Question

A 3-phase, 400 V, 4 pole, 50 Hz star connected induction motor has the following parameters referred to the stator:
\( R_r' = 1 \, \Omega \), \( X_s = X_r' = 2 \, \Omega \)
Stator resistance, magnetizing reactance and core loss of the motor are neglected.
The motor is run with constant \( V/f \) control from a drive. For maximum starting torque, the voltage and frequency output, respectively, from the drive, is closest to:

• A. 400 V and 50 Hz
• B. 200 V and 25 Hz
• C. 100 V and 12.5 Hz
• D. 300 V and 37.5 Hz

Hint:

For maximum starting torque in an induction motor, adjust the supply frequency such that the rotor resistance equals the total leakage reactance: \( R_r' = X_s + X_r' \). Since \( X \propto f \), reducing frequency lowers reactance and helps meet this condition.

Answer 1

The Correct Option is C

For maximum starting torque in an induction motor, the rotor resistance should equal the total leakage reactance: \[ R_r' = X_s + X_r' \] However, in this case, \[ X_s + X_r' = 2 + 2 = 4 \, \Omega \Rightarrow R_r' = 1 \, \Omega<4 \, \Omega \] To make the condition \( R_r' = X_{total} \) true for maximum torque, we must reduce the frequency since reactance is frequency dependent: \[ X \propto f \Rightarrow {Let } f' { be the new frequency such that } R_r' = X_s(f') + X_r'(f') \] \[ 1 = 2 \cdot \frac{f'}{50} + 2 \cdot \frac{f'}{50} = 4 \cdot \frac{f'}{50} \Rightarrow f' = \frac{50}{4} = 12.5 \, {Hz} \] Since \( V/f = \) constant, \[ V' = \frac{12.5}{50} \cdot 400 = 100 \, {V} \] Therefore, the required voltage and frequency are: \[ \boxed{100 \, {V and } 12.5 \, {Hz}} \]