Induction motor equivalent circuit

The stator supplies the magnetomotive force that sets up the air-gap flux, and this flux links the rotor exactly as a transformer's primary flux links its secondary. So three of the four elements in the induction motor's per-phase equivalent circuit are transformer elements, referred to the stator side by the effective turns ratio:

• $R_1, X_1$ — stator winding resistance and leakage reactance, in series with the supply.
• $R_0, X_0$ (equivalently $G_0, B_0$) — the magnetising branch, shunted across the supply, carrying the exciting current that sets up the mutual flux and accounts for core loss.
• $R_2', X_2'$ — rotor resistance and leakage reactance, referred to the stator.

The rotor branch is where the transformer analogy breaks. A cage rotor is short-circuited on itself (or, for a wound rotor, closed through external resistance) and turns at slip $s$, so its induced EMF and frequency are both $s$ times the stator's. Referring this rotor circuit to the stator at the supply frequency turns the plain resistance $R_2'$ into the slip-dependent term:

where: $R_2'$ is the stator-referred rotor resistance (ohms), $s$ is the slip (dimensionless, between 0 and 1 for motoring), and $R_2'/s$ is the effective resistance that the stator-referred rotor current $I_2'$ sees in the equivalent circuit, drawn as $R_2'$ in series with $jX_2'$ and this variable resistance. The split on the right matters physically. The $R_2'$ term is real rotor copper loss — heat, always present. The extra term $R_2'(1-s)/s$ is not a physical resistor; it is a bookkeeping resistance across which the gross mechanical power developed appears, electrically, in the circuit model. As $s \to 0$ (near synchronous speed, light load), $R_2'/s \to \infty$ and rotor current shrinks toward zero. As $s \to 1$ (standstill), $R_2'/s$ falls to its minimum, $R_2'$ itself, so rotor — and hence stator — current is largest. This is exactly the physical picture of a large starting current that tapers off as the motor runs up to speed.

• Treating the induction motor equivalent circuit as identical to a transformer's — the stator impedance and magnetising branch are analogous, but the slip-dependent $R_2'/s$ has no counterpart in a static transformer.
• Forgetting that $R_2'/s$ already includes $R_2'$ inside it; do not add $R_2'$ again on top of $R_2'/s$ when computing total rotor-branch resistance.
• Assuming the referral of rotor quantities to the stator is automatic — it uses an effective turns ratio, treated at an introductory level here without full derivation.
• Ignoring the magnetising branch when estimating stator current at low slip — the exciting current is not negligible compared with the shrinking load component of current near synchronous speed.

Problem. A three-phase induction motor has a stator-referred rotor resistance $R_2' = 0.4\ \Omega$ per phase and stator-referred rotor current $I_2' = 20\ \text{A}$ per phase at full load, where the slip is $s = 0.04$. Find (a) the effective rotor-branch resistance $R_2'/s$, (b) its split into copper-loss and mechanical-power terms, and (c) the rotor copper loss and mechanical power developed per phase.

Step 1. Compute the effective rotor-branch resistance at $s = 0.04$: $R_2'/s = 0.4/0.04 = 10\ \Omega$. Compare with standstill ($s=1$), where $R_2'/s = 0.4\ \Omega$ — twenty-five times smaller, which is why standstill current is so much larger than full-load current.

Step 2. Split $10\ \Omega$ using the derivation above: $R_2'\left(\dfrac{1-s}{s}\right) = 0.4\left(\dfrac{1-0.04}{0.04}\right) = 0.4\times24 = 9.6\ \Omega$. Check: $R_2' + 9.6 = 0.4+9.6 = 10\ \Omega = R_2'/s$, consistent.

Step 3. Rotor copper loss per phase, dissipated in $R_2'$ alone: $P_{cu,2} = I_2'^2R_2' = 20^2\times0.4 = 400\times0.4 = 160\ \text{W}$.

Step 4. Mechanical power developed per phase, dissipated (in the model) in the $9.6\ \Omega$ term: $P_{mech} = I_2'^2\times9.6 = 400\times9.6 = 3840\ \text{W}$.

Answer: $R_2'/s = 10\ \Omega$ at full load, splitting into $0.4\ \Omega$ (copper loss) and $9.6\ \Omega$ (mechanical power); the corresponding powers per phase are $P_{cu,2} = 160\ \text{W}$ and $P_{mech} = 3840\ \text{W}$ (multiply by 3 for the total three-phase values).

Competitive stretch

If instead $I_2' = 20\ \text{A}$ and $R_2' = 0.5\ \Omega$ at $s = 0.05$, the same method gives $R_2'/s = 0.5/0.05 = 10\ \Omega$, split as $R_2' = 0.5\ \Omega$ and $R_2'(1-s)/s = 0.5\times19 = 9.5\ \Omega$. The powers per phase are $P_{cu,2} = 20^2\times0.5 = 200\ \text{W}$ and $P_{mech} = 20^2\times9.5 = 3800\ \text{W}$. Notice how sensitive $P_{mech}$ is to slip even for a small change in $R_2'$ — always recompute $R_2'/s$ freshly for the given slip rather than reusing a value from a different operating point.

Ask the tutor: if the R2 prime over s split feels abstract, ask the AI tutor to walk through the algebra step by step for a slip value of your choosing.