Question

The induced EMF in a 3.3 kV, 4-pole, 3-phase, Y-connected synchronous motor is considered to be equal and in phase with the terminal voltage under no-load condition. On application of a mechanical load, the induced EMF phasor is deflected by an angle of 2 degrees with respect to the terminal voltage phasor. If the isochronous reactance is 2 ohms and the series resistance is negligible, the motor armature current magnitude, in amperes, during loaded condition, is closest to:

• A. 50.7
• B. 66.49
• C. 75.1
• D. 90.0

Hint:

For synchronous motors, the armature current can be found using the relationship between the induced EMF and the reactance.

Answer 1

The Correct Option is B

Step 1: For a synchronous motor, the armature current magnitude \( I_a \) can be found using the following formula: \[ I_a = \frac{E}{X_s} \] where \( E \) is the induced EMF, and \( X_s \) is the isochronous reactance.
Step 2: The induced EMF is related to the terminal voltage and the phase difference between the induced EMF and the terminal voltage: \[ E = V \cos(\theta) \] where \( V \) is the terminal voltage and \( \theta \) is the phase angle (2 degrees). The terminal voltage is 3.3 kV.
Thus: \[ E = 3.3 \times \cos(2^\circ) \]
Step 3: Substituting the values: \[ E \approx 3.3 \times 0.9994 \approx 3.299 \, \text{kV} \]
Now, the armature current is: \[ I_a = \frac{3.299}{2} \approx 66.49 \, \text{A} \]
Thus, the armature current magnitude is closest to 66.49 A.