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I assume your problem is that you want to remove field excitation by opening a switch. If that is the question, then you have a simple L-R, switch, voltage source circuit. With current flowing thru the inductor there is energy stored in the magnetic field. Upon opening the switch that energy has to go somewhere. That somewhere is a voltage breakdown at the weakest point from a very high voltage until that something breaks down. What breaks down may be the winding insulation as well as the switch air gap.
At the instant of opening the switch the inductor maintains exactly the same current magnitude and direction as just before opening. Thus, you pick a resistor that limits the peak terminal voltage to a safe level. Suppose you had 5 amps excitation current and you want to limit the terminal voltage to 500 V, then you need a 100 ohm resistor. If the normal excitation voltage is 100 V, and the resistor is a shunt across the terminals all the time, then the resistor dissipation is 100 W continuously. If the resistor is only connected to the circuit a short time before opening the switch, then a lower wattage resistor could be used.
Consider a different way. Connect a reverse biased diode in parallel with the terminals. I would use a 20 to 30 A diode with a 400 PIV rating just because they are cheap. However, a 1N5625 (3 A fwd, 400 PIV) (this is a little ball under 0.3" in dia) would probably be OK but you would want to anaylze the energy being dissipated. In a larger motor you would do a more detailed analysis than use my off hand approach of 20 to 30 A and 400 V because now the diode becomes more expensive than the analysis time.
Have I answered your question?
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