An induction motor draws five to eight times its full-load current when it starts — enough to dip the supply voltage, trip breakers and overheat the windings if the start is repeated too often. Yet the same motor runs happily on its normal full-load current once it is up to speed. The reason is in the equivalent circuit: at standstill the motor is, electrically, a short-circuited transformer with no back-EMF. This post works through that circuit and shows why the current collapses as the motor speeds up.
The equivalent circuit
The per-phase equivalent circuit of the induction motor, referred to the stator, gives the stator current as:
I1 ≈ I2′ = V1 / √[(R1 + R2′/s)² + (X1 + X2′)²]
Here the slip is s = (Ns − Nr)/Ns, and the rotor resistance referred to the stator splits into two parts:
R2′/s = R2′ + R2′(1−s)/s
The first term, R2′, is the rotor copper loss. The second term, R2′(1−s)/s, is a fictitious resistance that represents the mechanical load — the power converted into shaft work. It is this second term that makes all the difference between starting and running.
Figure 7(a): Why induction motor starting current is high — the per-phase equivalent circuit (left) and stator current vs slip (right). At standstill the load resistance is zero, so the current is limited only by the small winding impedance. (Schematic, not to scale.)
At standstill: a short-circuited transformer
At standstill the rotor is not turning (Nr = 0), so the slip is s = 1 and the load term vanishes: R2′(1−s)/s = 0. There is no back-EMF and no load resistance — the motor is simply a short-circuited transformer, and the only impedance in the circuit is the small stator and rotor winding resistance plus leakage reactance:
Zstart = √[(R1 + R2′)² + (X1 + X2′)²] ≈ 0.12–0.16 p.u.
With the full supply voltage applied across only 0.12–0.16 p.u. of impedance, the starting current is:
Istart = V1 / Zstart ≈ 5 to 8 × IFL at a low lagging power factor of cos φ ≈ 0.2–0.3
The power factor is low because the circuit at standstill is almost purely reactive — mostly leakage reactance, with very little resistance to absorb real power.
At running speed: the load resistance takes over
At normal running speed the slip is only s ≈ 0.02–0.04, so the effective rotor resistance R2′/s becomes 25 to 50 times larger than at standstill. That resistance now dominates the total impedance of the circuit, and the stator current falls to its normal full-load value IFL. The motor is no longer a short circuit — it is a loaded machine, and the load resistance is what limits the current.
That is the whole story in one line: the starting current is high because at standstill the load has not yet appeared in the circuit; as the motor speeds up, the load resistance grows 25–50 times and the current falls to normal.
Key points to remember
- Stator current: I1 ≈ I2′ = V1 / √[(R1 + R2′/s)² + (X1 + X2′)²], with s = (Ns − Nr)/Ns.
- R2′/s = R2′ (copper loss) + R2′(1−s)/s (mechanical load).
- At standstill (s = 1): load term is zero, no back-EMF — the motor is a short-circuited transformer with Zstart ≈ 0.12–0.16 p.u.
- Starting current: Istart ≈ 5–8 × IFL at cos φ ≈ 0.2–0.3 (low, lagging).
- At running speed (s ≈ 0.02–0.04): R2′/s grows 25–50×, dominates the impedance, and the current falls to IFL.
Based on Power System Protection & Switchgear, Paper II, Q7(a), Exam 11 June 2026.