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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] dev null> 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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