Power Flow In Synchronous Motor - MCQs with Answers
Q1. The expression for the mechanical power developed in terms of the load angle δ and the internal machine angle ∅, for constant voltageVph and constant excitation Eb isa. [{(E
bV
ph) / Z
s} * cos(θ - δ)] - [{(E
2b) / Z
s} * cos θ]
b. [{(E
bV
ph) / Z
s} * cos(δ - θ)] - [{(E
2b) / Z
s}* cos θ]
c. [{(E
2b) / Z
s} * cos θ] - [{(E
b V
ph) / Z
s} * cos(δ - θ)]
d. [{(E
2b) / Z
s} * cos θ] - [({E
b V
ph) / Z
s} * cos(θ - δ)]
View Answer / Hide AnswerANSWER: a. [{(Eb Vph) / Zs} * cos(θ - δ)] - [{(E2b) / Zs} * cos θ]
Q2. The value of load angle δ and the internal machine angle ∅ for maximum power developed in synchronous motor is equal toa. 0°electrical, 0° electrical
b. 0° electrical, 90° electrical
c. 90° electrical, 0° electrical
d. 90° electrical, 90° electrical
View Answer / Hide AnswerANSWER: d. 90° electrical, 90° electrical
Q3. A synchronous motor is running at a load angle of 25 degree at rated frequency with negligible armature resistance. Now, if the supply frequency is increased by 15% keeping the other parameters constant, then the new load angle will be equal toa. 26.89 degree
b. 29.07 degree
c. 32.05 degree
d. 38.20 degree
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