"Orchid XP v8" <voi### [at] devnull> wrote in message
news:47f29204$1@news.povray.org...
> Phil Cook wrote:
>
> > Compared to some of the tabloids the Daily Mash is the only one to make
> > any sense
> >
> > http://tinyurl.com/2t2bpa is topical for here :-P
>
> ...what the...? o_O

Hehehe. Forwarded on to my atheist friend. I know she'll enjoy it.

I don't know the context of this, but it's funny anyway
http://www.thedailymash.co.uk/sport/sport-headlines/scotland-dies-laughing-2
0071121549/

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Orchid XP v8 wrote:
>>> Hell, for that price it should also make me a cup of tea when I wake 
>>> up in the morning! :-S
>>
>> http://www.boingboing.net/2002/12/08/coffeemaking-pc-case.html
> 
> ....just when you thought the Internet couldn't get any larger... you 
> were wrong! o_O

http://www.odditycentral.com/pics/pc-deep-frier.html

-- 
   Darren New / San Diego, CA, USA (PST)
     "That's pretty. Where's that?"
          "It's the Age of Channelwood."
     "We should go there on vacation some time."

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Darren New escribió:
> http://www.odditycentral.com/pics/pc-deep-frier.html

Here's the real source, with the full story:
http://www.hardforum.com/showthread.php?t=1104341

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> 1.0 W (RMS) per speaker.

ARGHHHHHHH one of my pet hates, it's MEAN POWER not RMS.  Actually what is 
usually meant is "continuous sine wave average power", so you pump out the 
loudest sine wave you can, and measure the *MEAN* power (which is RMS 
voltage * RMS current).  That's the figure you put on the speaker.

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On Wed, 2 Apr 2008 09:47:50 +0200, "scott" <sco### [at] laptopcom> wrote:

>> 1.0 W (RMS) per speaker.
>
>ARGHHHHHHH one of my pet hates, it's MEAN POWER not RMS.  Actually what is 
>usually meant is "continuous sine wave average power", so you pump out the 
>loudest sine wave you can, and measure the *MEAN* power (which is RMS 
>voltage * RMS current).  That's the figure you put on the speaker.
> 

You try talking about Volt amps in a shop and I wish you well :)
-- 

Regards
     Stephen

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>> The speaker housings appear to be metal, which is unusual. (Mind you, 
>> they claim to be magnetically shielded...)
> 
> So if you put a floppy disk in front of the speaker, it won't get 
> affected? Cool :)

More so you can put it next to your CRT monitor and not magnetise it...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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> You try talking about Volt amps in a shop and I wish you well :)

Hehe no way :-)  It took me until I was at my power electronics course in 
1st year University before I finally found out why things had power written 
on them in VA rather than W.  I don't expect the person selling me speakers 
in a shop to have done an Engineering degree.

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Stephen wrote:

> You try talking about Volt amps in a shop and I wish you well :)

Trying to explain to me how Volt Amps is different from Watts would 
probably be a bigger challenge. :-P

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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And lo on Tue, 01 Apr 2008 21:51:01 +0100, Gail Shaw sa dot com>  
<"<initialsurname"@sentech> did spake, saying:

>
> "Orchid XP v8" <voi### [at] devnull> wrote in message
> news:47f29204$1@news.povray.org...
>> Phil Cook wrote:
>>
>> > Compared to some of the tabloids the Daily Mash is the only one to >  
>> make any sense
>> >
>> > http://tinyurl.com/2t2bpa is topical for here :-P
>>
>> ...what the...? o_O
>
> Hehehe. Forwarded on to my atheist friend. I know she'll enjoy it.

Some of the stories can be a bit crude, but you get the odd gem.

-- 
Phil Cook

--
I once tried to be apathetic, but I just couldn't be bothered
http://flipc.blogspot.com

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>> So if you put a floppy disk in front of the speaker, it won't get 
>> affected? Cool :)
> 
> More so you can put it next to your CRT monitor and not magnetise it...

What's a CRT monitor? Seriously, haven't seen one of them for years now.

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>> You try talking about Volt amps in a shop and I wish you well :)
> 
> Trying to explain to me how Volt Amps is different from Watts would 
> probably be a bigger challenge. :-P

Read this then ask questions ;-)

http://en.wikipedia.org/wiki/AC_power

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On Wed, 02 Apr 2008 09:05:21 +0100, Invisible <voi### [at] devnull> wrote:

>Stephen wrote:
>
>> You try talking about Volt amps in a shop and I wish you well :)
>
>Trying to explain to me how Volt Amps is different from Watts would 
>probably be a bigger challenge. :-P

Volt amps takes into account the phase difference between the volts
and the current. As you know then you pass an ac current through a
capacitor or an inductor there is a phase difference. Volt amps is V *
cos(phase) *A.
Simple, What?
-- 

Regards
     Stephen

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On Wed, 2 Apr 2008 10:05:53 +0200, "scott" <sco### [at] laptopcom> wrote:

>> You try talking about Volt amps in a shop and I wish you well :)
>
>Hehe no way :-)  It took me until I was at my power electronics course in 
>1st year University before I finally found out why things had power written 
>on them in VA rather than W.  I don't expect the person selling me speakers 
>in a shop to have done an Engineering degree.
> 
To true and why should they :)
-- 

Regards
     Stephen

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> 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.

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scott wrote:

> What's a CRT monitor? Seriously, haven't seen one of them for years now.

In my universe, CRT is still infinitely more popular than LCD. This 
might be related to price...

[Obviously, where I work most of the monitors are so ancient that the 
plastic is turning interesting colours, so they're mostly CRT. My 
monitor at home is also CRT, because this gives it the property that it 
can display *any* resolution correctly. Useful for games that won't run 
fast enough at high res...]

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Stephen wrote:

> Volt amps takes into account the phase difference between the volts
> and the current. As you know then you pass an ac current through a
> capacitor or an inductor there is a phase difference. Volt amps is V *
> cos(phase) *A.
> Simple, What?

That makes no sense at all.

Since Amps are directly proportional to Volts, how can there possibly be 
a phase difference?

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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> Since Amps are directly proportional to Volts, how can there possibly be
> a phase difference?

"Much to learn you have!"

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, however a large current (of charges) is flowing in, 
accumulating on the cap's plates thus creating an electric field in the 
cap. Which is proportional to the plate distance, charges sitting on them 
and die dielectric. This field prevents further flow of charges into the 
cap and finally you have the full voltage across the cap and no current 
flowing. Now repeat the whole thing by applying the opposite of the 
voltage, the cap gets charged the other way around, but with the same 
principal timing. Repeat, repeat, repeat... you get the picture -> phase 
shift (90 degrees).

Same applies to inductors (your speakers' driver coils), only this time 
the energy is stored in the magnetic field produced by the current, so I 
and V are exchanged in the timing, but still a phase shift.

That's very basic electrical engineering, you know... but maybe your 
universe is different after all. SCNR

Michael

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> That makes no sense at all.
>
> Since Amps are directly proportional to Volts,

You know Ohm's law, V = I*R, well think what happens if V and I are complex 
numbers ... as soon as R becomes complex, the phase of V and I will be 
different.

And yes you guessed it, the "resistance" of capacitors and inductors are 
complex numbers ;-)

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>> Since Amps are directly proportional to Volts,
> 
> You know Ohm's law, V = I*R, well think what happens if V and I are 
> complex numbers ... as soon as R becomes complex, the phase of V and I 
> will be different.
> 
> And yes you guessed it, the "resistance" of capacitors and inductors are 
> complex numbers ;-)

That doesn't make any sense.

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.

Obviously, electrons don't just move around for the hell of it. They 
move in direct response to a potential difference. The very concept of a 
current with no force driving it is obviously ridiculous.

The equations might make perfect sense in the presence of complex 
numbers, but that doesn't mean they match the real world.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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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?

> That's very basic electrical engineering, you know... but maybe your 
> universe is different after all. SCNR

Ah, maybe...

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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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] laptopcom> 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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> 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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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] 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

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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/
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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...

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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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Am Wed, 02 Apr 2008 10:45:08 +0100 schrieb Invisible:

> volition anyway; it's not like it requires a *force* to drive them or
> anything...

Well, strictly spoken no. Newtons first law applies to electrons too... 
(Think of Vacuum tubes, particle accelerators, synchrotron storage ring 
although with these electrons lose energy simply by getting accelerated 
*radially* [to remain on cirular course], and by some magic not covered 
by mechanics, they lose energy through radiation, which is the whole 
point of synchrotrons)

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On 2 Apr 2008 06:34:59 -0500, Michael Zier <mic### [at] mirizide> wrote:

>Am Wed, 02 Apr 2008 10:45:08 +0100 schrieb Invisible:
>
>> volition anyway; it's not like it requires a *force* to drive them or
>> anything...
>
>Well, strictly spoken no. Newtons first law applies to electrons too... 
>(Think of Vacuum tubes, particle accelerators, synchrotron storage ring 
>although with these electrons lose energy simply by getting accelerated 
>*radially* [to remain on cirular course], and by some magic not covered 
>by mechanics, they lose energy through radiation, which is the whole 
>point of synchrotrons)

I think Invisible was being ironic.
But why do you say "synchrotron storage ring although with these
electrons lose energy simply by getting accelerated *radially*"?
Why lose energy not gain energy? They are being accelerated is it
considered a negative acceleration?

-- 

Regards
     Stephen

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scott wrote:

> Maybe trying to 
> teach you about complex AC circuit theory and power electronics was a 
> bad starting point :-)

Hmm... maybe.

> This explains why on old cars the radio switches off momentarily as you 
> start your car.

What do you mean "old cars"?

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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scott wrote:

> Capacitors and inductors can STORE ENERGY!

No - really?

> 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?

No - because there is positive pressure at the value you just opened, 
and this pressure is what is pushing the water along.

Similarly, a capacitor can store charge such that when you turn off your 
external current source, the capacitor is still presenting a voltage to 
the system, and hence current is still flowing. There is no mystery in 
that. The mystery is in saying that there is actually "no voltage", and 
the current is just magically flowing all by itself.

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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Am Wed, 02 Apr 2008 12:50:58 +0100 schrieb Stephen:

> But why do you say "synchrotron storage ring although with these
> electrons lose energy simply by getting accelerated *radially*"? Why
> lose energy not gain energy? They are being accelerated is it considered
> a negative acceleration?

I said "radially" not "tangentially". They must be accelerated towards 
the center, otherwise  they'd fly straight-line (it's done by bending 
magnets, so it's not a circle but a N-gon with rounded corners). In spite 
of having the radial force perpendicular to the flight path, yielding 
zero for the scalar product Work=Force \times distance and therefore 
can't change the energy of the electrons, the will lose energy in form of 
electromagnetic radiation (depends on the facility, but most will produce 
X-Rays for all kinds of useful experiments). That comes from 
electrodynamics: accelerated charges emit radiation.

That caused great confusion about how atoms could be stable and resulted 
in the advent of quantum mechanics. How could electrons circle around the 
atom's core, without radiation emission and finally plunging into the 
core? Bohr's model of standing particle waves as electrons was the first 
explanation of atoms, that didn't involve accelerated charges (and as we 
know today, was not quite correct).

Yes, I admit it, I'm a physicist. Now you know it...

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Am Wed, 02 Apr 2008 07:03:37 -0500 schrieb Michael Zier:

> Am Wed, 02 Apr 2008 12:50:58 +0100 schrieb Stephen:
> 
>> lose energy not gain energy? They are being accelerated is it
>> considered a negative acceleration?

Ahh, and yes: I consider a body being accelerates whenever it's velocity 
vector changes. That includes cases, where only the magnitude changes 
(either + or -) or where only the direction changes (and of course 
combinations).

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On 2 Apr 2008 07:05:43 -0500, Michael Zier <mic### [at] mirizide> wrote:

>Am Wed, 02 Apr 2008 07:03:37 -0500 schrieb Michael Zier:
>
>> Am Wed, 02 Apr 2008 12:50:58 +0100 schrieb Stephen:
>> 
>>> lose energy not gain energy? They are being accelerated is it
>>> considered a negative acceleration?
>
>Ahh, and yes: I consider a body being accelerates whenever it's velocity 
>vector changes. That includes cases, where only the magnitude changes 
>(either + or -) or where only the direction changes (and of course 
>combinations).

As an engineer I agree, so it must be true :)
Does an electron have less energy in a circular path than a straight
one. You are pumping energy into the system so I would have thought
that it would be a gain in energy for the electron and a loss of
energy to the magnetic field.
BTW I also have problems with credit and debit when working on
financial systems ;)
-- 

Regards
     Stephen

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