Re: R = V/I ?

[Mark Sylvester <msylvest@SPIN.IT>]

At 16.43 09/05/00 -0700, Leigh Palmer wrote:
> >I'll respond just to this bit and then go to bed!
> >
> >At 15.10 09/05/00 -0700, Leigh Palmer wrote:
> >>Let's not complicate things. My system is a Mixmaster and a bowl of
> >>cake batter. It has two terminals. Mixmasters can be run perfectly
> >>well off DC mains because they have universal (AC/DC) motors, as
> >>anyone watching an old TV next to a running Mixmaster will readily
> >>perceive. Let's say I run my Mixmaster on DC. My system dissipates
> >>energy, it has an IV characteristic, and the power dissipated P = IV.
> >>I can define R for this system by your causal relation: R = P/I^2.
> >>Like the light bulb however, this relation will yield a variable
> >>resistance. Unlike the light bulb, electrical resistivity plays no
> >>important part in dissipating energy in this system.
> >
> >Of course one would not apply R = V/I to the motor. From "my viewpoint" the
> >reason is clear. There is an emf in the motor, symptomatic of an energy
> >conversion which is not (directly) dissipative. One can always choose to
> >mix eggs and flour, dissipating the energy converted. Or instead one could
> >use it to accelerate a vehicle. And then perhaps apply the brakes... but
> >this doesn't produce a resistance. For resistance to be present the charge
> >carriers must dissipate the work done on them by the E field, directly,
> >within the device, heating it. Jam a spoon between the Mixmaster paddles to
> >stop them from turning. This will demonstrate the difference between the
> >dissipation due to its resistance, and the energy being converted by
> >working on the pancake mix!
>
> There's no back emf detectable at the terminals. A potential difference
> is a potential difference, regardless of the mechanism which produces it.
>
> If I put a black box around the system it will be a two-terminal device
> with a simple (i.e. non hysteretic) IV characteristic. In steady state
> the energy dissipated will appear at the surface of the box manifest as
> increased tempreature. Are you saying that you would not say the device
> has a resistance? Why not? No fair peeking inside black boxes!

If you want a black box, OK. I thought you wanted a Mixmaster stirring
batter. A black box with current I and p.d. V, heating itself and its
surroundings at a rate P = IV, has resistance R = V/I.

If the black box has a torque-exerting shaft sticking out of it, which can
do work when there is a current I, then R is not given by V/I unless the
shaft is blocked so that work = 0.

Mark




_____________________________________
Mark Sylvester
United World College of the Adriatic,
34013 Duino TS, Italy.
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