On 20 Jun 2017 at 0:52, Robert Laird wrote:
> So you need a rheostat that can take the field current. 2 amps? (I'm too
> lazy to look it up right now.) But now we run into a problem. When you first
> put the rheostat inline, it has to take just about the entire field current.
> (not much limiting is taking place yet.) A 50 ohm rheostat for 2 amps is 200
> watts. Kind of big. But, if you simply switch in a 20 ohm resistor, the
> current (about) gets cut in half, and you only need a 20-ish watt resistor.
Well, sure, if you connect a 100 ohm resistor across 100 volts, it will
dissipate P = E^2 / R = 100 watts. If you connect a 50 ohm resistor across
100 volts, it will dissipate 200 watts.
But that's not what we're doing here! When you're using the resistor IN
SERIES with a load to control the current through that load, the situation
is quite different.
Let me illustrate with an example, using your 2a field at our ETs' nominal
36 volts. For the sake of simplicity, I'm going to treat the field as if it
were a pure resistance.
By Ohm's law, a 2 amp field at 36v would have to have a resistance of R = E
/ I = 36 / 2 = 18 ohms. Powered at 36 volts, it will dissipate (again as a
pure resistance) P = I E = 36v * 2a = 72 Watts.
So do we need a 72 Watt rheostat to control that field -- or worse yet, a
144 Watt one? Not at all.
Suippose we want to the reduce the field current to half, or 1 amp. System
voltage is still 36v. R = E / I = 36v / 1a = 36 ohms. This means that to
reduce current to 1 amp, our circuit has to have a total resistance of 36
ohms. Since our motor field is fixed at 18 ohms, we need to add resistance
in series with it. This resistance (your rheostat, or a fixed resistor like
the ET uses) will have to be [36 ohms (total needed) - 18 (field
resistance)] = 18 ohms.
In a series circuit, current is everywhere the same. The voltage drop
across each circuit element is determined by its resistance as a proportion
of the total circuit resistance. The voltage drop across the motor field
is 18 / 36 * 36v, or 18v. The voltage drop across the resistor is also 18 /
36 * 36v, or 18v. P = I E. So our series resistor only has to dissipate
18v * 1a = 18 watts, not 72 watts. That's one-quarter of what your
intuition might tell you to expect. A 20 watt wirewound resistor would do
just fine, though I'd probably go higher than that for reliability.
Now let's look at a few more cases, if you have the patience. I promise you
the numbers we get will be interesting, so if you're in a hurry, just skip
to the last sentence in each paragraph.
Say we want 3/4 field current, 1.5 amps. Again remembering that the current
is equal everywhere in a series circuit, our total circuit resistance must
be R = E / I = 36 / 1.5 = 24 ohms. So our series resistor (rheostat
setting, if you will) has to be 24 - 18 = 6 ohms. The voltage across the
field will be 18 / 24 * 36 = 27 volts and across the resistor will be 6 / 24
* 36 = 9 volts. The resistor has to dissipate only 9v * 1.5a = 13.5 Watts!
Or say we want the field current to be 0.5 amps. For this, we need a
circuit resistance of R = E / I = 36 / 0.5 = 72 ohms. Our series resistor
is 72 - 18 = 54 ohms. Field voltage will be 18 / 72 * 36 = 9v. Voltage
across the resistor will be 54 / 72 * 36v = 27v. The series resistor for
1/4 field has to dissipate 27v * 0.5a = 13.5 Watts, exactly what the series
resistor dissipated at 3/4 field.
Now let's see if this relationship holds up. We'll make the field current
0.25 amps, one-eighth of the normal 2a strength. Now at 36v we need a total
resistance of R = E / I = 36 / 0.25 = 144 ohms. Our series resistor
(rheostat) will have to be 144 - 18 ohms = 126 ohms. Now the field gets 18
/ 144 * 36v = 4.5v. The series resistance gets 126 / 144 * 36v = 31.5v.
The series resistor has to dissipate even less, only 31.5v * 0.25a = 7.9
Watts.
Finally, how about 7/8 strength, or 1.75a? R = 36 / 1.75 = 20.57 ohms.
Series resistor = 20.57 - 18 = 2.57 ohms. Resistor voltage = 4.5v and field
voltage = 31.5v. Series resistor power dissipation at 7/8 field is exactly
the same as it was when we were at 1/8 field, 7.9 Watts.
I know, lots of numbers, sorry! The takeaway is that your field control
resistors or rheostat won't have to be nearly as beefy as it might at first
appear.
David Roden - Akron, Ohio, USA
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