"Don Foreman" wrote in message
...
On Wed, 30 Mar 2005 17:35:23 -0800, "Mike Fields"
wrote:


"Don Foreman" wrote in message
.. .
On Wed, 30 Mar 2005 13:21:49 -0800, Jim Stewart
wrote:

Ken Finney wrote:
Off-topic, but the learned denizens here are pretty good at asking
bizarre
questions.

I need an inductive load to switch a 28 VDC 12 A circuit into. Note
that
this is a one time test circuit, and I don't want to break the bank.
My
initial thought was to parallel a couple of 28 VDC relay coils to
get
the 12
A current. What I'm finding is that most of the relay coils draw
less
than
an amp, and I don't want have to parallel 16 relays to draw that
much
current. Does anyone have a better suggestion for me? I'm having
zero
luck
finding out how much an automotive starter solenoid draws, if I
could
find a
24 VDC diesel starter solenoid that requires 3 or more amps, that
would
probably be doable.

Just a quick back-of-an-envelop calculation
shows you'd need about 500 feet of 16 gauge
wire for a dc resistance of 2.3 ohms in your
coil. I'd not want to use anything less than
16 gauge and even then for just a short time.
It's going to have to dissipate 336 watts once
the core is saturated.

That's a big coil, either to wind or to find
in an existing device.

The induction motor winding will have some resistance. The rest can
be made up with 1/16" dia stainless TIG rod. I measured one such 36"
piece to be about 0.313 ohms. I don't think it'd overheat at 12
amps. It's probably settle out at about 315C. .

The flaw with that theory is that while a "real inductor" is indeed
inductance
in series with the resistance of the winding, putting a resistor external
to
the
winding takes away from the amount of inductance you would have if that
was all in the coil -- taken to the extreme, you could say 1 turn of wire
and
a 2.3 ohm resistor is the same as the 500 feet of 16 gauge wire -- from
a DC viewpoint, that would be true, but for the purposes of an inductive
load
with a changing field, they are definitely NOT the same circuit. Of
course it also depends if it is wound on an iron core or air core or ...
Your best bet to meet your requirements as stated would probably be to
find a large transformer with 2-2.5 ohms dc resistance and use that.
There is still the part about liability that you need to make sure you
are
covered there -- as another poster said, probably UL, or others.

I specified a lumped inductance able to handle 12 amps without
saturating: an induction motor. Whatever the value of L is, the
circuit here must have 2.3 ohms of resistance for steady-state
current to be the specified 12 amps at 28 volts. It doesn't matter
where this resistance is located in the circuit., and adding series R
does not diminish inductance.

I'm not making this up. You'll find it in any first course in AC
circuits for undergraduate EE candidates.


That was not my point -- the question had been asked if the available
inductor had too low a resistance was it legit to add series resistance.
My answer stands -- you are correct that series resistance does not
change the value of the inductance, HOWEVER, if you simply pick
some coil and add series resistance to limit the current, you do NOT
have anywhere near the inductance that you would have if you had
enough wire in the coil to limit the current with it's own resistance.
Look at the case I gave -- taking it to the extreme, 1 loop of wire with
a series resistor is NOT the same as a large inductor with 2+ ohms worth
of wire in it. Go back and READ YOUR BOOKS.