Equivalent Circuit of Induction Motors

Introduction

An induction motor can be represented using an equivalent circuit model. In this representation, the stator elements appear on the left side of the circuit, while the rotor elements appear on the right side. Because induction motors operate through electromagnetic induction, a transformer-like relationship naturally exists between the stator and rotor sides.

Induction Motor Transformer-Like Equivalent Circuit

R₁: Resistance of the stator winding
X₁: Stator flux leakage reactance
R_m / R_c: Core loss resistance
X_m: Magnetising reactance
R_R: Resistance of the rotor at standstill
X_R: Rotor reactance at standstill (frequency dependent)
a_eff: Effective stator-to-rotor turns ratio

The Challenge of Slip-Dependent Modeling

In a standard static transformer, we can easily form a single, fixed equivalent circuit because the induced voltages and currents always operate at a constant supply frequency, keeping electrical conditions steady. In an induction motor, however, this is not the case.

The rotor’s induced electromotive force (EMF) directly depends on the slip (s). Since the slip varies dynamically with mechanical load, the rotor frequency and the total magnitude of the induced EMF continuously change. Because the rotor does not rotate at synchronous speed, the relative motion between the physical rotor and the rotating stator magnetic field shifts with load, causing the rotor EMF to scale directly with slip:

EMF = s × E_R

Rotor Equivalent Circuit with Slip Dependent Voltage and Reactance

The core challenge is that both the operating rotor reactance (sX_R) and the induced EMF (sE_R) are strictly slip-dependent. This makes it impossible to treat the rotor side as a fixed impedance or to simply combine it directly with the stator side as we would in a conventional transformer. Instead, the equivalent circuit model must be algebraically adapted to account for this variation.

Deriving a Stationary Model

We can resolve this slip dependency by analyzing the rotor current loop using Ohm’s Law:

I = V / Z = sE_R / (R_R + jsX_R)

By dividing both the numerator and the denominator by the slip variable (s), the expression rearranges into a highly useful equivalent form:

I = V / Z = E_R / ((R_R / s) + jX_R)

Using this rearrangement, the rotor circuit can be redrawn to have a constant induced voltage (E_R) and a fixed standstill reactance (X_R), shifting the entire effect of slip into a variable rotor resistance element (R_R / s):

Redrawn Rotor Circuit with Constant Voltage and Variable Resistance

Now that the voltage and reactance values have been transformed into fixed-frequency equivalent terms, we can safely refer the rotor parameters across the air gap and combine the two sides together, yielding the final consolidated equivalent circuit model:

Final Unified Equivalent Circuit Model of an Induction Motor

Example Performance Problem

Motor Specifications

A 460 V, 25 hp, 60 Hz, four-pole, Y-connected induction motor features the following per-phase impedance parameters:

R1 = 0.521 Ω
R2 = 0.432 Ω
X1 = 1.301 Ω
X2 = 0.563 Ω
XM = 25.8 Ω

The total rotational losses (including core losses) are 1200 W and assumed constant. At rated voltage and frequency with a rotor slip of 3.1%, we will determine the full operating characteristics.

Equivalent Circuit Model Populated with Problem Parameters

Step-by-Step Performance Calculations

1. Motor Speed Calculation

Using the slip relationship equation:

s = (n_s – n_r) / n_s → 0.031 = (60 – n_r) / 60

This yields a mechanical rotor speed of:

n_r = 29.07 rev/sec = 1744.2 rpm

2. Phase Voltage and Input Impedance

For a Y-connected system, the per-phase line voltage value is calculated as:

V_φ = 460 / √3 = 265.6 V

The dynamic rotor resistance component expands under load to:

R₂ / s = 0.432 / 0.031 = 13.94 Ω

Combining series and parallel branches gives the total network input impedance (Z_T):

Z_T = 0.521 + j1.301 + (1 / j25.8 + 1 / (13.94 + j0.563))^-1 = 13.2023 ∠ 33.93° Ω

3. Current, Power Factor, and Power Input

Stator Current (I₁):

I₁ = V / Z = 265.6 / (13.2023 ∠ 33.93°) = 20.117 ∠ -33.9° A

Operating Power Factor (PF):

PF = cos(33.9°) = 0.830 lagging

Total Active Input Power (P_in):

P_in = 3 × V_φ × I₁ × cosθ = 3 × (460 / √3) × 20.1 × 0.830 = 13,292 W

4. Power Flow Diagrams and Loss Stages

Induction Motor Power Flow Loss Stages Tree Diagram

Stator Copper Loss (P_SCL):
P_SCL = 3 × I² × R₁ = 3 × (20.1)² × 0.521 = 632 W
Air-Gap Power (P_AG):
P_AG = P_in – P_SCL = 13,292 W – 632 W = 12,661 W
Converted Mechanical Power (P_conv):
P_conv = P_AG × (1 – s) = 12,661 × (1 – 0.031) = 12,268 W
Net Output Power (P_out):
P_out = P_conv – P_rotational = 12,268 W – 1200 W = 11,068 W

5. Torque Calculations

Induced Torque (τ_ind):

τ_ind = P_conv / (2π × n_r) = 12,268 / (2π × 29.07) = 67.2 Nm

Net Useful Load Torque (τ_load):

τ_load = P_out / (2π × n_r) = 11,068 / (2π × 29.07) = 60.6 Nm

6. System Efficiency

Efficiency (η) = P_out / P_in = 11,068 / 13,292 = 0.833 (83.3%)

By transforming slip-dependent values into a stationary reference frame, the model accurately calculates that this 25 hp induction motor operates at an overall electrical efficiency of 83.3%, delivering 60.6 Nm of torque to the load.