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> That doesn't make any sense.
Oh yes it does, it's the basis for almost all electronic circuit design.
> More precisely, if we assume that V and I are not necessarily in phase,
> this immediately allows me to derive a contradiction.
>
> Apparently, all I have to do is generate a sufficiently low-frequency
> wave, with V and I sufficiently far out of phase, and we arrive at an
> impossible situation. I could have a system with an arbitrarily large
> current passing through it, for an arbitrarily long time, despite the
> entire system having zero potential difference.
Note that the complex version of Ohm's Law only holds for steady-state
continuous sinusoidal operation, it won't explain what happens when you
first turn on a circuit or suddenly introduce voltage or current
spikes/steps.
In your example, if you have previously charged up a capacitor an arbitrary
amount, then yes, you can then get an arbitrary amount of current out with
no voltage. Just charge up a capacitor with a constant voltage, then short
the terminals ;-) But the voltage must be in a sine wave cycle though for
Ohm's law to hold.
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On Wed, 2 Apr 2008 10:22:50 +0200, "scott" <sco### [at] laptop com> wrote:
>> To true and why should they :)
>
>A friend of mine worked in a mobile phone shop for a bit. During their
>training the trainer asked some technical question about a phone which my
>friend answered. Then she told everyone they were here to sell phones and
>not to answer technical questions like that.
>
One wonders why she asked if not to show the others how
unknowledgeable they were. I hate bosses like that.
--
Regards
Stephen
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On Wed, 02 Apr 2008 10:19:21 +0100, Invisible <voi### [at] devnull> wrote:
>Michael Zier wrote:
>
>> "Much to learn you have!"
>
>True enough...
>
>> Take a capacitor: initially it's discharged. Now connect it to a voltage
>> source, the first instant you do, the voltage across the capacitor is
>> still zero
>
>How on earth do you work that one out?
>
>If you connect a capacitor to a 9 V source, then the potential
>difference across the capacitor is... exactly 9 V. In which universe is
>that 0 V?
You don't do it directly you use a resistor in series to limit the
current. Michael left this step out for simplicity. Probably not
wanting to overload your brain :)
>> That's very basic electrical engineering, you know... but maybe your
>> universe is different after all. SCNR
>
>Ah, maybe...
--
Regards
Stephen
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On Wed, 02 Apr 2008 10:14:46 +0100, Invisible <voi### [at] devnull> wrote:
>That doesn't make any sense.
Also the rules are different for dc and ac. The ac rules take into
account the frequency of the waveform. This is how the crossover
filters in your speakers work, assuming you have a bass and a tweeter.
It works, trust me man.
--
Regards
Stephen
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>> Take a capacitor: initially it's discharged. Now connect it to a voltage
>> source, the first instant you do, the voltage across the capacitor is
>> still zero
>
> How on earth do you work that one out?
>
> If you connect a capacitor to a 9 V source, then the potential difference
> across the capacitor is... exactly 9 V. In which universe is that 0 V?
A 9V source has an internal resistance, which will initially take all the
voltage as the capacitor draws a huge amount of current (even if it doesn't,
the wires will have some resistance). So across the capacitor terminals
will be a very low voltage, but a very high current through it. Once the
capacitor is fully charged, there will be no current flowing, and so no
voltage drop across the internal resistance, and then the full 9 V across
the capacitor.
So you see, in this simple case, voltage across the capacitor is definitely
not proportional to current flowing through it. Quite the opposite to a
resistor.
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scott wrote:
> Note that the complex version of Ohm's Law only holds for steady-state
> continuous sinusoidal operation, it won't explain what happens when you
> first turn on a circuit or suddenly introduce voltage or current
> spikes/steps.
...or rather, it does after you take those spikes and analyse their
component harmonics. (?)
> In your example, if you have previously charged up a capacitor an
> arbitrary amount, then yes, you can then get an arbitrary amount of
> current out with no voltage. Just charge up a capacitor with a constant
> voltage, then short the terminals ;-)
But if the capacitor is charged then it *has* a voltage! 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.
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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Am Wed, 02 Apr 2008 10:19:21 +0100 schrieb Invisible:
> Michael Zier wrote:
>
>> "Much to learn you have!"
>
> True enough...
>
>> Take a capacitor: initially it's discharged. Now connect it to a
>> voltage source, the first instant you do, the voltage across the
>> capacitor is still zero
>
> How on earth do you work that one out?
Out of continuity?
> If you connect a capacitor to a 9 V source, then the potential
> difference across the capacitor is... exactly 9 V. In which universe is
> that 0 V?
That's why I said "the first instant". dt (that's a differential "d")
before you contact the cap's terminal, the voltage across the terminals
is zero, right? Why should the potential be different one dt later
(except for a dV)? We live in a continuous world, perhaps you dont, IDK,
where most real-world measures are differentiable. And even if there was
no (ohmic) resistance at all, at t=0 (contact the terminals) there must
be a current into the cap first before the voltage across the cap can
increase (that's causality!). And only if there's no (ohmic) resistance,
you get the full 9V across the cap in finite time (namely 0+dt, and that
may be mathematically sloppy to say so). In the real world, where wires
have ohmic resistance, you'll *never* reach 9V across the cap in finite
time: V(cap)=V(source)*exp(-t/RC)
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scott wrote:
> A 9V source has an internal resistance, which will initially take all
> the voltage as the capacitor draws a huge amount of current (even if it
> doesn't, the wires will have some resistance). So across the capacitor
> terminals will be a very low voltage, but a very high current through
> it. Once the capacitor is fully charged, there will be no current
> flowing, and so no voltage drop across the internal resistance, and then
> the full 9 V across the capacitor.
>
> So you see, in this simple case, voltage across the capacitor is
> definitely not proportional to current flowing through it. Quite the
> opposite to a resistor.
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?
Right. Sure. Makes perfect sense. Electrons just move of their own
volition anyway; it's not like it requires a *force* to drive them or
anything...
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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>
> But if the capacitor is charged then it *has* a voltage! 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.
What about solar wind? The sun emits charged particles (of really high
energy). That's a stream of charges, meaning a current. Unless they
dissipate their energy by scattering events, they will move on forever in
the (mostly) empty universe (unless they reach the end of the universe,
may it be spatial or temporal, whereby in the latter case the can visit
Milliway's) or the wrap around when the space/time position overflows...
You could argue that the sun emits isotropically in all directions,
meaning no net current, but where's the difference to a closed electical
circuit (battery, cables, lamp i.e.), from outside the circuit, all
current vectors at all positions sum to zero too...
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> time: V(cap)=V(source)*exp(-t/RC)
bah, thats BS, V(cap)=V(source)*(1-exp(-t/RC)) of course.
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