A DC Motor develops a torque of 150 N-m. A 10 percent reduction in the field flux causes a 50 percent increase in armature current. The new value of torque is: A) 102.5 N-m B) 202.5 N-m C) 172.5 N-m D) 232.5 N-m
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A DC Motor develops a torque of 150 N-m. A 10 percent reduction in the field flux causes a 50 percent increase in armature current. The new value of torque is: 202.5 N-m

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T1=KIf,T2=K1.5I*0.9f

T1=150Nm.  T2=T1*1.5*0.9=150*1.35=202.5Nm
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Good answer broo

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