How rotor resistance affects starting and maximum torque of an induction motor

Differentiate T with respect to s, or look at the slip at which maximum torque occurs.

Model answer.

Using the approximate equivalent circuit (ignoring stator impedance), the torque at slip s is
T = k × s E₂² R₂ / (R₂² + (s X₂)²),
where R₂ and X₂ are the rotor resistance and standstill rotor reactance per phase, and E₂ is the standstill rotor EMF.

Maximum torque. Setting dT/ds = 0 gives maximum torque at slip s_m = R₂ / X₂. Substituting this back gives T_max = k E₂² / (2X₂). So the value of maximum torque does not depend on R₂; only the slip at which it occurs does.

Starting torque. At standstill s = 1, so T_st = k E₂² R₂ / (R₂² + X₂²). This is largest when s_m = 1, i.e. when R₂ = X₂. In that case the motor produces its maximum torque at starting.

Slip-ring (wound-rotor) motors. External resistors are connected to the rotor through slip rings. At starting, the extra resistance is chosen so that R₂ is close to X₂, giving high starting torque with reduced starting current. As the motor speeds up, the resistance is cut out step by step, so the motor runs with low R₂ and low slip. This is useful for cranes, hoists, mills and other high-inertia loads.

Why not a permanently high-resistance cage? At normal running slip, rotor copper loss = s × (air-gap power). High R₂ means higher full-load slip, so more rotor loss, lower efficiency, more heating and poorer speed regulation. The cage motor's fixed resistance is therefore a compromise. Double-cage and deep-bar rotors get around it: at starting the rotor frequency is high, current crowds into the high-resistance part, and at running speed the rotor frequency is low and current flows through the low-resistance part.

Common mistake. Claiming that adding rotor resistance increases the maximum torque. It only shifts where the maximum occurs.

Likely follow-up. "What happens to T_max if supply voltage drops by 10%?" T_max ∝ V², so it drops to 0.81 of its original value, about 19% lower.