Electrical Machines Armature resistance control method is also called as-----------? constant torque control constant voltage control constant power control constant current control constant torque control constant voltage control constant power control constant current control ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines If a transformer has turns ratio K, the primary and secondary current are I1 and I2 respectively and magnetizing current and core loss component of no load current Iμ and Iw respectively, then I1 = KI2 + Iw + Iμ.ok I1 = KI2. I1 = KI2 + Iw. I1 = KI2 + Iμ. I1 = KI2 + Iw + Iμ.ok I1 = KI2. I1 = KI2 + Iw. I1 = KI2 + Iμ. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines A 400 V, 50 Hz, 30 Hp, three phase IM is drawing 50 A current at 0.8 p.f lagging. The stator and rotor copper losses are 1.5 KW and 900 W respectively. The friction and windage losses are 1050 W and core losses are 1200 W. The air gap power of the motor will be 23.06 kW. 24.11 kW. 25 kW. 26.21 kW. 23.06 kW. 24.11 kW. 25 kW. 26.21 kW. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines In normal dc machines operating at full-load conditions, the most powerful electromagnet is field winding interpole and compensating windings together interpole winding armature winding field winding interpole and compensating windings together interpole winding armature winding ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines Compensating winding are used to rectify the flux density waveform at throughout the air gap. interpole region. region under main pole. None of these throughout the air gap. interpole region. region under main pole. None of these ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines If input of a transformer is square wave the output will be square wave. pulsed wave. triangular wave. sine wave. square wave. pulsed wave. triangular wave. sine wave. ANSWER DOWNLOAD EXAMIANS APP