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On Wed, 02 Apr 2008 10:14:46 +0100, Invisible <voi### [at] dev null> 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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> 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.
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.
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Michael Zier wrote:
>> 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 my brain is really failing to comprehend reality, wouldn't that
mean there's a high concentration of such charged particles near the
sun, and a lower concentration of them everywhere else, and hence a net
difference in electrical potential?
--
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*
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> 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?
Exactly. A totally uncharged capacitor appears as a short-circuit. For a
split second it is exactly the same as if you had connected a wire between
the + and - on the battery.
> 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...
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. 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...
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