What causes locked rotor amps on a motor?

[fourthindiana]

TL;DR

What causes locked rotor amps on a motor to be approximately 5X higher than the regular amps when the motor is running?

When an electric motor first starts running, the motor will draw approximately 5X the normal amp draw of the motor for approximately the first half second that the motor runs. The amps that an electric motor draws for about half a second that are about 5X the normal amp draw of the motor are called locked rotor amps.

It has occurred to me that maybe the reason that an electric motor draws locked rotor amps for the first half second that the motor runs is that when a motor first starts running, a motor has no momentum. Does an electric motor pull locked rotor amps when it first starts due to the lack of momentum?

When an electric motor starts, why does an electric motor pull locked rotor amps for part of the first second that the electric motor runs?

 

[davenn]

When an electric motor first starts running, the motor will draw approximately 5X the normal amp draw of the motor for approximately the first half second that the motor runs. The amps that an electric motor draws for about half a second that are about 5X the normal amp draw of the motor are called locked rotor amps.

Regarding DC motors, the voltage across the motor, is proportional to the RPM of the motor. The current through the motor is proportional to the torque produced by the motor. Under normal conditions, the back EMF created by the spin of the motor opposes the voltage being applied and limits the current the motor draws and therefore it's torque. As more load is applied to the motor, it's rotation slows reducing the back EMF which allows more current to flow, therefore producing more torque.
So the process is the same for a motor starting up as it is being more heavily loaded or where the armature locks - ceases up.

 

[fourthindiana]

What about AC motors?

 

[fourthindiana]

Regarding DC motors, the voltage across the motor, is proportional to the RPM of the motor. The current through the motor is proportional to the torque produced by the motor. Under normal conditions, the back EMF created by the spin of the motor opposes the voltage being applied and limits the current the motor draws and therefore it's torque. As more load is applied to the motor, it's rotation slows reducing the back EMF which allows more current to flow, therefore producing more torque.
So the process is the same for a motor starting up as it is being more heavily loaded or where the armature locks - ceases up.

This doesn't even explain what causes locked rotor amps on a DC motor.

 

[hutchphd]

This doesn't even explain what causes locked rotor amps on a DC motor.

A locked motor is just a coil with very low electrical resistance. When connected to DC, there will be a very short period where the back emf (Lenz's Law) reduces the current but after that it pulls a lot of current (=V/R). When the motor is rotating the transient effects are renewed with every revolution and this limits the current, The faster the turning the more back emf.

 

[fourthindiana]

A locked motor is just a coil with very low electrical resistance.

Is that why a motor pulls 5X the normal running amps for the first half second or so that a motor runs? If so, why does the coil on a motor have very low electrical resistance when the motor first starts?

When connected to DC, there will be a very short period where the back emf (Lenz's Law) reduces the current but after that it pulls a lot of current (=V/R). When the motor is rotating the transient effects are renewed with every revolution and this limits the current, The faster the turning the more back emf.

What about when connected to AC?

 

[jim hardy]

Think simple.
A motor is just a transformer with a rotating secondary.

Think of the motor when its rotor is locked.
It's just a transformer with a shorted secondary..
The squirrel cage rotor is the secondary and it's just bars tied together at their ends.
So there's high current in the rotor.
As with any transformer, high secondary current means high primary current too.

The mystery is more one of 'why does current drop off as the motor approaches tuning oops- running speed ?' .

 

[davenn]

This doesn't even explain what causes locked rotor amps on a DC motor.

yes it does !
I am sorry you didn't understand the answer

 

[fourthindiana]

Think simple.
A motor is just a transformer with a rotating secondary.

I will think of a motor this way.

Think of the motor when its rotor is locked.

What locks the rotor?

It's just a transformer with a shorted secondary..
The squirrel cage rotor is the secondary and it's just bars tied together at their ends.
So there's high current in the rotor.
As with any transformer, high secondary current means high primary current too.

Interesting.

The mystery is more one of 'why does current drop off as the motor approaches tuning speed ?' .

I'm assuming you meant to write turning speed, not tuning speed. Correct me if I am wrong.

Why does current drop off as the motor approaches turning speed? Does the current drop off as the motor approaches turning speed because the resistance increases?

 

[fourthindiana]

yes it does !
I am sorry you didn't understand the answer

If I had enough knowledge to understand your answer, I wouldn't have needed to ask what causes locked rotor amps in the first place.

 

[jim hardy]

What locks the rotor?

Actually it's only locked until it starts to turn
If it's a small motor you can hold it with a pair of vise-grip pliers , or just disconnect the start capacitor.

How fast it accelerates depends on the total inertia of the rotor and load.
Watch your squirrel cage blower - it accelerates slow enough to watch..
so, to your question - inertia 'locks' the rotor but not for long.
To measure Locked Rotor Amps(LRA) we actually clamp the rotor with something mechanical.
Here's current and torque vertical plotted versus speed horizontal.
(image courtesy https://www.researchgate.net/figure/Typical-induction-motor-torque-speed-current-curve_fig2_3171041)_{orange mine jh}

I'm assuming you meant to write turning speed, not tuning speed. Correct me if I am wrong.

absolutely. Fixed it. Thanks.

Why does current drop off as the motor approaches turning speed?

There's no short yet complete answer that i know of.
Here's part of it to ponder for now...
Recall from motor theory that the applied AC voltage makes a magnetic field that rotates around the stator.
Your squirrel cage sits in that magnetic field.
As its speed of rotation approaches that of the stator's magnetic field, the lines of flux cut the rotor bars ever more slowly.
The rate at which flux cuts the rotor bars dictates the voltage induced in them,
...so...
as speed of the rotor approaches that of the stator field - voltage in the rotor bars decreases, and so does current .

I warn you that is an incomplete answer... a complete aswer has to go ito the numbers describing magnetic fluxes and currents in stator and rotor both

what is important now is for you to grasp that concept of speed difference between stator field and rotor. It's called Slip and represented as either % or a number between 0 and 1.
Locked rotor is slip of 100% or 1.0
Synchronous speed is slip of 0% or 0.0.
Typical fully loaded slip is maybe 5% to 15%
If you watch an unloaded induction motor(not permanent capacitor type just a capacitor start) with a stroboscope or under fllourescent lights you can see the keyway turn at slip speed due to the strobe effect.
A good one might take a whole minute to slip one complete turn. That'd be slip of maybe 1/1800 = .05%

old jim

 

[jim hardy]

Wow just noticed a blooper in that graph i posted.

At rated torque , 1.0 pu, current should be also 100% rated not the 280% shown.

I don't know why it shows current so high - perhaps author did something that doesn't show up in that excerpt he posted and i copied.
Anyhow

normal operation is very near synchronous speed and at 100% rated torque or less, the very bottom right of that curve.

.

notice rated torque occurs at around 96% synchronous speed, that's 4% slip. Not unusual.

old jim