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Eero Ahonen <aer### [at] removethis zbxtnetinvalid> wrote:
> Why would there be more actual power with greater frequency? The
> phase-angle won't change, the current won't change and RMS voltage won't
> change, so U*I*cos(fi) won't change.
You need to invest extra power to get the higher frequency at the same AC
voltage and amps, right?
So where does that extra power end up if you have the otherwise same AC running
through your body?
EM field I'd guess - which may be just the extra "kick" you'd rather do without.
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clipka wrote:
>
> You need to invest extra power to get the higher frequency at the same AC
> voltage and amps, right?
Hmm, possibly.
> So where does that extra power end up if you have the otherwise same AC running
> through your body?
At least part of it goes to heating the generator, which runs faster :).
> EM field I'd guess - which may be just the extra "kick" you'd rather do without.
That actually sounds reasonable. My head is atm too empty to remember
how to calculate energies of EM fields, but it would make sense, while
the field sure is frequency-dependent.
-Aero
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Invisible wrote:
> Mike Raiford wrote:
>
>> This is why battery operated electronics can run until the battery is
>> almost flat, even though the battery's voltage continually drops as
>> it's drained.
>
> Really? That's interesting. I always thought it was the battery outputs
> full voltage until just before it actually dies (which is why battery
> meters are never, ever, under any circumstances, actually accurate).
Here's an example of a charge pump voltage regulator:
http://www.ti.com/lit/gpn/reg711-33
Note the input voltage range; +1.8V to +5.5V vs. the output voltage; +3.3V
for the REG711-3.3.
--
Tor Olav
http://subcube.com
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Tor Olav Kristensen wrote:
> Invisible wrote:
>> Mike Raiford wrote:
>>
>>> This is why battery operated electronics can run until the battery is
>>> almost flat, even though the battery's voltage continually drops as
>>> it's drained.
>>
>> Really? That's interesting. I always thought it was the battery
>> outputs full voltage until just before it actually dies (which is why
>> battery meters are never, ever, under any circumstances, actually
>> accurate).
>
> Here's an example of a charge pump voltage regulator:
>
> http://www.ti.com/lit/gpn/reg711-33
>
> Note the input voltage range; +1.8V to +5.5V vs. the output voltage; +3.3V
> for the REG711-3.3.
I also recommend reading this:
http://www.maxim-ic.com/appnotes.cfm/an_pk/737/
--
Tor Olav
http://subcube.com
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Invisible wrote:
> This (I presume)
> is why my phone tells me it's on maximum charge for months on end, until
> the exact moment when I try to make a phonecall...
Well, for that you need to consult Murphy's Law, not physics :)
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> Of course you can generate an out-of-phase signal. It's just an AC
> generator like everything else. What happens is that if you are out of
> sync with the other generators, there will be an energy loss because the
> out-of-sync signal dampens the others.
1) That "energy loss" will be enough to annihilate your generator
2) Even if you built your generator to withstand it, the resultant torques
and currents would pull it into sync with the others
3) It's not even relevant in practise, because there are protection
mechanisms to avoid exactly the above happening
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> I have no idea what internal resistance is.
A pretty accurate model of a battery is a perfect voltage source in series
with an "internal" resistor. This then nicely explains how the more current
you try to draw the lower the apparent voltage will be at the terminals of
the battery.
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scott wrote:
>> I have no idea what internal resistance is.
>
> A pretty accurate model of a battery is a perfect voltage source in
> series with an "internal" resistor. This then nicely explains how the
> more current you try to draw the lower the apparent voltage will be at
> the terminals of the battery.
As far as I'm aware, resistors affect only current, not potential. But
hey, I'm sure you know more about this than me...
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Eero Ahonen <aer### [at] removethiszbxtnetinvalid> wrote:
> clipka wrote:
> >
> > You need to invest extra power to get the higher frequency at the same AC
> > voltage and amps, right?
>
> Hmm, possibly.
>
> > So where does that extra power end up if you have the otherwise same AC running
> > through your body?
>
> At least part of it goes to heating the generator, which runs faster :).
>
> > EM field I'd guess - which may be just the extra "kick" you'd rather do without.
>
> That actually sounds reasonable. My head is atm too empty to remember
> how to calculate energies of EM fields, but it would make sense, while
> the field sure is frequency-dependent.
>
> -Aero
electrical impulses to the heart are interrupted and fibrillation occurs. IIRC
there are parts of the heart that receive the electrical impulses from the
brain then transmit secondary impulses to the muscle tissue to contract and
relax in an organised way. The electrical shock interrupts this process and you
get a disorganised high frequency oscillation in different parts of the heart. A
defibrillator is used to stop this effectively stopping the heart and the
natural sinus rhythm can be resumed. Before defibrillators became commonplace
the manual method was to strike the victims chest above the heart with a
controlled blow using the pinkie side of your fist. This was supposed to have
Stephen
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Invisible wrote:
> scott wrote:
>>> I have no idea what internal resistance is.
>>
>> A pretty accurate model of a battery is a perfect voltage source in
>> series with an "internal" resistor. This then nicely explains how the
>> more current you try to draw the lower the apparent voltage will be at
>> the terminals of the battery.
>
> As far as I'm aware, resistors affect only current, not potential. But
> hey, I'm sure you know more about this than me...
Try putting 2 resistors in series, connect the power (you'll have eg.
+12V on R1+, R1- on R2+ and R2- on ground) and measure voltages between
R1+/R2-, R1+/R1-, R2+/R2-, you might be surprised
+
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R1
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R2
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-
-Aero
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