> 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.

Post a reply to this message

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*

Post a reply to this message

> 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...

Post a reply to this message

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*

Post a reply to this message

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*

Post a reply to this message

> 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

Post a reply to this message

>> 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.

Post a reply to this message

On Wed, 02 Apr 2008 10:45:08 +0100, Invisible <voi### [at] devnull> 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

Post a reply to this message

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*

Post a reply to this message

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*

Post a reply to this message