Torque-speed characteristics

Every induction-motor torque–speed curve comes from one relation, obtained (at the level required here, without derivation) from the motor's equivalent circuit: torque is proportional to power delivered to the rotor-resistance branch, expressed per phase in terms of rotor resistance $R_2$, standstill rotor reactance $X_2$, supply/stator voltage $V_1$, and slip $s$, all referred consistently to one side of the circuit.

where: $s$ is the fractional slip between synchronous speed and rotor speed (dimensionless, $0$ at synchronous speed, $1$ at standstill), $R_2$ and $X_2$ are the per-phase rotor resistance and standstill reactance, and $V_1$ is the applied per-phase voltage.

Two limiting regimes shape the curve. Near synchronous speed, slip is small, so $(sX_2)^2$ is negligible next to $R_2^2$ and the expression collapses to $T \propto sV_1^2/R_2$ — torque grows roughly linearly with slip. This linear stretch is the motor's normal running region: as load increases, slip increases just enough to supply more torque. Near standstill, slip is close to $1$ and $(sX_2)^2$ dominates $R_2^2$, so $T \propto V_1^2/(sX_2^2)$ — torque now falls as slip increases further toward $1$. Between these two regimes the torque must pass through a maximum.

The peak is the pull-out torque. Differentiating the expression with respect to $s$ (carried out in the derivation below) shows the maximum occurs where the two terms of the denominator are equal, $sX_2 = R_2$, giving $s_{max} = R_2/X_2$, and the maximum torque value itself turns out to be independent of $R_2$. This peak is called the pull-out or breakdown torque — the largest torque the motor can develop at rated voltage. At $s=1$ (standstill) the same formula gives the starting torque, generally well below pull-out.

The curve has a stable half and an unstable half. Plotting $T$ against slip from $s=0$ (synchronous speed, zero torque, since zero slip means no induced rotor current) up through $s=1$ traces a curve that rises to the pull-out peak and then falls back toward the starting-torque value. The region from $s=0$ to $s_{max}$ is the stable operating region: here, torque increases with slip, so if load torque rises the motor slows slightly, slip increases, and the induction motor automatically supplies more torque to match — a self-correcting equilibrium. Beyond $s_{max}$, torque decreases as slip increases further, so any extra load slows the motor further, which only lowers available torque — the motor cannot find a new balance and stalls. Practical machines are therefore always operated to the left of the pull-out point.

Voltage sensitivity. Because $T \propto V_1^2$, torque is far more sensitive to supply voltage than to almost anything else in the equation — a fact with real consequences whenever supply voltage sags.

Problem. An induction motor has per-phase rotor resistance $R_2 = 0.3\ \Omega$ and standstill rotor reactance $X_2 = 1.5\ \Omega$ (referred consistently). Find (a) the slip at which pull-out torque occurs, and (b) the ratio of pull-out torque to starting torque, taking $V_1$ constant.

Step 1. Use the derived condition $s_{max} = R_2/X_2$ from the derivation above: $s_{max} = 0.3/1.5 = 0.2$, so pull-out occurs at $20\%$ slip.

Step 2. Write $T \propto sR_2/(R_2^2+(sX_2)^2)$, dropping the common factor $V_1^2$, and evaluate at standstill, $s=1$: $T_{start} \propto \dfrac{1\times0.3}{0.3^2+(1\times1.5)^2} = \dfrac{0.3}{0.09+2.25} = \dfrac{0.3}{2.34} = 0.1282$.

Step 3. Evaluate the same expression at $s_{max}=0.2$: $T_{max} \propto \dfrac{0.2\times0.3}{0.3^2+(0.2\times1.5)^2} = \dfrac{0.06}{0.09+0.09} = \dfrac{0.06}{0.18} = 0.3333$. This matches the closed-form check $1/(2X_2) = 1/(2\times1.5) = 0.3333$ from Step 4 of the derivation, confirming the algebra.

Step 4. Take the ratio: $T_{max}/T_{start} = 0.3333/0.1282 \approx 2.6$.

Answer: pull-out torque occurs at $s = 0.2$ and is about $2.6$ times the starting torque for this motor — a sizeable margin between what the motor produces at switch-on and its true ceiling, which is why cage motors, despite having positive starting torque, often still need starting aids to limit inrush current rather than to boost torque.

Competitive stretch

If a supply voltage sag of $10\%$ occurs while this motor is starting, the starting torque falls to $(0.9)^2 = 0.81$, i.e. about $81\%$ of its rated value — a $19\%$ torque loss from only a $10\%$ voltage drop, since $T \propto V_1^2$. The slip at pull-out, $s_{max}=R_2/X_2=0.2$, is unaffected by voltage, because it depends only on $R_2$ and $X_2$; only the magnitude of torque at every slip scales with $V_1^2$, not the shape of the curve.

• Assuming maximum torque occurs at maximum slip ($s=1$, standstill) — pull-out actually occurs at the intermediate slip $s_{max}=R_2/X_2$, and starting torque is generally well below pull-out torque, as the worked example shows.
• Forgetting $T \propto V_1^2$: a modest supply voltage sag reduces available torque by a much larger fraction (a $10\%$ sag costs about $19\%$ of torque), which is why induction motors are especially sensitive to voltage dips at starting.
• Treating the high-slip side of the curve (beyond pull-out, toward $s=1$) as usable — that region is unstable, since torque falls as slip rises further, so any extra load there stalls the motor rather than settling to a new equilibrium.
• Confusing "starting torque" with "pull-out (breakdown) torque" — they are different points on the same curve with generally different numerical values, and only one of them (pull-out) is a true maximum.

Ask the tutor: want to see how adding external rotor resistance in a wound-rotor motor shifts the pull-out slip toward standstill while leaving the peak torque unchanged — the idea behind rotor-resistance starters covered next?