzled that after you switched off your force water kept
coming out for a bit longer?
That is exactly what capacitors and inductors are doing in an electrical
circuit, they store up energy during parts of the AC cycle, and let it out
at others.
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