The input current of a 16 hp, 400V, 25 Hz, 2-pole single-phase series motor at full-load is 60A. Motor equivalent resistance and reactionance are 0.3 Q and 0.525 Q, respectively. Knowing that at this full load current the equivalent current to iron losses is 20 A. Use phasor diagram to calculate: a. Find motor speed, torque and power factor. b. If the motor is connected to 267 V, 16.667 Hz source, find motor speed, torque, the output horsepower and power factor at the rated current

1) A 230-V, 50-Hz. 6-pole, single-phase induction motor has the following constants. r₁=0.12 2.₂ = 0.14 52 x₁=x₂=0.25 2.x = 15 52. If the slip is given as 0.05, find the mechanical power output developed.

A 30 kW, 460 V, 1180 rpm, 3 -phase, 60 Hz, Y-connected induction motor has a full-load efficiency of 93.6% and a power factor of 0.83 lagging. The resistance of the stator winding per phase is 0.14 N/phase. Determine: (a) The full-load active power drawn by the motor (b) The full-load line current (c) The full-load rotor winding losses (d) The rotational losses

Determine (a) the output power. and (b) the efficiency of a three-phase, 400 V induction motor running on load with a fractional slip of 0.04 and taking a current of 50 A at a power- factor of 0.86. When running light, without any load, at 400 V, the motor has an input current of 15 A and the power taken is 2000 W, of which 650 W represents the friction-and-windage loss. The resistance between any two terminals of the delta-connected winding is 0.333 N. Do not ignore the stator impedance and assume that when the motor is running light, the rotor current is negligible.