- #1
lazypast
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Homework Statement
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
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)