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Homework Help: Shunt motor

  1. May 17, 2008 #1
    1. The problem statement, all variables and given/known data
    the questions quite big, but ill stick to the relevant parts

    power at the load, 35000W
    Voltage at load, 220V

    cable resistance = 0.1ohm
    field resistance = 60V

    3. The attempt at a solution

    the current at the load is P (load)/ V (load) = 35000/220 = 159A

    the voltage drop from resistance, IR = 159x0.1 = 15.9V
    terminal voltage is 15.9 + 220 = 235.9V

    and so across the inductor - current (field) = V/R(field) = 235.9/60 = 3.93A

    so now at this point, it asks for the armature current. which is calculated to be
    159A + 3.93A = 163A

    so here the current in armature is 163A yet the current at the load is 159A
    there is only 1 cable the current can travel down, and to me it can either be 1 or the other.

    can someone please explain any of this. any help greatly appreciated.

    (by the way, these are the values calculated by my lecturer. he would give 100% on if you answered with these values)
     

    Attached Files:

  2. jcsd
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