boss_notes/T-15c_Torque_Speed_Curves.md

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← T-15b: Running & Max Torque | 🏠 Index | T-16: Power Flow →

T-15c: Torque-Speed Characteristics

Section: B | Priority: 🟠 HIGH | Exam Frequency: 4/7 years Sources: Theraja Ch-34 (Art. 34.26-34.30), VK Mehta Ch-8 (Art. 8.16), Slides L-04

Why This Topic Matters

"Draw and explain the torque-speed/torque-slip curve" appeared in 4 out of 7 papers. It is a 3-5 mark question every time. You need to draw the curve accurately, label all key points, and explain the stable vs unstable regions. The effect of $R_{2}$ on the curve shape is also tested.

📝 Key Definitions

Torque-Slip Characteristic: "The curve showing the relation between the torque and the slip (or speed) of an induction motor is called its torque-slip (or torque-speed) characteristic." — Theraja, Art. 34.26

The Torque-Slip Curve

From the torque equation: $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$

Key points on the curve:

Point	Slip	Speed	Torque	Notes
Synchronous	$s = 0$	$N_{s}$	$T = 0$	No relative motion
Full load	$s_{f} \approx 0.03$	$\approx 0.97 N_{s}$	$T_{f}$	Normal operating point
Breakdown	$s_{m T} = R_{2} / X_{2}$	$N_{s} (1 - s_{m T})$	$T_{m a x}$	Peak torque
Standstill	$s = 1$	$0$	$T_{s t}$	Starting torque

Complete torque-speed characteristic showing all operating regions

Two Operating Regions

Low-Slip Region ( $0 < s < s_{m T}$ ): STABLE

When $s$ is small, $(s X_{2})^{2} ≪ R_{2}^{2}$ , so:

$T \approx \frac{k s E _{2}^{2} R _{2}}{R _{2}^{2}} = \frac{k s E _{2}^{2}}{R _{2}} \propto s$

Torque is approximately proportional to slip. The curve is nearly linear.

Why this region is stable: If load increases, the motor slows down slightly. Slip increases. Torque increases to match the new load. A self-correcting equilibrium.

High-Slip Region ( $s_{m T} < s \leq 1$ ): UNSTABLE

When $s$ is large, $(s X_{2})^{2} ≫ R_{2}^{2}$ , so:

$T \approx \frac{k s E _{2}^{2} R _{2}}{s ^{2} X _{2}^{2}} = \frac{k E _{2}^{2} R _{2}}{s X _{2}^{2}} \propto \frac{1}{s}$

Torque is inversely proportional to slip. The curve falls.

Why this region is unstable: If load increases, the motor slows down. Slip increases. But now torque DECREASES. The motor slows further. Torque drops more. The motor stalls.

Complete Torque-Speed Curve (All Three Regions)

Complete torque-speed curve showing motoring, generating, and braking regions

Region	Speed Range	Slip Range	Description
Motoring	$0 < N < N_{s}$	$0 < s < 1$	Normal operation
Generating	$N > N_{s}$	$s < 0$	Driven above $N_{s}$ , feeds power back
Braking (Plugging)	$N < 0$	$s > 1$	Rotor rotates against stator field

Effect of Rotor Resistance on Torque-Speed Curve

Family of torque-slip curves for varying R2

Increasing $R_{2}$ (by adding external resistance in wound-rotor motor):

$T_{m a x}$ stays the same ( $k E_{2}^{2} /2 X_{2}$ , no $R_{2}$ in formula)
$s_{m T}$ increases ( $s_{m T} = R_{2} / X_{2}$ , peak shifts to higher slip = lower speed)
Starting torque $T_{s t}$ increases (up to $R_{2} = X_{2}$ , where $T_{s t} = T_{m a x}$ )
Beyond $R_{2} = X_{2}$ : $T_{s t}$ decreases again

Practical use: In wound-rotor motors, external resistance gives maximum starting torque and smooth acceleration of heavy loads.

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Derive torque-slip characteristics of a 3-phase IM and explain.

Appeared: 2020 Q8(b) — (3 marks)

Full Answer:

Torque equation: $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$

At $s = 0$ : $T = 0$ (synchronous speed, no relative motion).

As $s$ increases from 0: Torque increases. In the low-slip region, $T \propto s$ (linear).

At $s = s_{m T} = R_{2} / X_{2}$ : $T = T_{m a x}$ (breakdown torque).

For $s > s_{m T}$ : Torque decreases. $(s X_{2})^{2}$ term dominates the denominator. $T \propto 1/ s$ .

At $s = 1$ : $T = T_{s t}$ (starting torque, typically 1.5-2 times $T_{f}$ ).

The motor operates stably only in the region $0 < s < s_{m T}$ . In this region, if load increases, slip increases and torque increases to match. Beyond $s_{m T}$ , the motor is unstable and will stall if load exceeds $T_{m a x}$ .

🎯 Q2: Draw the complete torque-speed curve of an induction motor.

Appeared: 2021 Q3(c) — (3 marks)

Full Answer:

The complete curve covers three regions:

Motoring region ( $0 < N < N_{s}$ , $0 < s < 1$ ):

At $N = 0$ ( $s = 1$ ): Starting torque $T_{s t}$
Maximum torque $T_{m a x}$ at $s_{m T} = R_{2} / X_{2}$
At $N = N_{s}$ ( $s = 0$ ): Torque = 0

Generating region ( $N > N_{s}$ , $s < 0$ ):

Rotor driven above synchronous speed by external prime mover
Negative torque: machine delivers power back to supply

Braking/Plugging region ( $N < 0$ , $s > 1$ ):

Rotor rotates opposite to stator RMF direction
Large braking torque developed
Used for rapid stopping

🎯 Q3: Explain the effect of rotor resistance on the torque-speed characteristic curve.

Appeared: 2024 Q7(b) — (3 marks)

Full Answer:

From $s_{m T} = R_{2} / X_{2}$ and $T_{m a x} = k E_{2}^{2} / (2 X_{2})$ :

Increasing rotor resistance $R_{2}$ :

$s_{m T}$ increases: The peak torque shifts to higher slip (lower speed).
$T_{m a x}$ remains unchanged: No $R_{2}$ in the formula.
Starting torque $T_{s t}$ increases as $R_{2}$ increases, up to $R_{2} = X_{2}$ where $T_{s t} = T_{m a x}$ .

By selecting appropriate external resistance, the wound-rotor motor can develop maximum torque at any desired speed. This is used for smooth starting of heavy loads and step-speed control.

🎯 Q4: Derive the torque expression $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$ .

Appeared: 2023 Q5(a) — (8 marks)

Full Answer:

At running slip $s$ , per-phase rotor quantities:

Rotor EMF: $E_{2 s} = s E_{2}$ . Rotor reactance: $X_{2 s} = s X_{2}$ .

Rotor current: $I_{2} = s E_{2} / R_{2}^{2} + s^{2} X_{2}^{2}$

Rotor power factor: $cos ϕ_{2} = R_{2} / R_{2}^{2} + s^{2} X_{2}^{2}$

Air-gap power using $R_{2} / s$ model:

$P_{g} = 3 I_{2}^{2} \cdot \frac{R _{2}}{s} = 3 \cdot \frac{s ^{2} E _{2}^{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}} \cdot \frac{R _{2}}{s} = \frac{3 s E _{2}^{2} R _{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}}$

Torque = air-gap power / synchronous angular speed:

$T = \frac{P _{g}}{ω _{s}} = \frac{P _{g}}{2 π n _{s}} = \frac{3 s E _{2}^{2} R _{2}}{( R _{2}^{2} + s ^{2} X _{2}^{2} ) \cdot 2 π n _{s}}$

$T = \frac{k s E _{2}^{2} R _{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}}, k = \frac{3}{2 π n _{s}}$

🎯 Q5: What happens to torque and speed if supply frequency increases suddenly?

Appeared: 2021 Q6(a) — (4 marks)

Full Answer:

If supply frequency $f$ increases suddenly (with voltage $V$ unchanged):

$N_{s}$ increases ( $N_{s} = 120 f / P$ ). Rotor cannot follow instantly, so slip increases momentarily.
$X_{2} = 2 π f L_{2}$ increases proportionally with $f$ .
$T_{m a x} = k E_{2}^{2} / (2 X_{2}) \propto V^{2} / f^{2}$ decreases. Since $X_{2} \propto f$ and $V$ is constant, $T_{m a x}$ drops significantly.
Operating slip increases because the motor must develop the same load torque with reduced $T_{m a x}$ . The motor operates closer to pull-out: less stable.
Speed may change: New $N_{s}$ is higher, but the rotor only partially catches up.

This is why VFDs change $V$ and $f$ together (constant V/f ratio) to maintain constant flux and torque capability.

🎯 Q6: 6-pole, 400V, 50 Hz, star. $R_{2}^{'} = 0.5 Ω$ , $X_{2}^{'} = 2.0 Ω$ . $s_{f} = 4%$ . Find $T_{s t}$ , $T_{f}$ , $T_{m a x}$ , efficiency (mech losses = 500 W).

Appeared: 2024 Q6(b) — (5 marks)

Full Answer:

$N_{s} = 1000$ rpm $= 50/3$ rps. $k = 3/ (2 π \times 50/3) = 0.02865$

$V_{ϕ} = 400/ 3 = 231.0$ V. $E_{2}^{'} \approx V_{ϕ} = 231.0$ V.

Starting torque ( $s = 1$ ): $T_{s t} = \frac{0.02865 \times 1 \times 23 1 ^{2} \times 0.5}{0.25 + 4.0} = \frac{764.7}{4.25} = 179.9 N-m$

Full-load torque ( $s = 0.04$ ): $T_{f} = \frac{0.02865 \times 0.04 \times 53361 \times 0.5}{0.25 + 0.0064} = \frac{30.55}{0.2564} = 119.2 N-m$

Maximum torque: $T_{m a x} = \frac{0.02865 \times 53361}{4.0} = 382.2 N-m$

Efficiency: $P_{g} = T_{f} \times ω_{s} = 119.2 \times 104.72 = 12483$ W

$P_{m} = (1 - s) P_{g} = 0.96 \times 12483 = 11983$ W

$P_{o u t} = 11983 - 500 = 11483$ W

$η \approx \frac{11483}{12483} \times 100 = 92.0%$

Exam Variants

Year	Question	Key Focus
2020 Q8(b)	Derive and explain T-s characteristics	Theory
2021 Q3(c)	Draw complete T-speed curve (all 3 regions)	Drawing
2021 Q6(a)	Effect of frequency change on torque/speed	Conceptual
2023 Q5(a)	Derive torque expression (8 marks)	Full derivation
2024 Q6(b)	6-pole numerical with efficiency	Numerical
2024 Q7(b)	Effect of $R_{2}$ on T-speed curve	Theory

⚡ Exam Tips & Common Mistakes

Label ALL key points on the curve: $T_{s t}$ , $T_{m a x}$ , $T_{f}$ , $s_{m T}$ , $N_{s}$ . Missing labels lose marks.
Stable region is LEFT of the peak (low slip side). Unstable is RIGHT of the peak.
Draw the curve starting from $s = 0$ (right side) going to $s = 1$ (left side) if plotting vs speed. Or $s = 0$ (left) to $s = 1$ (right) if plotting vs slip. Be consistent.
When sketching effect of $R_{2}$ : Draw 3-4 curves with the SAME $T_{m a x}$ but peaks at different slips.

T-15a: Starting Torque — The $s = 1$ end of the curve
T-15b: Running & Max Torque — The equations behind the curve
T-20: Speed Control — Practical use of $R_{2}$ variation

← T-15b: Running & Max Torque | 🏠 Index | T-16: Power Flow →

boss_notes/T-15c_Torque_Speed_Curves.md

9.7 KB · commit d2f3b6e

← T-15b: Running & Max Torque | 🏠 Index | T-16: Power Flow →

T-15c: Torque-Speed Characteristics

Section: B | Priority: 🟠 HIGH | Exam Frequency: 4/7 years Sources: Theraja Ch-34 (Art. 34.26-34.30), VK Mehta Ch-8 (Art. 8.16), Slides L-04

Why This Topic Matters

📝 Key Definitions

Torque-Slip Characteristic: "The curve showing the relation between the torque and the slip (or speed) of an induction motor is called its torque-slip (or torque-speed) characteristic." — Theraja, Art. 34.26

The Torque-Slip Curve

From the torque equation: $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$

Key points on the curve:

Point	Slip	Speed	Torque	Notes
Synchronous	$s = 0$	$N_{s}$	$T = 0$	No relative motion
Full load	$s_{f} \approx 0.03$	$\approx 0.97 N_{s}$	$T_{f}$	Normal operating point
Breakdown	$s_{m T} = R_{2} / X_{2}$	$N_{s} (1 - s_{m T})$	$T_{m a x}$	Peak torque
Standstill	$s = 1$	$0$	$T_{s t}$	Starting torque

Complete torque-speed characteristic showing all operating regions

Two Operating Regions

Low-Slip Region ( $0 < s < s_{m T}$ ): STABLE

When $s$ is small, $(s X_{2})^{2} ≪ R_{2}^{2}$ , so:

$T \approx \frac{k s E _{2}^{2} R _{2}}{R _{2}^{2}} = \frac{k s E _{2}^{2}}{R _{2}} \propto s$

Torque is approximately proportional to slip. The curve is nearly linear.

Why this region is stable: If load increases, the motor slows down slightly. Slip increases. Torque increases to match the new load. A self-correcting equilibrium.

High-Slip Region ( $s_{m T} < s \leq 1$ ): UNSTABLE

When $s$ is large, $(s X_{2})^{2} ≫ R_{2}^{2}$ , so:

$T \approx \frac{k s E _{2}^{2} R _{2}}{s ^{2} X _{2}^{2}} = \frac{k E _{2}^{2} R _{2}}{s X _{2}^{2}} \propto \frac{1}{s}$

Torque is inversely proportional to slip. The curve falls.

Why this region is unstable: If load increases, the motor slows down. Slip increases. But now torque DECREASES. The motor slows further. Torque drops more. The motor stalls.

Complete Torque-Speed Curve (All Three Regions)

Complete torque-speed curve showing motoring, generating, and braking regions

Region	Speed Range	Slip Range	Description
Motoring	$0 < N < N_{s}$	$0 < s < 1$	Normal operation
Generating	$N > N_{s}$	$s < 0$	Driven above $N_{s}$ , feeds power back
Braking (Plugging)	$N < 0$	$s > 1$	Rotor rotates against stator field

Effect of Rotor Resistance on Torque-Speed Curve

Family of torque-slip curves for varying R2

Increasing $R_{2}$ (by adding external resistance in wound-rotor motor):

$T_{m a x}$ stays the same ( $k E_{2}^{2} /2 X_{2}$ , no $R_{2}$ in formula)
$s_{m T}$ increases ( $s_{m T} = R_{2} / X_{2}$ , peak shifts to higher slip = lower speed)
Starting torque $T_{s t}$ increases (up to $R_{2} = X_{2}$ , where $T_{s t} = T_{m a x}$ )
Beyond $R_{2} = X_{2}$ : $T_{s t}$ decreases again

Practical use: In wound-rotor motors, external resistance gives maximum starting torque and smooth acceleration of heavy loads.

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Derive torque-slip characteristics of a 3-phase IM and explain.

Appeared: 2020 Q8(b) — (3 marks)

Full Answer:

Torque equation: $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$

At $s = 0$ : $T = 0$ (synchronous speed, no relative motion).

As $s$ increases from 0: Torque increases. In the low-slip region, $T \propto s$ (linear).

At $s = s_{m T} = R_{2} / X_{2}$ : $T = T_{m a x}$ (breakdown torque).

For $s > s_{m T}$ : Torque decreases. $(s X_{2})^{2}$ term dominates the denominator. $T \propto 1/ s$ .

At $s = 1$ : $T = T_{s t}$ (starting torque, typically 1.5-2 times $T_{f}$ ).

🎯 Q2: Draw the complete torque-speed curve of an induction motor.

Appeared: 2021 Q3(c) — (3 marks)

Full Answer:

The complete curve covers three regions:

Motoring region ( $0 < N < N_{s}$ , $0 < s < 1$ ):

At $N = 0$ ( $s = 1$ ): Starting torque $T_{s t}$
Maximum torque $T_{m a x}$ at $s_{m T} = R_{2} / X_{2}$
At $N = N_{s}$ ( $s = 0$ ): Torque = 0

Generating region ( $N > N_{s}$ , $s < 0$ ):

Rotor driven above synchronous speed by external prime mover
Negative torque: machine delivers power back to supply

Braking/Plugging region ( $N < 0$ , $s > 1$ ):

Rotor rotates opposite to stator RMF direction
Large braking torque developed
Used for rapid stopping

🎯 Q3: Explain the effect of rotor resistance on the torque-speed characteristic curve.

Appeared: 2024 Q7(b) — (3 marks)

Full Answer:

From $s_{m T} = R_{2} / X_{2}$ and $T_{m a x} = k E_{2}^{2} / (2 X_{2})$ :

Increasing rotor resistance $R_{2}$ :

$s_{m T}$ increases: The peak torque shifts to higher slip (lower speed).
$T_{m a x}$ remains unchanged: No $R_{2}$ in the formula.
Starting torque $T_{s t}$ increases as $R_{2}$ increases, up to $R_{2} = X_{2}$ where $T_{s t} = T_{m a x}$ .

By selecting appropriate external resistance, the wound-rotor motor can develop maximum torque at any desired speed. This is used for smooth starting of heavy loads and step-speed control.

🎯 Q4: Derive the torque expression $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$ .

Appeared: 2023 Q5(a) — (8 marks)

Full Answer:

At running slip $s$ , per-phase rotor quantities:

Rotor EMF: $E_{2 s} = s E_{2}$ . Rotor reactance: $X_{2 s} = s X_{2}$ .

Rotor current: $I_{2} = s E_{2} / R_{2}^{2} + s^{2} X_{2}^{2}$

Rotor power factor: $cos ϕ_{2} = R_{2} / R_{2}^{2} + s^{2} X_{2}^{2}$

Air-gap power using $R_{2} / s$ model:

$P_{g} = 3 I_{2}^{2} \cdot \frac{R _{2}}{s} = 3 \cdot \frac{s ^{2} E _{2}^{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}} \cdot \frac{R _{2}}{s} = \frac{3 s E _{2}^{2} R _{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}}$

Torque = air-gap power / synchronous angular speed:

$T = \frac{P _{g}}{ω _{s}} = \frac{P _{g}}{2 π n _{s}} = \frac{3 s E _{2}^{2} R _{2}}{( R _{2}^{2} + s ^{2} X _{2}^{2} ) \cdot 2 π n _{s}}$

$T = \frac{k s E _{2}^{2} R _{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}}, k = \frac{3}{2 π n _{s}}$

🎯 Q5: What happens to torque and speed if supply frequency increases suddenly?

Appeared: 2021 Q6(a) — (4 marks)

Full Answer:

If supply frequency $f$ increases suddenly (with voltage $V$ unchanged):

$N_{s}$ increases ( $N_{s} = 120 f / P$ ). Rotor cannot follow instantly, so slip increases momentarily.
$X_{2} = 2 π f L_{2}$ increases proportionally with $f$ .
$T_{m a x} = k E_{2}^{2} / (2 X_{2}) \propto V^{2} / f^{2}$ decreases. Since $X_{2} \propto f$ and $V$ is constant, $T_{m a x}$ drops significantly.
Operating slip increases because the motor must develop the same load torque with reduced $T_{m a x}$ . The motor operates closer to pull-out: less stable.
Speed may change: New $N_{s}$ is higher, but the rotor only partially catches up.

This is why VFDs change $V$ and $f$ together (constant V/f ratio) to maintain constant flux and torque capability.

🎯 Q6: 6-pole, 400V, 50 Hz, star. $R_{2}^{'} = 0.5 Ω$ , $X_{2}^{'} = 2.0 Ω$ . $s_{f} = 4%$ . Find $T_{s t}$ , $T_{f}$ , $T_{m a x}$ , efficiency (mech losses = 500 W).

Appeared: 2024 Q6(b) — (5 marks)

Full Answer:

$N_{s} = 1000$ rpm $= 50/3$ rps. $k = 3/ (2 π \times 50/3) = 0.02865$

$V_{ϕ} = 400/ 3 = 231.0$ V. $E_{2}^{'} \approx V_{ϕ} = 231.0$ V.

Starting torque ( $s = 1$ ): $T_{s t} = \frac{0.02865 \times 1 \times 23 1 ^{2} \times 0.5}{0.25 + 4.0} = \frac{764.7}{4.25} = 179.9 N-m$

Full-load torque ( $s = 0.04$ ): $T_{f} = \frac{0.02865 \times 0.04 \times 53361 \times 0.5}{0.25 + 0.0064} = \frac{30.55}{0.2564} = 119.2 N-m$

Maximum torque: $T_{m a x} = \frac{0.02865 \times 53361}{4.0} = 382.2 N-m$

Efficiency: $P_{g} = T_{f} \times ω_{s} = 119.2 \times 104.72 = 12483$ W

$P_{m} = (1 - s) P_{g} = 0.96 \times 12483 = 11983$ W

$P_{o u t} = 11983 - 500 = 11483$ W

$η \approx \frac{11483}{12483} \times 100 = 92.0%$

Exam Variants

Year	Question	Key Focus
2020 Q8(b)	Derive and explain T-s characteristics	Theory
2021 Q3(c)	Draw complete T-speed curve (all 3 regions)	Drawing
2021 Q6(a)	Effect of frequency change on torque/speed	Conceptual
2023 Q5(a)	Derive torque expression (8 marks)	Full derivation
2024 Q6(b)	6-pole numerical with efficiency	Numerical
2024 Q7(b)	Effect of $R_{2}$ on T-speed curve	Theory

⚡ Exam Tips & Common Mistakes

Label ALL key points on the curve: $T_{s t}$ , $T_{m a x}$ , $T_{f}$ , $s_{m T}$ , $N_{s}$ . Missing labels lose marks.
Stable region is LEFT of the peak (low slip side). Unstable is RIGHT of the peak.
Draw the curve starting from $s = 0$ (right side) going to $s = 1$ (left side) if plotting vs speed. Or $s = 0$ (left) to $s = 1$ (right) if plotting vs slip. Be consistent.
When sketching effect of $R_{2}$ : Draw 3-4 curves with the SAME $T_{m a x}$ but peaks at different slips.

T-15a: Starting Torque — The $s = 1$ end of the curve
T-15b: Running & Max Torque — The equations behind the curve
T-20: Speed Control — Practical use of $R_{2}$ variation

← T-15b: Running & Max Torque | 🏠 Index | T-16: Power Flow →

T-15c: Torque-Speed Characteristics

Why This Topic Matters

📝 Key Definitions

The Torque-Slip Curve

Two Operating Regions

Low-Slip Region (0<s<smT​): STABLE

High-Slip Region (smT​<s≤1): UNSTABLE

Complete Torque-Speed Curve (All Three Regions)

Effect of Rotor Resistance on Torque-Speed Curve

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Derive torque-slip characteristics of a 3-phase IM and explain.

🎯 Q2: Draw the complete torque-speed curve of an induction motor.

🎯 Q3: Explain the effect of rotor resistance on the torque-speed characteristic curve.

🎯 Q4: Derive the torque expression T=ksE22​R2​/(R22​+s2X22​).

🎯 Q5: What happens to torque and speed if supply frequency increases suddenly?

🎯 Q6: 6-pole, 400V, 50 Hz, star. R2′​=0.5Ω, X2′​=2.0Ω. sf​=4%. Find Tst​, Tf​, Tmax​, efficiency (mech losses = 500 W).

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

T-15c: Torque-Speed Characteristics

Why This Topic Matters

📝 Key Definitions

The Torque-Slip Curve

Two Operating Regions

Low-Slip Region (0<s<smT​): STABLE

High-Slip Region (smT​<s≤1): UNSTABLE

Complete Torque-Speed Curve (All Three Regions)

Effect of Rotor Resistance on Torque-Speed Curve

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Derive torque-slip characteristics of a 3-phase IM and explain.

🎯 Q2: Draw the complete torque-speed curve of an induction motor.

🎯 Q3: Explain the effect of rotor resistance on the torque-speed characteristic curve.

🎯 Q4: Derive the torque expression T=ksE22​R2​/(R22​+s2X22​).

🎯 Q5: What happens to torque and speed if supply frequency increases suddenly?

🎯 Q6: 6-pole, 400V, 50 Hz, star. R2′​=0.5Ω, X2′​=2.0Ω. sf​=4%. Find Tst​, Tf​, Tmax​, efficiency (mech losses = 500 W).

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

Low-Slip Region ( $0 < s < s_{m T}$ ): STABLE

High-Slip Region ( $s_{m T} < s \leq 1$ ): UNSTABLE

🎯 Q4: Derive the torque expression $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$ .

🎯 Q6: 6-pole, 400V, 50 Hz, star. $R_{2}^{'} = 0.5 Ω$ , $X_{2}^{'} = 2.0 Ω$ . $s_{f} = 4%$ . Find $T_{s t}$ , $T_{f}$ , $T_{m a x}$ , efficiency (mech losses = 500 W).

Low-Slip Region ( $0 < s < s_{m T}$ ): STABLE

High-Slip Region ( $s_{m T} < s \leq 1$ ): UNSTABLE

🎯 Q4: Derive the torque expression $T = k s E_{2}^{2} R_{2} / (R_{2}^{2} + s^{2} X_{2}^{2})$ .

🎯 Q6: 6-pole, 400V, 50 Hz, star. $R_{2}^{'} = 0.5 Ω$ , $X_{2}^{'} = 2.0 Ω$ . $s_{f} = 4%$ . Find $T_{s t}$ , $T_{f}$ , $T_{m a x}$ , efficiency (mech losses = 500 W).