A 230-V series motor takes 50 A at a speed of 1200 rpm. The armature current intake is 20 A. What is the percent change in the torque? What is the speed of the motor at the reduced load?
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Torque (T) = (Power (P) / Angular velocity (ω)) Power (P) = Voltage (V) * Current (I) Given: Voltage (V) = 230 V Current (I) = 50 A Speed (N) = 1200 rpm Power (P) = 230 V * 50 A = 11500 W Angular velocity (ω) = 2π * Speed (N) / 60 ω = 2π * 1200 / 60 = 40π Show more…
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