Re: Dow does this work?
- From: "ian henden" <ian@xxxxxxxxxxxx>
- Date: Sat, 25 Feb 2006 08:02:30 GMT
"Ben Barker" <dbrb2@xxxxxxxxx> wrote in message
news:dtie6q$37c$1@xxxxxxxxxxxxxxxxxxxxxxx
Faraday's law states that: V=-Ndphi/dt
In english, that means that a voltage is generated across a coil of N
turns that is proportional
to the rate of change of phi with time, where phi are the lines of
magnetic flux - as are seen when you put iron filings
on a piece of paper on top of a bar magnet.
So the faster you pedal, the higher the V generated.
True.... but terminal emf is regulated somewhat by the reactive inductance
of the generator (proprtional to frequency, and therefore proprtional to
velocity).
This is one reason why you don't get twice the voltage across your lamps,
with a traditional bike dynamo, when you go twice as fast.
(Pedants will waffle on about solid-state regulators etc, zener diodes and
similar which are apparently employed in some modern dynamo setups.)
A nice way of thinking about it is this:
Electrons are charged particles, and experience a force in a magnetic
field if there is relative motion between
the field and the charge. If you pedal with nothing connected, electrons
pile up at one end of your
conductor, making it more negative than the other end of the conductor.
Opposite charges
attract, so in this situation, any self respecting
electron would fall back to the positive end, but for every one that does,
it is replaced by another forced to the
negative end by the magnetic field.But like charges repel each other, so
eventually the force due to the magnetic
field is balanced out by the repulsion of the elctrons. No more electrons
move. In reality this build up of voltage
happens almost instantaneously, so is not noticed.
Because the electrons are somewhere, but would rather be somewhere else,
they have potential energy - like a child at the top of a slide.
This energy can be used to do things - like move, or generate heat.
If the circuit is now connected to something, then electrons can flow from
the negtaive to the positive end of the
generator, just as children flow down a slide. On the way, they will hit
things (ok, so the slide analogy has just fallen
apart). They loose some energy, and the things they hit vibrate, and
generate heat.
The amount of heat generated is a function of the amount of current
flowing (electrons per second), and the resiatance of the wire - how many
things there are to bump into. As long as the flow of electrons is less
than that the magnetic field can "pump" back up, then the voltage stays
constant. As more current flows, more heat is generated. If the resistance
of the circuit is too low however, the voltage will fall, as, to reuse the
analogy, children are sliding down faster than they can climb back up.
Now that IS a dodgy explanation!!!
W = Isqurd *R ..... heat is proprtional to square of current.
Voltage loss (at pioint of usage, ie the lamps) in conductors is proprtional
to the resistance of the circuit.
Total current drawn is inversely proportional to the total resistance
(I = V/R in a DC circuit, and I=V/Z in an AC circuit, where Z= square root
of (R squared plus X squared) ... Fred Pythagoras has a lot to answer for!)
It's the AC formula which explains the "regulation" of a bog standard dynamo
....since X= 2*Pi*f*L wher f is frequency, L is inductance.
HTH HAND
Bit of a dodgy explanation, but reasonably right. As for the "1 wire"
circuit, it is very common to use the bike frame as part of the dynamo
circuit -
unless you have a bike made of something very hi-tech, it is almost
certainly going to be steel or aluminum. Both of these are good
conductors- in fact the resistance od the return path through the frame is
likely to
be lower than that along the wire, since the effective surface area of the
wire is larger -
the bigger the wire, the lower it's resistance per m. This makes sense
when you thing about an electron trying to get from A to B hitting as few
things
as possible -easier the larger the cross sectional area of the conductor.
Hope this helps,
Ben
The return path bit is absolutely correct
.
- References:
- Dow does this work?
- From: Tom Crispin
- Re: Dow does this work?
- From: Ben Barker
- Dow does this work?
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