Re: R = V/I ?

[Leigh Palmer <palmer@SFU.CA>]

> 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!

Leigh