boss_notes/T-14_IM_Equivalent_Circuit.md

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← T-13: Slip & Basics | 🏠 Index | T-15a: Starting Torque →

T-14: IM Equivalent Circuit

Section: B | Priority: 🟠 HIGH | Exam Frequency: 3/7 years Sources: Theraja Ch-34 (Art. 34.47), VK Mehta Ch-8 (Art. 8.8-8.9), Chapman Ch-7, Slides L-03

Why This Topic Matters

The equivalent circuit question appeared in 3 out of 7 papers (2017, 2020, 2023). It is worth 3-6 marks per appearance. More importantly, every torque and power calculation in Section B relies on this circuit. If you can draw the 6-step equivalent circuit development, you can answer any IM analysis question.

📝 Key Definitions

Equivalent Circuit of IM: "An induction motor can be treated as a rotating transformer i.e. one in which primary winding is stationary but the secondary is free to rotate." — Theraja, Art. 34.2. The equivalent circuit represents the per-phase electrical model of this rotating transformer.

Load Resistance $R_{L}$ : "The resistance $R_{2} (1 - s) / s$ represents the electrical equivalent of the gross mechanical power developed by the motor." — VK Mehta, Art. 8.9

Step-by-Step Equivalent Circuit Development

Step 1: Transformer Model at Standstill ( $s = 1$ )

At standstill, the IM acts exactly like a transformer. The stator is the primary. The rotor is the short-circuited secondary.

Stator: resistance $R_{1}$ , leakage reactance $X_{1}$
Shunt branch: core loss resistance $R_{c}$ , magnetizing reactance $X_{m}$
Rotor: resistance $R_{2}$ , standstill reactance $X_{2}$

Step 1: Transformer model at standstill

Step 2: Rotor at Running Slip $s$

When running at slip $s$ , the rotor frequency changes to $s f$ . The rotor EMF and reactance scale with slip:

$E_{2 s} = s E_{2}, X_{2 s} = s X_{2}$

Rotor current:

$I_{2} = \frac{s E _{2}}{R _{2} + j s X _{2}} = \frac{s E _{2}}{R _{2}^{2} + ( s X _{2} ) ^{2}}$

Step 2: Rotor circuit at slip s

Step 3: Frequency Transformation

Divide numerator and denominator of the rotor current equation by $s$ :

$I_{2} = \frac{E _{2}}{R _{2} / s + j X _{2}}$

This transforms the rotor to the stator frequency $f$ . The rotor resistance appears as a variable resistance $R_{2} / s$ at the stator frequency.

Step 3: Frequency transformation to stator frequency

Step 4: Power Separation

Split $R_{2} / s$ into two parts:

$\frac{R _{2}}{s} = R_{2} + R_{2} (\frac{1 - s}{s})$

$R_{2}$ represents actual rotor copper loss (heat)
$R_{L} = R_{2} (1 - s) / s$ represents the mechanical load (gross mechanical power)

Step 4: Power separation into copper loss and mechanical load

Step 5: Exact Equivalent Circuit (Referred to Stator)

Refer all rotor quantities to the stator using the effective turns ratio $a = N_{1} / N_{2}$ :

$R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, R_{L}^{'} = R_{2}^{'} (\frac{1 - s}{s})$

Step 5: Exact equivalent circuit referred to stator

Step 6: Approximate Equivalent Circuit

Since the stator voltage drop is small (typically < 5%), the shunt branch ( $R_{c}$ , $X_{m}$ ) can be moved to the input terminals. This simplifies calculations without much loss of accuracy.

Step 6: Approximate equivalent circuit

Key Equations from the Equivalent Circuit

Air-gap power (per phase):

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

Total air-gap power (3-phase):

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

Rotor copper loss:

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

Mechanical power:

$P_{m} = P_{g} - P_{C u, r} = (1 - s) P_{g} = 3 I_{2}^{2} R_{2} \frac{( 1 - s )}{s}$

🏆 Golden Questions (Past Exam Archive)

🎯 Q1: Draw the step-by-step equivalent circuit of a 3-phase induction motor.

Appeared: 2017 Q3(a), 2020 Q5(c) — (3-6 marks)

Full Answer:

See Step-by-Step Equivalent Circuit Development above for the complete 6-step development:

Standstill transformer model: Stator ( $R_{1}, X_{1}$ ) + shunt ( $R_{c}, X_{m}$ ) + rotor ( $R_{2}, X_{2}$ )
Running rotor: EMF becomes $s E_{2}$ , reactance becomes $s X_{2}$
Frequency transformation: Divide by $s$ to get $R_{2} / s$ at stator frequency
Power separation: $R_{2} / s = R_{2} + R_{2} (1 - s) / s$ , where $R_{2} (1 - s) / s$ is mechanical load
Refer to stator: $R_{2}^{'} = a^{2} R_{2}$ , $X_{2}^{'} = a^{2} X_{2}$
Approximate circuit: Move shunt branch to input terminals

🎯 Q2: How can the equivalent circuit model of an induction motor be obtained?

Appeared: 2020 Q5(c) — (6 marks)

Full Answer:

The equivalent circuit of an IM is obtained by treating it as a rotating transformer.

Starting point: At standstill ( $s = 1$ ), the IM is identical to a transformer. The stator is the primary with impedance $Z_{1} = R_{1} + j X_{1}$ . The shunt branch models core losses ( $R_{c}$ ) and magnetizing current ( $X_{m}$ ). The rotor is the short-circuited secondary with impedance $Z_{2} = R_{2} + j X_{2}$ .

Running modification: When the rotor spins at slip $s$ , two things change: (a) rotor EMF reduces to $s E_{2}$ , (b) rotor reactance reduces to $s X_{2}$ .

Key trick: The rotor current equation $I_{2} = s E_{2} / R_{2}^{2} + (s X_{2})^{2}$ can be rewritten as $I_{2} = E_{2} / (R_{2} / s)^{2} + X_{2}^{2}$ by dividing top and bottom by $s$ . This transforms the rotor circuit to stator frequency. The resistance $R_{2} / s$ splits as $R_{2} + R_{2} (1 - s) / s$ , where $R_{2}$ accounts for copper loss and $R_{2} (1 - s) / s$ represents mechanical power.

Final step: Refer all rotor quantities to the stator using $a^{2}$ scaling. Move shunt branch to input for the approximate circuit.

Exam Variants

Year	Question	Marks	Focus
2017 Q3(a)	Draw step-by-step equivalent circuit	3	All 6 steps
2020 Q5(c)	How to obtain eq. circuit model?	6	Derivation + explanation
2023 Q7(a)	Explain NL and BR tests, find parameters	9	Testing + circuit

⚡ Exam Tips & Common Mistakes

Draw ALL six steps. Don't jump to the final circuit. Each step carries marks.
Label $R_{2} (1 - s) / s$ clearly. This is the mechanical load, not just a resistor.
$R_{2} / s$ is NOT the rotor resistance. It is the total impedance that accounts for both copper loss and mechanical power.
The shunt branch is NOT negligible. In the exact circuit, it sits between $Z_{1}$ and $Z_{2}^{'}$ . Only move it for the approximate circuit.
Know both exact and approximate circuits. Some questions ask specifically for one or the other.

T-13: Slip & Basics — Slip determines all rotor quantities
T-15a: Starting Torque — Torque from the equivalent circuit
T-16: Power Flow — Power stages from the circuit
T-17a: No-Load Test — Finding shunt branch parameters
T-17b: Blocked Rotor Test — Finding series branch parameters

← T-13: Slip & Basics | 🏠 Index | T-15a: Starting Torque →

boss_notes/T-14_IM_Equivalent_Circuit.md

7.1 KB · commit d2f3b6e

← T-13: Slip & Basics | 🏠 Index | T-15a: Starting Torque →

T-14: IM Equivalent Circuit

Section: B | Priority: 🟠 HIGH | Exam Frequency: 3/7 years Sources: Theraja Ch-34 (Art. 34.47), VK Mehta Ch-8 (Art. 8.8-8.9), Chapman Ch-7, Slides L-03

Why This Topic Matters

📝 Key Definitions

Equivalent Circuit of IM: "An induction motor can be treated as a rotating transformer i.e. one in which primary winding is stationary but the secondary is free to rotate." — Theraja, Art. 34.2. The equivalent circuit represents the per-phase electrical model of this rotating transformer.

Load Resistance $R_{L}$ : "The resistance $R_{2} (1 - s) / s$ represents the electrical equivalent of the gross mechanical power developed by the motor." — VK Mehta, Art. 8.9

Step-by-Step Equivalent Circuit Development