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Induction motor equivalent circuit

Treat the rotor as a transformer secondary; the R_2/s trick reveals mechanical power as a fictitious resistor.

Freshman ~9 min

Step 1 — Recall the transformer's per-phase equivalent circuit

Animation speed 0.55×

s 0.04 R₂/s — P_mech/P_g —

Reference notes

Use Next → on the narrator above to step through how the transformer equivalent circuit is extended to the induction motor by introducing the famous R₂/s "load resistor".

The induction motor as a generalised transformer

An induction motor is electromagnetically equivalent to a transformer with the rotor acting as the secondary. The differences:

The rotor's "load" is mechanical, not electrical — but it can be modelled as an equivalent resistor.
The rotor EMF and reactance scale with slip s (from the previous lesson).
The "secondary" is short-circuited internally; the load shows up as power dissipated in the equivalent R₂/s.

The R₂/s trick

Start from the rotor circuit at slip s:

I₂ = (s · E₂) / (R₂ + j · s · X₂)

Divide top and bottom by s:

I₂ = E₂ / (R₂/s + j · X₂)

This is now exactly the form of a transformer secondary at the supply frequency, with an effective rotor impedance R₂/s + jX₂. R₂/s is the equivalent per-unit-slip rotor resistance — a function of operating speed.

Splitting R₂/s — where does the power go?

Algebra:

R₂ / s = R₂ + R₂·(1 − s) / s

So the R₂/s "resistor" naturally splits into TWO physical pieces:

R₂ — the real rotor-circuit resistance. Power dissipated here is the rotor copper loss.
R₂·(1−s)/s — a fictitious resistor. Power dissipated here is the mechanical power developed on the rotor shaft.

Power flow and the 1:s:(1−s) ratio

Let P_g = air-gap power (real power crossing the air gap from stator to rotor) = 3·I₂'²·(R₂'/s). Then:

P_cu,rotor = s · P_g P_mech = (1 − s) · P_g

So three ratios you'll use constantly:

Stator → Rotor copper loss → Mechanical power ::: 1 : s : (1 − s)

At rated load with s = 0.04: of every 100 W crossing the air gap, 4 W is dissipated in rotor copper and 96 W is delivered as mechanical power. Lovely.

Per-phase equivalent circuit referred to the stator

The full per-phase circuit (with all quantities referred to the stator side):

V₁ — r₁ — jx₁ — [ R_c ‖ jX_m ] — jx₂' — R₂'/s

Identical to the transformer equivalent circuit, except the rotor's "load" is R₂'/s. As s changes (different operating point), R₂'/s changes — that's the only thing the operator sees. The rest is constant per-machine.

The key insight. The whole magic of induction-machine analysis is hidden in dividing by s. Once you accept that R₂/s carries both the rotor copper loss AND the mechanical power, every formula for efficiency, torque, and power factor falls out of the standard transformer machinery.

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Reference notes

The induction motor as a generalised transformer

The R2/s trick

Splitting R2/s — where does the power go?

Power flow and the 1:s:(1−s) ratio

Per-phase equivalent circuit referred to the stator

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The R₂/s trick

Splitting R₂/s — where does the power go?