What a Locked Rotor Condition Actually Is

Why this matters

"Locked rotor" sounds like something is jammed, so techs go looking for a seized bearing or a blocked wheel, find nothing, and put the trip down to a weak overload. The phrase does not describe a mechanical condition at all. It describes an electrical state that every motor passes through on every single start, and the difference between a normal start and the one that cooks a winding is how long the machine sits in that state, not how much current it draws while it is there. Get that backwards and you condemn good starting components, reset the overload one more time, and hand the customer a burned motor a month later.

Before you attempt another start

A motor that failed to accelerate is already hot, and each retry adds far more heat than a normal start does. Limit attempts to the number the equipment documentation permits per hour; where that figure is not available, take one attempt with instruments on it and then stop rather than cycling the disconnect to see if it catches.

The frame and windings can be hot enough to burn skin after a failed start, so measure surface temperature with a non-contact instrument rather than a hand. If the smell of burnt varnish is present, do not re-energize: overheated winding insulation and varnish off-gas irritant decomposition products, which is an inhalation hazard, so ventilate the space, work from outside the plume, and treat persistent fumes as requiring respiratory protection rather than gloves.

Before hands go near a coupling, belt, sheave or fan wheel, the driven equipment needs mechanical isolation and stored-energy release under 29 CFR 1910.147, including waiting out or blocking a wheel that can still coast. The electrical side is a separate duty with a separate home: de-energize, lock and tag under 29 CFR 1910.333(b)(2), because 1910.147(a)(1)(ii)(C) expressly carves electrical utilization work out of the lockout standard and sends it to Subpart S. On construction work the electrical counterpart is 29 CFR 1926.417. Prove dead with the live-dead-live sequence in NFPA 70E-2021, 120.5.

Any current or voltage capture on a running start is energized work permitted under the troubleshooting exception at 29 CFR 1910.333(a)(1), with instruments, leads and clamps rated for the system under 29 CFR 1910.334(c)(2) and an arc-flash risk assessment done first under NFPA 70E-2021, 130.5.

Slip is the whole story

An induction motor's rotor is not connected to anything electrically. Current appears in it because the stator's rotating field sweeps past it, and induction requires relative motion. Slip is the name for that relative motion: the difference between the field's speed and the rotor's speed, expressed as a fraction of the field speed.

At rated load, slip is small. A four-pole motor on 60 Hz has a synchronous field speed of 1800 rpm; a nameplate reading 1745 rpm means slip is 55 divided by 1800, about 3.1 percent. Read your own nameplate rather than assuming a figure, because slip varies by design class and rating.

At standstill, the rotor is not moving at all, so slip equals 1. That is the definition of the locked-rotor condition, and it has nothing to do with whether the shaft is free. The instant you close the contactor, before the shaft has moved a degree, the machine is in a locked-rotor condition. It leaves that condition as it accelerates and slip falls toward its rated few percent.

What "locked" looks like at the terminals

At slip of 1 the rotor behaves as a shorted secondary of a transformer, and the stator sees a low reflected impedance. Two things follow.

The motor draws its maximum current. Locked-rotor current commonly lands somewhere around five to eight times full-load current on general-purpose designs, but the nameplate code letter is what actually defines it for the machine in front of you, so read it rather than applying a multiple.

And the motor produces no back-EMF from a spinning rotor to oppose the applied voltage. As the rotor comes up to speed, that opposition builds, reflected impedance rises, and current falls. The current decay during a start is therefore a direct readout of the rotor accelerating. A current that decays is a rotor that is turning. A current that holds flat at the locked-rotor figure until something opens is a rotor that is not.

Torque during that acceleration falls off hard with voltage: available torque in an induction motor varies roughly with the square of the applied voltage. Ten percent low on voltage is close to 20 percent low on torque, at exactly the moment the machine needs the most of it.

The heat that actually kills the winding

Winding heating is proportional to current squared times time. That squared term is why a start is thermally expensive and why a slow start is catastrophic rather than merely slow.

Take a motor with 14.2 A full-load current. A start that peaks at 88 A and settles in 1.4 seconds deposits, if you treat the current as holding at its peak for the whole acceleration, an amount of heat equal to (88 divided by 14.2) squared, which is 38.4, times 1.4 seconds, or about 54 seconds of full-load heating. That model overstates the real figure because the current decays throughout the start, and it is deliberately the pessimistic version so that the comparison below is not flattered.

Now the same motor on a start that never completes, averaging roughly 60 A over 8.2 seconds before the overload opens: (60 divided by 14.2) squared is 17.9, times 8.2 seconds, or about 146 seconds of full-load heating. That single failed attempt costs about 2.7 times the winding heat of the healthy start, at less than half the peak current, entirely because of the time term. Three retries in a few minutes deposit roughly 439 equivalent seconds against 162 for three healthy starts, and the motor's own shaft-mounted fan is barely turning throughout, so almost none of it is being carried away.

This is also why a protective device cannot simply compare current to a threshold; it has to integrate heating over time. That mechanism belongs to a sibling article on how a protective device decides to open, and this article does not re-derive it.

The start-event record

One start, instrumented once, answers more than three resets do. The fields below are the artifact; each earns its place by ruling something in or out.

The record filled in, two machines

Same equipment type, adjacent installations, nameplate full-load 14.2 A, nameplate locked-rotor 91 A, which is 6.4 times full load.

Unit A, the one that works. No-load supply 238 V. Minimum during start 209 V. Peak current 88 A, which is 97 percent of the nameplate locked-rotor figure. Current settled at 13.9 A after 1.4 seconds. Running current at five minutes 13.9 A. Outcome: accelerated.

Unit B, the one that trips. No-load supply 238 V. Minimum during start 196 V. Peak current 86 A. Current decayed to 41 A and stopped falling; the overload opened at 8.2 seconds with the machine still at 41 A. Driven-load condition noted: discharge valve found nearly closed. Outcome: tripped.

Work the record. The peak of 86 A is within a couple of amps of Unit A's 88 A and just under the nameplate figure, so the winding is presenting normal impedance at standstill and there is no shorted-turn or shorted-connection story. The current decayed rather than holding flat at 86 A, so the rotor was turning; nothing is seized. Both of those are eliminations you cannot make from a trip alone.

The voltage sag is real but small as a cause. Torque scales with the square of applied voltage, so Unit B at 196 V has (196 divided by 238) squared, about 68 percent of the torque it would have at its no-load supply voltage, while Unit A at 209 V has about 77 percent. Unit B has roughly 88 percent of Unit A's starting torque, a 12 percent shortfall - and note that the 77 percent baseline is itself measured under sag, not at nameplate voltage, so comparing Unit B's 68 percent against a nameplate-voltage figure would badly overstate what the supply contributed.

A 12 percent torque shortfall does not turn a 1.4 second start into a stall. What does is the load: current that decays and then plateaus at 41 A, well above the 13.9 A running value, is a machine that reached the speed where its torque and the load's torque balanced and could go no further. The throttled discharge valve was the load. Reset with the valve in its design position, the same motor accelerated in 1.6 seconds with an 87 A peak.

What the record rules in and what it leaves open

The filled record settles the three families of cause cleanly. A motor-side fault shows as an abnormal or unbalanced peak current, because the winding's standstill impedance is wrong. A supply-side cause shows as a deep sag with a suppressed peak, because the source cannot deliver locked-rotor current. A load-side cause shows as a normal peak, a normal sag, and an acceleration that plateaus, which is what happened here.

Two things it does not settle. It cannot distinguish a rotor-bar problem from a load problem on a single start, because both extend acceleration with a normal peak; that separation needs a no-load run with the coupling broken, which is mechanical work and therefore isolation work first. And it cannot tell you what the winding has already lost. A motor that has taken several failed starts has accumulated heat that no post-repair reading reveals, so a machine that starts correctly after the load is fixed may still have a shortened life, and that belongs in the customer conversation rather than in the "repaired" line of the invoice.

References

• 29 CFR 1910.147 including (a)(1)(ii)(C) - mechanical isolation and stored energy, and its exclusion of electrical utilization work
• 29 CFR 1910.333(a)(1) and (b)(2) - the troubleshooting exception and safe work practices for electrical work; 29 CFR 1926.417 for the construction counterpart
• 29 CFR 1910.334(c)(2) - test instruments and equipment rated for the circuits to which they are connected
• NFPA 70E-2021, 120.5 (establishing an electrically safe work condition) and 130.5 (arc flash risk assessment)
• Motor nameplate data and manufacturer documentation for code letter, permitted starts per hour and acceleration limits
• See related: How a Protective Device Decides to Open; How Heat and Current Relate in a Conductor; How to Read Current as a Diagnostic