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Finding out Kv motor velocity constant

  1. Jun 25, 2015 #1

    If I take a DC motor and put it in generator mode, what will happen if I connect a resistance at the wire ending of the DC motor? Will the motor turn slower compared to when the electric wire of the motor is connected to nothing?
    What effect will the resistance have on the current flowing in the motor coil?

    I know that in generator mode, mechanical energy is being converted to electrical energy. In this case, mechanical power = torque x angular velocity and electrical power=voltage x current.

    I also know that V=RI+L(dI/dt)+E, and E=Kv.W. By neglecting the coil resistance R, is the voltage in the electrical power equation equals to E in generator mode?

    I am kinda confused here. Please help me out.
  2. jcsd
  3. Jun 25, 2015 #2


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    Gold Member

    Yes, because there will be a voltage drop in the resistor, thus the back-emf of the motor must also be decreased ( emf = ω * Km ). Km stands for motor-constant, and that's a better name because this motor-constant also have another role in the motor:

    Torque = Km * I

    So maybe it's easier to measure this Km by measuring the torque as a function of motor-current? Load the motor by some known torque and adjust the current so that the motor can just turn ( slowly ).
    Last edited: Jun 25, 2015
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