GATE EE 2024 Set 1 — Question 54

For a 3-phase induction motor, the condition for maximum torque at starting (where slip $s = 1$) is that the total rotor resistance referred to the stator must be equal to the magnitude of the Thevenin equivalent impedance looking from the rotor terminals.Neglecting the magnetizing branch ($X_m \to \infty$) and core losses, the equivalent circuit simplifies to a series combination of stator and rotor impedances. The condition for maximum starting torque is:$$R_{r,total}' = \sqrt{R_s^2 + (X_s + X_r')^2}$$Given the parameters referred to the stator:

• Stator resistance, $R_s = 3\ \Omega$
• Stator reactance, $X_s = 2\ \Omega$
• Rotor reactance referred to stator, $X_r' = 2\ \Omega$
• Rotor resistance referred to stator, $R_r' = 2.5\ \Omega$

Calculating the required total rotor resistance referred to the stator:$$R_{r,total}' = \sqrt{3^2 + (2 + 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\ \Omega$$The total rotor resistance referred to the stator is the sum of the inherent rotor resistance and the added external resistance:$$R_{r,total}' = R_r' + R_{ext}'$$$$5 = 2.5 + R_{ext}' \implies R_{ext}' = 2.5\ \Omega$$The external resistance referred to the rotor side ($R_{ext}$) is calculated by dividing the value referred to the stator by the square of the effective turns ratio ($a = N_s/N_r$):$$R_{ext} = \frac{R_{ext}'}{a^2}$$Given the turns ratio $a = 3$:$$R_{ext} = \frac{2.5}{3^2} = \frac{2.5}{9} \approx 0.2778\ \Omega$$Rounding to two decimal places, the value is $0.28\ \Omega$. The official range provided is $0.26$ to $0.29$.