# Power Flow In Synchronous Motor - MCQs with Answers

## 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 is

a. [{(EbVph) / Zs} * cos(θ - δ)] - [{(E2b) / Zs} * cos θ]
b. [{(EbV ph) / Zs} * cos(δ - θ)] - [{(E2b) / Zs}* cos θ]
c. [{(E2b) / Zs} * cos θ] - [{(Eb Vph) / Zs} * cos(δ - θ)]
d. [{(E2b) / Zs} * cos θ] - [({Eb Vph) / Zs} * cos(θ - δ)]

ANSWER: 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 to

a. 0°electrical, 0° electrical
b. 0° electrical, 90° electrical
c. 90° electrical, 0° electrical
d. 90° electrical, 90° electrical

ANSWER: 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 to

a. 26.89 degree
b. 29.07 degree
c. 32.05 degree
d. 38.20 degree