Electrical Machines In above question, the output power delivered by generator G1 and G2 will be 9.6 kW, 12 kW. 9.3 kW, 9.3 kW. 5.7 kW, 8.7 kW. 6.3 kW, 9.3 kW. 9.6 kW, 12 kW. 9.3 kW, 9.3 kW. 5.7 kW, 8.7 kW. 6.3 kW, 9.3 kW. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines The voltmeter connected across a generator reads voltage same at no load and at full load (rated). The generator is of the type short - shunt generator. level compound. shunt generator. series generator. short - shunt generator. level compound. shunt generator. series generator. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines A 250 volt dc series motor having an armature resistance of 0.2 Ω takes a load current of 60 A. What will be the torque produced at the shaft of the motor? Consider brush drop of 1V. can not be determined. 133.22 N-m 234.76 N-m. 266.20 N-m. can not be determined. 133.22 N-m 234.76 N-m. 266.20 N-m. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines In a DC generator, if P be the number of poles and N be the rpm of the rotor, then frequency of magnetic reversals will be PN/120. PN/2. PN/60. PN/3000. PN/120. PN/2. PN/60. PN/3000. ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines Consider the following statements : For a level compounded dc generator to run at constant voltage, the series field mmf must effectively compensate 1. armature reaction mmf 2.armature resistance drop 3. brush contact voltage drop Which of these statements is/are correct ? 1, 2 and 3 1 and 2 1 and 3 2 alone 1, 2 and 3 1 and 2 1 and 3 2 alone ANSWER DOWNLOAD EXAMIANS APP
Electrical Machines Crawling in an induction motor is due to slip ring rotor. space harmonics produced by winding currents. time harmonics in supply. insufficient starting torque. slip ring rotor. space harmonics produced by winding currents. time harmonics in supply. insufficient starting torque. ANSWER DOWNLOAD EXAMIANS APP
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