How Does a Rheostat Work?

The last post might look confusing without knowing what rheostats are and what they are used for. So this post will explain that without going into too much detail.

First, a simple reminder from the electric motor theory: any electric motor is a reversible machine, meaning that it can be used for both converting electricity into mechanical force and for converting mechanical energy back into electric energy.

Taking for example the train acceleration, as train starts to gain velocity its motors start acting as electric generators, generating counter-current that negates the externally applied voltage. If one would simply connect an electric motor to the DC power, it would simply spin up to a certain finite (and fixed) RPM value – at that point the generated voltage would be equal to external voltage.

The Ohm law applies for electric motors as well. A motor has very low internal electric resistance. Applying a high voltage would quite literally burn the motor. This is why when DC motor is starting, it requires only a fraction of voltage. One of the easiest way to adjust voltage is to dissipate the excess energy using extra resistance.

As train accelerates, the resistance should be decreased to compensate for constantly decreasing voltage (which is caused by motors generating electric voltage in the opposite direction of externally applied voltage). A rheostat is simply a variable electric resistance. The 81-717 train we simulate in Subtransit uses a rheostat controller to switch contacts through preset positions, resulting in parts of rheostat being shorted out and total current adjusted as needed (the schematic is a simplified version just for illustration):
(black square indicates closed contact, empty square indicates contact is open)

So assuming the motor has an internal resistance of 0.2 Ohm and 750 V of voltage is applied to it, without a rheostat it would experience 750/0.2 = 3,750 A of electric current (far more than it could possibly sustain). Adding extra 2.4 Ohm of resistance produces a starting current of 750/(0.2+2.4) = 288 A, a well acceptable value (for the motors of this class, operating current is between 200 and 400 A).

A special electric relay is sensitive to the current in power network and will move to next rheostat controller position every time the total current drops below a threshold due to train speed increasing. For the rheostat above, assuming the threshold is set to 200 A, the electric current applied can be found using the graph below:

The graph should be read left-to-right, from 0 km/h to 40 km/h as the train keeps accelerating. As you can see, the train starts acceleration with electric current of 288 A @ 0 km/h. As the speed keeps increasing, at 4 km/h rheostat controller turns to position 2. This increases electric current from 200 amps to 300 amps.

As the train continues its motion, the rheostat controller will move to position 3 at 8 km/h, position 4 at 12 km/h, position 5 at 16 km/h, position 6 at 21 km/h, and finally to position 7 at 24 km/h. At this point rheostat is fully shorted out and the train continues to gain speed with quickly decreasing electric current and torque.

At 30 km/h the electric current is about 280 amps, but by 40 km/h it would have dropped down to less than 70 amps!

Of course, rheostat is hardly the most efficient way to start the motor. All the excess power simply dissipates as heat, plus as it can be evident from graph above, if rheostat does not have a sufficient number of steps the train would experience sudden increases in electric current, causing jerking. Though, spiking from 200 to 350 amps is not as intense as it may sound – due to high inductivity of the electric engine, such spikes are actually smoothed out significantly, leading to a more smooth acceleration curve.

So far this imaginary train has only managed to reach about 30 km/h. That is just about the speed at which 81-717 rheostat exhausts all of its resistance as well. So how does the train go faster? More on that in one of the next posts! The keywords for the answer are “parallel connection” and “field reduction”.