I a R I Vs Example 6.4 14.75-120 -0.75 Твол O I= 14-15-0·75 14 A 120-V, 2400-rpm shunt motor has an armature resistance of 0.4 2 and a shunt field resistance of 160 2. The motor operates at its rated speed at full load and takes 14.75 A. The no-load current is 2 A. When an external resistance of 3.6 Q is inserted in the armature circuit with no change in the torque developed, calculate the motor speed, the power loss in the external resistance, and the efficiency of the motor. Assume that the rotational loss is proportional to the speed. No = Enfi 14.75A = Ean 2400 120-1470.4 Eur - 120 - 14 (604 +31 6 + oh results • • · = -X2400 114.4 134208 Ppx=14x36 705.60 114. +.4V It is evident from this example that (a) a large percentage of the power supplied to the motor is lost in the control resistance, (b) the efficiency has decreased considerably, and, (c) the speed has been reduced to one-half of its rated value. 2-0.75-1.25A Eq=1200-125A (0.4) = It is left for the reader to verify that there is no change in the torque developed by the motor. P₁ = Us I 159.50 119.44×2400=2507pm Num= 114.4 n = 1775 -X100-818 - 120x1475=1775"