boss_notes/T-16_Power_Flow.md

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← T-15c: Torque-Speed Curves | 🏠 Index | T-17a: No-Load Test →

T-16: Power Flow & Rotor Power Division

Section: B | Priority: 🟡 MEDIUM | Exam Frequency: 2/7 years Sources: Theraja Ch-34 (Art. 34.33-34.38), VK Mehta Ch-8, Slides L-06

Why This Topic Matters

The power flow proof ( $P_{g} : P_{C u, r} : P_{m} = 1 : s : (1 - s)$ ) appeared in 3 out of 7 papers (2018, 2023, 2024). The rotor efficiency proof ( $η_{r} = 1 - s$ ) appeared in 2021 and 2023. The synchronous watt definition appeared in 2021. These are short, high-scoring proofs (3-4 marks each). The power flow numerical appeared in 2020.

📝 Key Definitions

Air-Gap Power ( $P_{g}$ ): "The power transferred from the stator to the rotor across the air gap is called the air-gap power. It is given by $P_{g} = 3 I_{2}^{2} R_{2} / s$ ." — Theraja, Art. 34.34

Synchronous Watt: "A synchronous watt is the torque which, at the synchronous speed of the machine under consideration, would give a power dissipation of 1 watt. Torque in synchronous watts equals the air-gap power in watts." — Theraja, Art. 34.35

Rotor Efficiency: "The rotor efficiency is the ratio of the mechanical power developed to the air-gap power: $η_{rotor} = (1 - s)$ ." — VK Mehta, Ch-8

Power Flow Diagram

$Stator Input P_{1} - P_{s, C u} - P_{F e} Air-gap P_{g} - P_{r, C u} Mech Power P_{m} - P_{f & w} Output P_{o u t}$

Power flow diagram of an induction motor

From the equivalent circuit:

Stator input: $P_{1} = 3 V_{1} I_{1} cos ϕ_{1}$
Stator copper loss: $P_{s, C u} = 3 I_{1}^{2} R_{1}$
Stator iron loss: $P_{F e} \approx 3 E_{1}^{2} / R_{c}$
Air-gap power: $P_{g} = P_{1} - P_{s, C u} - P_{F e} = 3 I_{2}^{2} \cdot R_{2} / s$
Rotor copper loss: $P_{r, C u} = 3 I_{2}^{2} R_{2}$
Mechanical power: $P_{m} = P_{g} - P_{r, C u} = 3 I_{2}^{2} R_{2} (1 - s) / s$
Output: $P_{o u t} = P_{m} - P_{f & w} - P_{stray}$

The Golden Power Ratio

This is the most important relationship in IM power analysis.

Proof: $P_{C u, r} = s P_{g}$

From the equivalent circuit:

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

$P_{C u, r} = 3 I_{2}^{2} R_{2} = s \cdot 3 I_{2}^{2} \cdot \frac{R _{2}}{s} = s \cdot P_{g}$

$P_{C u, r} = s P_{g}$

Mechanical power:

$P_{m} = P_{g} - P_{C u, r} = P_{g} - s P_{g} = (1 - s) P_{g}$

The ratio:

$P_{g} : P_{C u, r} : P_{m} = 1 : s : (1 - s)$

Power stages block diagram

Physical meaning: For every 1 watt crossing the air gap, $s$ watts become rotor heat and $(1 - s)$ watts become mechanical power. At $s = 0.04$ (typical full load), 96% of air-gap power becomes mechanical, 4% becomes heat.

Rotor Efficiency

$η_{rotor} = \frac{P _{m}}{P _{g}} = \frac{( 1 - s ) P _{g}}{P _{g}} = (1 - s)$

At $s = 0.04$ : $η_{rotor} = 96%$ . This is why induction motors run at low slip.

Synchronous Watt

One synchronous watt is the torque that develops one watt of power at synchronous speed.

$T (synchronous watts) = P_{g} (watts)$

To convert to N-m:

$T = \frac{P _{g}}{ω _{s}} = \frac{P _{g}}{2 π N _{s} /60} N-m$

Example: $P_{g} = 5000$ W, $N_{s} = 1000$ rpm:

$T = \frac{5000}{2 π \times 1000/60} = \frac{5000}{104.7} = 47.75 N-m$

Worked Example (PYQ 2020)

2020 Q6(d): 400V, 50 Hz, 6-pole IM. $P_{g} = 75$ kW. Rotor EMF makes 100 alternations per minute. Find slip, rotor speed, Cu losses per phase, mechanical power.

Rotor frequency: $f_{r} = 100/60 = 5/3$ Hz

(i) Slip: $s = f_{r} / f = (5/3) /50 = 1/30 = 0.0333 = 3.33%$

(ii) Rotor speed: $N_{s} = 120 \times 50/6 = 1000$ rpm

$N = N_{s} (1 - s) = 1000 (1 - 0.0333) = 966.7 rpm$

(iii) Rotor Cu losses (total): $P_{C u, r} = s P_{g} = 0.0333 \times 75000 = 2500$ W

Per phase: $P_{C u, r} /3 = 833.3 W$

(iv) Mechanical power: $P_{m} = (1 - s) P_{g} = 0.9667 \times 75000 = 72500 W = 72.5 kW$

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

Appeared: 2018 Q5(b), 2023 Q6(a), 2024 Q5(b) — (4 marks)

Full Answer:

From the equivalent circuit, air-gap power: $P_{g} = 3 I_{2}^{2} \cdot R_{2} / s$

Rotor copper loss: $P_{C u, r} = 3 I_{2}^{2} R_{2} = s \cdot 3 I_{2}^{2} \cdot R_{2} / s = s P_{g}$

$P_{C u, r} = s P_{g}$

Mechanical power: $P_{m} = P_{g} - P_{C u, r} = P_{g} - s P_{g} = (1 - s) P_{g}$

$P_{m} : P_{C u, r} : P_{g} = (1 - s) P_{g} : s P_{g} : P_{g} = (1 - s) : s : 1$

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .

Appeared: 2021 Q6(c), 2023 Q6(c) — (3-4 marks)

Full Answer:

Rotor efficiency is the ratio of mechanical power developed to electrical power input to the rotor (air-gap power).

$η_{rotor} = \frac{P _{m}}{P _{g}} = \frac{( 1 - s ) P _{g}}{P _{g}} = (1 - s)$

At $s = 0.04$ (typical full load): $η_{rotor} = 96%$ .

For every unit of power crossing the air gap, fraction $s$ is wasted as rotor copper heat and fraction $(1 - s)$ becomes mechanical work. Low slip means high rotor efficiency. This is why induction motors are designed to operate at small slip.

🎯 Q3: Define synchronous watt with an example.

Appeared: 2021 Q8(a) — (3 marks)

Full Answer:

Synchronous watt: A unit of torque used in induction motor analysis. One synchronous watt is the torque that develops one watt of power at synchronous speed.

$T (synchronous watts) = P_{g} (watts)$

Conversion: $T (N-m) = P_{g} / ω_{s} = P_{g} / (2 π N_{s} /60)$

Example: Motor has $P_{g} = 5000$ W, $N_{s} = 1000$ rpm ( $ω_{s} = 104.7$ rad/s):

$T = 5000/104.7 = 47.75 N-m$

The torque is 5000 synchronous watts, which equals 47.75 N-m at this synchronous speed.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

Appeared: 2020 Q6(b) — (4 marks)

Full Answer:

From the per-phase equivalent circuit:

Stator input: $P_{1} = 3 V_{1} I_{1} cos ϕ_{1}$

Stator copper loss: $P_{s, C u} = 3 I_{1}^{2} R_{1}$

Stator iron loss: $P_{F e} \approx 3 V_{1}^{2} / R_{c}$

Air-gap power: $P_{g} = P_{1} - P_{s, C u} - P_{F e} = 3 I_{2}^{2} \cdot R_{2} / s$

Rotor copper loss: $P_{C u, r} = 3 I_{2}^{2} R_{2} = s P_{g}$

Mechanical power: $P_{m} = P_{g} - P_{C u, r} = (1 - s) P_{g} = 3 I_{2}^{2} R_{2} (1 - s) / s$

Output: $P_{o u t} = P_{m} - P_{f & w}$

Summary: $P_{g} : P_{C u, r} : P_{m} = 1 : s : (1 - s)$

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Appeared: 2021 Q6(b) — (4 marks)

Full Answer:

For small slip (low-slip approximation): $T \approx k E_{2}^{2} s / R_{2} \propto s V^{2} / R_{2}$

At constant torque: $s_{1} V_{1}^{2} = s_{2} V_{2}^{2}$

$s_{2} = s_{1} (V_{1} / V_{2})^{2} = s_{1} \times (1/0.9)^{2} = s_{1} \times 1.2346$

Rotor Cu loss: $P_{C u} = s P_{g}$ , and $P_{g} = T ω_{s}$ (constant).

$P_{C u, new} / P_{C u, old} = s_{2} / s_{1} = 1.2346$

Increase: $(1.2346 - 1) \times 100% = 23.46%$

Cu losses increase by about 23.5% when voltage drops to 90%.

Exam Variants

Year	Question	Key Result
2018 Q5(b)	Show $P_{C u, r} = s P_{g}$	Proof
2020 Q6(b)	Derive power equations	Full chain
2020 Q6(d)	Power flow numerical	$s = 3.33%$ , $P_{m} = 72.5$ kW
2021 Q6(b)	90% voltage, Cu loss increase	23.46% increase
2021 Q6(c)	Rotor efficiency	$η_{r} = 1 - s$
2021 Q8(a)	Synchronous watt	Definition + example
2023 Q6(a)	Show power ratio	$1 : s : (1 - s)$
2023 Q6(c)	Rotor efficiency proof	$η_{r} = 1 - s$
2024 Q5(b)	Show $P_{C u, r} = s P_{g}$ + ratio	Same proof

⚡ Exam Tips & Common Mistakes

The ratio is $1 : s : (1 - s)$ , not $1 : (1 - s) : s$ . Air gap : Cu loss : Mechanical. Don't swap the last two.
Rotor efficiency is $(1 - s)$ , not $(1 - s) %$ . Express as per-unit or convert explicitly.
Synchronous watt is numerically equal to $P_{g}$ in watts. The conversion to N-m requires dividing by $ω_{s}$ .
For power flow numericals: Always start by finding slip from rotor frequency ( $s = f_{r} / f$ ).

T-14: IM Equivalent Circuit — Source of power equations
T-15b: Running & Max Torque — Torque from air-gap power
T-17a: No-Load Test — Measures fixed losses

← T-15c: Torque-Speed Curves | 🏠 Index | T-17a: No-Load Test →

boss_notes/T-16_Power_Flow.md

8.4 KB · commit d2f3b6e

← T-15c: Torque-Speed Curves | 🏠 Index | T-17a: No-Load Test →

T-16: Power Flow & Rotor Power Division

Section: B | Priority: 🟡 MEDIUM | Exam Frequency: 2/7 years Sources: Theraja Ch-34 (Art. 34.33-34.38), VK Mehta Ch-8, Slides L-06

Why This Topic Matters

📝 Key Definitions

Air-Gap Power ( $P_{g}$ ): "The power transferred from the stator to the rotor across the air gap is called the air-gap power. It is given by $P_{g} = 3 I_{2}^{2} R_{2} / s$ ." — Theraja, Art. 34.34

Synchronous Watt: "A synchronous watt is the torque which, at the synchronous speed of the machine under consideration, would give a power dissipation of 1 watt. Torque in synchronous watts equals the air-gap power in watts." — Theraja, Art. 34.35

Rotor Efficiency: "The rotor efficiency is the ratio of the mechanical power developed to the air-gap power: $η_{rotor} = (1 - s)$ ." — VK Mehta, Ch-8

Power Flow Diagram

$Stator Input P_{1} - P_{s, C u} - P_{F e} Air-gap P_{g} - P_{r, C u} Mech Power P_{m} - P_{f & w} Output P_{o u t}$

Power flow diagram of an induction motor

From the equivalent circuit:

Stator input: $P_{1} = 3 V_{1} I_{1} cos ϕ_{1}$
Stator copper loss: $P_{s, C u} = 3 I_{1}^{2} R_{1}$
Stator iron loss: $P_{F e} \approx 3 E_{1}^{2} / R_{c}$
Air-gap power: $P_{g} = P_{1} - P_{s, C u} - P_{F e} = 3 I_{2}^{2} \cdot R_{2} / s$
Rotor copper loss: $P_{r, C u} = 3 I_{2}^{2} R_{2}$
Mechanical power: $P_{m} = P_{g} - P_{r, C u} = 3 I_{2}^{2} R_{2} (1 - s) / s$
Output: $P_{o u t} = P_{m} - P_{f & w} - P_{stray}$

The Golden Power Ratio

This is the most important relationship in IM power analysis.

Proof: $P_{C u, r} = s P_{g}$

From the equivalent circuit:

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

$P_{C u, r} = 3 I_{2}^{2} R_{2} = s \cdot 3 I_{2}^{2} \cdot \frac{R _{2}}{s} = s \cdot P_{g}$

$P_{C u, r} = s P_{g}$

Mechanical power:

$P_{m} = P_{g} - P_{C u, r} = P_{g} - s P_{g} = (1 - s) P_{g}$

The ratio:

$P_{g} : P_{C u, r} : P_{m} = 1 : s : (1 - s)$

Power stages block diagram

Rotor Efficiency

$η_{rotor} = \frac{P _{m}}{P _{g}} = \frac{( 1 - s ) P _{g}}{P _{g}} = (1 - s)$

At $s = 0.04$ : $η_{rotor} = 96%$ . This is why induction motors run at low slip.

Synchronous Watt

One synchronous watt is the torque that develops one watt of power at synchronous speed.

$T (synchronous watts) = P_{g} (watts)$

To convert to N-m:

$T = \frac{P _{g}}{ω _{s}} = \frac{P _{g}}{2 π N _{s} /60} N-m$

Example: $P_{g} = 5000$ W, $N_{s} = 1000$ rpm:

$T = \frac{5000}{2 π \times 1000/60} = \frac{5000}{104.7} = 47.75 N-m$

Worked Example (PYQ 2020)

2020 Q6(d): 400V, 50 Hz, 6-pole IM. $P_{g} = 75$ kW. Rotor EMF makes 100 alternations per minute. Find slip, rotor speed, Cu losses per phase, mechanical power.

Rotor frequency: $f_{r} = 100/60 = 5/3$ Hz

(i) Slip: $s = f_{r} / f = (5/3) /50 = 1/30 = 0.0333 = 3.33%$

(ii) Rotor speed: $N_{s} = 120 \times 50/6 = 1000$ rpm

$N = N_{s} (1 - s) = 1000 (1 - 0.0333) = 966.7 rpm$

(iii) Rotor Cu losses (total): $P_{C u, r} = s P_{g} = 0.0333 \times 75000 = 2500$ W

Per phase: $P_{C u, r} /3 = 833.3 W$

(iv) Mechanical power: $P_{m} = (1 - s) P_{g} = 0.9667 \times 75000 = 72500 W = 72.5 kW$

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

Appeared: 2018 Q5(b), 2023 Q6(a), 2024 Q5(b) — (4 marks)

Full Answer:

From the equivalent circuit, air-gap power: $P_{g} = 3 I_{2}^{2} \cdot R_{2} / s$

Rotor copper loss: $P_{C u, r} = 3 I_{2}^{2} R_{2} = s \cdot 3 I_{2}^{2} \cdot R_{2} / s = s P_{g}$

$P_{C u, r} = s P_{g}$

Mechanical power: $P_{m} = P_{g} - P_{C u, r} = P_{g} - s P_{g} = (1 - s) P_{g}$

$P_{m} : P_{C u, r} : P_{g} = (1 - s) P_{g} : s P_{g} : P_{g} = (1 - s) : s : 1$

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .

Appeared: 2021 Q6(c), 2023 Q6(c) — (3-4 marks)

Full Answer:

Rotor efficiency is the ratio of mechanical power developed to electrical power input to the rotor (air-gap power).

$η_{rotor} = \frac{P _{m}}{P _{g}} = \frac{( 1 - s ) P _{g}}{P _{g}} = (1 - s)$

At $s = 0.04$ (typical full load): $η_{rotor} = 96%$ .

🎯 Q3: Define synchronous watt with an example.

Appeared: 2021 Q8(a) — (3 marks)

Full Answer:

Synchronous watt: A unit of torque used in induction motor analysis. One synchronous watt is the torque that develops one watt of power at synchronous speed.

$T (synchronous watts) = P_{g} (watts)$

Conversion: $T (N-m) = P_{g} / ω_{s} = P_{g} / (2 π N_{s} /60)$

Example: Motor has $P_{g} = 5000$ W, $N_{s} = 1000$ rpm ( $ω_{s} = 104.7$ rad/s):

$T = 5000/104.7 = 47.75 N-m$

The torque is 5000 synchronous watts, which equals 47.75 N-m at this synchronous speed.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

Appeared: 2020 Q6(b) — (4 marks)

Full Answer:

From the per-phase equivalent circuit:

Stator input: $P_{1} = 3 V_{1} I_{1} cos ϕ_{1}$

Stator copper loss: $P_{s, C u} = 3 I_{1}^{2} R_{1}$

Stator iron loss: $P_{F e} \approx 3 V_{1}^{2} / R_{c}$

Air-gap power: $P_{g} = P_{1} - P_{s, C u} - P_{F e} = 3 I_{2}^{2} \cdot R_{2} / s$

Rotor copper loss: $P_{C u, r} = 3 I_{2}^{2} R_{2} = s P_{g}$

Mechanical power: $P_{m} = P_{g} - P_{C u, r} = (1 - s) P_{g} = 3 I_{2}^{2} R_{2} (1 - s) / s$

Output: $P_{o u t} = P_{m} - P_{f & w}$

Summary: $P_{g} : P_{C u, r} : P_{m} = 1 : s : (1 - s)$

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Appeared: 2021 Q6(b) — (4 marks)

Full Answer:

For small slip (low-slip approximation): $T \approx k E_{2}^{2} s / R_{2} \propto s V^{2} / R_{2}$

At constant torque: $s_{1} V_{1}^{2} = s_{2} V_{2}^{2}$

$s_{2} = s_{1} (V_{1} / V_{2})^{2} = s_{1} \times (1/0.9)^{2} = s_{1} \times 1.2346$

Rotor Cu loss: $P_{C u} = s P_{g}$ , and $P_{g} = T ω_{s}$ (constant).

$P_{C u, new} / P_{C u, old} = s_{2} / s_{1} = 1.2346$

Increase: $(1.2346 - 1) \times 100% = 23.46%$

Cu losses increase by about 23.5% when voltage drops to 90%.

Exam Variants

Year	Question	Key Result
2018 Q5(b)	Show $P_{C u, r} = s P_{g}$	Proof
2020 Q6(b)	Derive power equations	Full chain
2020 Q6(d)	Power flow numerical	$s = 3.33%$ , $P_{m} = 72.5$ kW
2021 Q6(b)	90% voltage, Cu loss increase	23.46% increase
2021 Q6(c)	Rotor efficiency	$η_{r} = 1 - s$
2021 Q8(a)	Synchronous watt	Definition + example
2023 Q6(a)	Show power ratio	$1 : s : (1 - s)$
2023 Q6(c)	Rotor efficiency proof	$η_{r} = 1 - s$
2024 Q5(b)	Show $P_{C u, r} = s P_{g}$ + ratio	Same proof

⚡ Exam Tips & Common Mistakes

The ratio is $1 : s : (1 - s)$ , not $1 : (1 - s) : s$ . Air gap : Cu loss : Mechanical. Don't swap the last two.
Rotor efficiency is $(1 - s)$ , not $(1 - s) %$ . Express as per-unit or convert explicitly.
Synchronous watt is numerically equal to $P_{g}$ in watts. The conversion to N-m requires dividing by $ω_{s}$ .
For power flow numericals: Always start by finding slip from rotor frequency ( $s = f_{r} / f$ ).

T-14: IM Equivalent Circuit — Source of power equations
T-15b: Running & Max Torque — Torque from air-gap power
T-17a: No-Load Test — Measures fixed losses

← T-15c: Torque-Speed Curves | 🏠 Index | T-17a: No-Load Test →

azazarfin60/Electrical_Machine_I_Semester_Final

boss_notes/T-16_Power_Flow.md

T-16: Power Flow & Rotor Power Division

Why This Topic Matters

📝 Key Definitions

Power Flow Diagram

The Golden Power Ratio

Rotor Efficiency

Synchronous Watt

Worked Example (PYQ 2020)

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .

🎯 Q3: Define synchronous watt with an example.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Exam Variants

⚡ Exam Tips & Common Mistakes

Report a problem

azazarfin60/Electrical_Machine_I_Semester_Final

boss_notes/T-16_Power_Flow.md

T-16: Power Flow & Rotor Power Division

Why This Topic Matters

📝 Key Definitions

Power Flow Diagram

The Golden Power Ratio

Rotor Efficiency

Synchronous Watt

Worked Example (PYQ 2020)

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .

🎯 Q3: Define synchronous watt with an example.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Exam Variants

⚡ Exam Tips & Common Mistakes

Report a problem

T-16: Power Flow & Rotor Power Division

Why This Topic Matters

📝 Key Definitions

Power Flow Diagram

The Golden Power Ratio

Rotor Efficiency

Synchronous Watt

Worked Example (PYQ 2020)

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = s× air-gap power. Also show Pm​:PCu,r​:Pg​=(1−s):s:1.

🎯 Q2: What is rotor efficiency? Show that ηrotor​=(1−s).

🎯 Q3: Define synchronous watt with an example.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

T-16: Power Flow & Rotor Power Division

Why This Topic Matters

📝 Key Definitions

Power Flow Diagram

The Golden Power Ratio

Rotor Efficiency

Synchronous Watt

Worked Example (PYQ 2020)

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Show that rotor copper loss = s× air-gap power. Also show Pm​:PCu,r​:Pg​=(1−s):s:1.

🎯 Q2: What is rotor efficiency? Show that ηrotor​=(1−s).

🎯 Q3: Define synchronous watt with an example.

🎯 Q4: Derive power equations of an induction motor from equivalent circuit.

🎯 Q5: Motor driving constant-torque load. Voltage drops to 90%. Find increase in Cu losses.

Exam Variants

⚡ Exam Tips & Common Mistakes

🔗 Related Topics

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .

🎯 Q1: Show that rotor copper loss = $s \times$ air-gap power. Also show $P_{m} : P_{C u, r} : P_{g} = (1 - s) : s : 1$ .

🎯 Q2: What is rotor efficiency? Show that $η_{rotor} = (1 - s)$ .