GATE 2025 EE Q.45: solution and answer

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Neglecting stator resistance, the magnetizing branch and core loss, the per-phase equivalent circuit reduces to a single series loop: supply phasor in series with the total leakage reactance and the referred rotor resistance $R_r^{\prime }/s$. The torque is $$T=\frac{3}{{\omega }_s}\cdot \frac{V^2\left(R_r^{\prime }/s\right)}{{\left(R_r^{\prime }/s\right)}^2+X_{20}^2}$$ where $X_{20}=X_s+X_r^{\prime }$ is the total leakage reactance referred to the stator.

2. Maximizing $T$ with respect to slip (for fixed $V$, $f$) gives the standard result $$s_{maxT}=\frac{R_r^{\prime }}{X_{20}}$$

3. At rated frequency (50 Hz): $X_{20}=X_s+X_r^{\prime }=2+2=4\Omega$.

4. Under constant V/f (scalar) control, the leakage reactance scales linearly with the drive's output frequency, while $R_r^{\prime }$ (a resistance) does not: $$X_{20}(f)=4\times \frac{f}{50}\Omega$$

5. Starting always corresponds to slip $s=1$. For maximum torque to occur exactly at starting, the peak-torque slip must equal 1: $$s_{maxT}=1\Rightarrow R_r^{\prime }=X_{20}(f)$$ $$1=4\times \frac{f}{50}$$

6. Solving: $$f=\frac{50}{4}=12.5\text{Hz}$$

7. Constant V/f ratio, taken from the rated nameplate values (the ratio is the same whether line or phase quantities are used): $$\frac{V}{f}=\frac{400}{50}=8\text{V/Hz}$$

8. Therefore $$V=8\times 12.5=100\text{V}$$

9. This matches option (C): 100 V and 12.5 Hz.