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A separately excited dc motor having 2 ohm armature
resistsance draws a current of 10A and torque developed is
20 Nm under rated field condtion. the speed of the motor
under the rated field condtion , when it is connected to a
100V DC source and delivering 10Nm torque....
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Answer Posted / dharannya
Basic separately excited motor equations are
V=E+Ia*Ra-------------(1)
E=K*W-----------------(2)
T=K*Ia----------------(3)
from the first set of data ie Ra=2, Ia=10 T=20,
and from equation (3)
20=K*10
therefore K=2.
with this knowledge of K
continuing to 2nd set of data ie V=100 and T=10
again from eqn(3) we get Ia as 5A
knowing Ia Ra and V from eqn (1) we get E as 90V
from eqn(2) knowing the values of E and K we obtain w(omega,
ie speed) as 45 rad/sec or 416.67 rpm
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