Current draw of a motor, under different situations

[maximiliano]

So, I'm kind of in a debate, and before I say something that makes me look like a clown , I thought I'd check with the group.

Someone said that a "dying or low battery in car or motorcycle will destroy the starter motor...because the motor will respond to reduced voltage by increasing the AMP draw...thus causing too much heat".

I don't think this is true...well, not exactly. (BTW- I also think heat is a function of WATTS not amps per say...am I wrong?) I don't think that a motor cares how many volts are coming in, so long as they are adequate to allow the motor to spin at its designed speed, and also are not too much. However, if LUGGING (trying to get up to speed but can't), then I think the amps are not increased per say...but they do stay at "peak" or "startup current" (due to the load) for an extended period. I guess what I'm saying is that I THINK it's not the low voltage that kills a motor (via heat from extra current)...it's the LOAD that has increased because the motor is not able to spin at full speed, thus increasing the amps (no higher than they were with a "good" battery...but keeping them there for extended period) ??

So, if a motor runs on 10-12 volts, and spikes at a current draw of 200amps for 1 second and then drops to 30 amps once up to speed...IF the battery is getting low and is only at 8 volts, then, the motor still won't draw more than the 200amps (right or wrong?)...but it may continually pull that kind of current (the 200a), since the motor is "lugging" and unable to attain speed. Am I out in left field on this one??

Maybe someone can take what I said...and make it more clear in terms of amp draw of a motor. Hope that's slightly more clear than mud...but I think maybe it isn't?

 

[psparky]

In General, motors are desinged for their specific voltage and amperage ratings. Anytime you lower the voltage...you generally have problems with starting the motor...etc. And yes it damages over time. If you lower the voltage...you will also lower the amperage...the motor does NOT like this.

I don't think that lowering the voltage raises the amps. V=I*R. The "R" or resistance of the starter motor is set.

The watts of the motor (P=I*V) is determined by the voltage and amps. It will only deliver the specified watts if the voltage and amps are the ones specified.

So if you say a motor has 100 watts for example...you can't say if I connect half the voltage to it, it will automatically double the amps. Doesn't work like that. The actual voltage and amps given to the motor determined by the RESISTANCE of the motor will determin your power (P=I*V) or ((V^2/R)=P) or (I^2*R=P)

The first of the formulas are first and formost determinded by V=I*R. It rules it...it owns it. The other two formulas above have R in them...these you can use.

Take the second one ((V^2/R)=P) for example.
You just said you went from 12 volts to 8 volts.

Lets say for the sake of the argument...the ohms (resistance) of your motor is 1 ohm.
(12^2)/1=144 watts.

Lower it to 8 volts...
(8^2)/1= 64 watts.

Not the same!

The resistance of the motor can certainly change with different loads...but again it is the resistance controlling all the variables.

Get it?

 

[DragonPetter]

Psparky, I think OP is right and you showed why amps does not increase when voltage is dropped.

But you agree with him that the starting motor is damaged, and I didn't catch a reason why. Is there a reason why it would be damaged? I don't know much about engines/starting motors to understand why.

 

[psparky]

I'm not sure who I was backing up...I was just pointing out my beliefs...
Some of the things max wrote were not quite right.

I'm not sure why lower voltage kills motors...but it certainly does. Motors hate when you change the specified current through them. And they simply respond by dying.

This is exactly why you hook a capacitor in parallel when correcting power factor. The motor has no idea the capacitor is there...the motor receives the same voltage and current. It loves that. However, the coal burning power plants sees a different load...one with a better power factor and burns less coal simply because of that capacitor shifting the current and voltage closer together.

Someone else will come in and correct me and add some further things soon enough...but I'm on to the right track...I believe.

 

[DragonPetter]

I'm not sure who I was backing up...I was just pointing out my beliefs...
Some of the things max wrote were not quite right.

I'm not sure why lower voltage kills motors...but it certainly does. Motors hate when you change the specified current through them. And they simply respond by dying.

This is exactly why you hook a capacitor in parallel when correcting power factor. The motor has no idea the capacitor is there...the motor receives the same voltage and current. It loves that. However, the coal burning power plants sees a different load...one with a better power factor and burns less coal simply because of that capacitor shifting the current and voltage closer together.

Oh, sorry I just was referring that he is correct that a motor will not draw more current just because the motor voltage is dropped, which is what his friend told him happens.

 

[psparky]

Oh, sorry I just was referring that he is correct that a motor will not draw more current just because the motor voltage is dropped, which is what his friend told him happens.

I agree with this statement. However, I don't agree with some of his other statements.

 

[psparky]

All that being said...when a motor is trying to start...it is trying to build up a magnetic field. Now if for some reason the magnetic field does not get built up (not enough voltage for example)...the motor will behave more like a dead short...In this case the motor may actually pull way more amps. Why? Because the resitance of the motor has gone way done...almost to a short.

Transient is tricky business...

Again though...it is based off the "resistance". So come to think of it...there may be some valid points in saying that the motor pulls more amps based off a short circuit effect...not because of the "watts" of the motor.

So to answer the question above...how does it kill the motor...I think the answer is that it fries it from too much current due to the short ciruit effect. (magnetic field not being built up)

My originial response was based more off of steady state...my second response is more off of transient.

In starting a motorcycle...is that transient or steady state? Hard to say...depends how long you are cranking.

I don't know man...just throwing some stuff out here so we can all learn. I have learned just by writing all this crap! It has raised more questions than answers! Some one else please help!

 

[skeptic2]

With regard to a DC motor, the voltage across the motor, minus IR losses, is proportional to the speed or RPMs of the motor. The current through the motor is proportional to the torque produced by the motor. Normally the back EMF created by the speed of the motor opposes the voltage being applied and limits the amount of current the motor draws and thus it's torque. As more and more load is applied to the motor, the motor slows reducing its back EMF allowing more current to flow, producing more torque.

With a lower applied voltage to the motor not only will the motor be turning slower, which reduces the back EMF and allows more current to flow, the losses such as the internal battery resistance, the cable loss and the IR drop in the motor become more important. This may mean that the motor may not have the speed or torque to turn over the engine and stall. When the motor stalls there is no longer any back EMF and all the current goes towards heating up the motor windings and that can damage the motor.

 

[Windadct]

The damage will mostly occur if the motor does not start - or sits there not moving with current flowing. At a low voltage you are more likely to run the motor too long in this stalled condition - or it takes longer to get to speed. With a good battery = the motor starts quickly - ideally on the first try. Repeated attempts to start the motor are the problem.
The Resistance of the motor is only part of the equation - in a DC - permanent magnet motor, the motor provides Back EMF as it comes to speed ( actually generates torque). When you first energize the motor a large amount of current flows (limited principally by it's resistance) and decreases as it begins to apply torque to the load. It is the BACK EMV x Current which defines the power delivered to the load - if it was just the winding resistance - it would not be a motor - it would just be a heater.

 

[russ_watters]

Are we talking about electric cars or motorcycles? My car doesn't have a motor...

 

[SirAskalot]

In looking at the equations and equivalent diagram of a DC motor you can learn a lot(everything?) about its behavior. For car/mc applications the motor is probably a series or compound dc motor (the way the field winding is connected). Try to google it.

As stated by psparky, extensive heating is the main cause for breaking a dc motor. And as you know, heat is a result of current where as cooling usually comes from a fan connected to the motor shaft. Thus cooling is proportional to the motor speed.

As seen in product data sheets, without external cooling, the motor cannot supply nominal power below some rotor speed over a longer time span, due to heating.

With the equations, related to current, speed and voltage and heating you can probably find out for your self what is happening at low voltage.

 

[skeptic2]

Are we talking about electric cars or motorcycles? My car doesn't have a motor...

Regardless of what type of car you have it is difficult to believe it doesn't have a motor. Do you crank start it?

 

[OldEngr63]

The OP did not clearly specify what sort of motor he was interested in, but there was the implication that we were discussing an electric starter motor.

 

[NascentOxygen]

Someone said that a "dying or low battery in car or motorcycle will destroy the starter motor...because the motor will respond to reduced voltage by increasing the AMP draw...thus causing too much heat".

I agree that scenario is likely or possible. A sluggish starter is likely to be engaged for much longer than intended, so the motor (not rated for continuous use) is likely to overheat. And because it is not turning as fast as designers intend, the reduced back emf allows higher armature current and this can overheat the windings and brushes.

 

[NascentOxygen]

The OP did not clearly specify what sort of motor he was interested in, but there was the implication that we were discussing an electric starter motor.

Slightly stronger than an implication, by my reading of OP:

Someone said that a "dying or low battery in car or motorcycle will destroy the starter motor...because the motor will respond to reduced voltage by increasing the AMP draw...thus causing too much heat".

 

[Dhende]

How to calculate induction motors formula ?

 

[NascentOxygen]

How to calculate induction motors formula ?

Hi Dhende. It is best that you start a new thread with your topic, and you are likely to get a lot more responses that way. Induction motors are not used as starter motors in cars, which is what this thread turned out to be, despite its misleadingly general topic.

When you post your question in its own thread, I think you should explain which equation you are wanting to derive, too.

Good luck!

 

[I_am_learning]

If the motor has permanent magnet, then the torque will be directly proportional to Armature Current. (At the start when the motor is at 0 speed, there is no Back EMF so Starting Current = V/R ).
So, the starting Torque decreases in direct proportion to applied voltage.
Once started the motor will slowly pick-up speed and hence back Emf increases.
At any given speed, motor torque is proportional to Armature current I = (V-E)/R
where E is back-emf which is directly proportional to motor speed.
So, as the speed increases the motor torque decreases (because E increases and I decreases), and at equilibrium (i.e. final speed), the motor torque exactly equals the load torque.
If the applied Voltage is reduced, the starting torque will be less. So, the motor picks-up speed slowly. But the steady state speed will also be much less because,
At steady state,
I = (V-E)/R and we need same I as before to produce same torque (remember Steady State = Motor Torque = Load Torque). But we have reduced V, so a reduced E (i.e. reduced Speed) will be providing the required I.

I checked some maths (including motor Torque, acceleration and speed pick-up times) and found that, in either case, the Timer-Integral of Current is the same, i.e., whether you apply higher voltage or lower voltage, the heating effect on the wires is the same.

BUT, a big BUT, if the reduced Torque due to Reduced Voltage couldn't accelerate the motor to its steady state speed (i.e. it stalls), then It would draw high starting current all way long, and will Damage the motor. Although the starting current in this case is lower than the starting current due to regular applied voltage, this current may be still many (10) times higher than the steady state current on regular voltage.

I don't much know about starter motors and how long does they generally need to operate, so I was mostly talking about Generic DC motors in Generic Applications.

 

[FOIWATER]

the person is correct that the low voltage may damage the motor. I have found a very upsetting misconception with power, voltage, current, and resistance relates back to P = EI, and V = IR... people tend to assume that power is a constant, so then they assume voltage and current are inversely proportional but this isn't the case, unless you use a drive or something to correct the amount of power. motor's draw current (at a fixed voltage) based on ONE variable, the motor winding impedance. if the motor is say 24 volts, it will draw locked rotor current at start-up. This current will decrease as the motor speeds up, because of counter emf. This is good, the motor would burn up if it drew locked rotor current for longer than a little while.

situation 1: if you increase the voltage to a motor, it will draw MORE locked rotor current.

Situation 2: if you decrease the voltage to a motor, it will draw LESS locked rotor current.

if the low battery voltage in situation two doesn't allow your motor to draw locked rotor current sufficient enough to overcome the locked rotor torque requirements for turning your engine... the motor will not develope counter emf and will draw a reduced version of rated locked rotor current for extended period of time.

This current draw will be LESS than rated locked rotor current, but greater than it's nominal operating current. It could be enough current (and thus heat) at a specific voltage to ruin the motor.

So... he .she isright in a way... but let's say the battery was nearly dead... and could only supply 4 volts. The motor would not draw a inversely proportional amount of current to "compensate" for the reduced voltage to attain the same power, it isn't a smart mechanism

When I was trying to get a grasp on this, I asked some one about motor starting... and why motor's draw such a high starting current. They said... "it's like moving a shopping cart loaded, it's harder to overcome the friction at first, but when you do, it becomes easier, you need more work at the start" But... I am much smarter than a motor... I know I have to overcome the friction, the motor does not know what it has to overcome...

The motor tries no harder to move the engine shaft than it does my hand if i hold it, it exhibits one locked rotor current and torque for a specified amount of supply voltage.

 

[saskcat]

Hi. Very new to this forum. was just passing by and read most, not all reply.
I am a sparky by trade and believe the phrase "lower voltage will kill the motor" came from the use of AC powered devices. where the r turns to z. on an induction motor (lawn mower, power drill ?, furnace motor, fridge/freezer motor) when the less than designed voltage is applied to the device that is under load, the induction motor will try to = the situation by drawing more amps to make up for loss of voltage and over heat the windings. it still wants/tries to do the work called for and will destroy itself.
i believe in fact any motor, with not enough "energy" applied to it will over heat when trying to move a load. stalled I mean. $.02

 

[psparky]

Hi. Very new to this forum. was just passing by and read most, not all reply.
I am a sparky by trade and believe the phrase "lower voltage will kill the motor" came from the use of AC powered devices. where the r turns to z. on an induction motor (lawn mower, power drill ?, furnace motor, fridge/freezer motor) when the less than designed voltage is applied to the device that is under load, the induction motor will try to = the situation by drawing more amps to make up for loss of voltage and over heat the windings. it still wants/tries to do the work called for and will destroy itself.
i believe in fact any motor, with not enough "energy" applied to it will over heat when trying to move a load. stalled I mean. $.02

This is where most of the disagreement takes place. The motor can not "try" to make up power by making more amps.

This is where I believe the magnetic field is not quite built up enough so the windings in the motor act more like a short than a load...thus increasing amps. The voltage source feels less resitance and reacts with more amps. V=IR

Then I think the motor is bouncing back and forth between full magnetic field and less magnetic field. The extra amps eventually fry out the situation.

Perhaps we are saying the same thing...but in a different way.

 

[psparky]

Incidentally, I'm assuming that motors are built to run under load.

For example...what happens when you run a 100 HP AC motor with no load on it for days on end? Let's say it pulls 124 amps under full load at 480 three phase.

Does it pull 124 amps with no load on it and just free spinning? Is this bad for a motor?

It obviously does pull 124 amps with a full load. How about a full load that actually is more than the motor can handle and slows down the normal RPM...does it increase the amperage then?...I'm assuming it does. I'm assuming it again feels less of a magnetic field (resistance) and increases amps accordingly...obviously bad for motor...or then again the "overloads" in the motor will pop when the current reaches the bad news state.

Straighten me out.

 

[I_am_learning]

If we have started talking Induction motor, then, as is popularly believed, the motor can get damaged by applying reduced voltage.
See the Torque Curve of Induction motor for 2 different applied Voltage

The diagram isn't quite to scale, but please bear with it.
Now, as can be seen, reducing the voltage has only slight effect on the speed. (How much the effect will be is dependent on the design parameter of the motor). Since the Load Torque is assumed constant, it follows that the output power remains more-or-less constant.
Now, since the input voltage has been greatly reduce, the only way the output power is maintained is by consuming more currents.

Of course, this can be explained in terms of the motors equivalent circuit diagrams, and explaining how the effective resistance of the motor decreases with reduced voltage due to magnetic phenomenons etc etc but Energy Conservation Principle Gives a straight answer.

 

[Windadct]

Hello PSparky - the case you describe is typical for an induciton motor - important to note this is an AC machine - so we need to think of V and A in vector form. I know complicated, but necessary. If the V and I waveforms are 90Deg out of phase (as would happen with a pure inductor or capacitor) - you can have full current and full voltage BUT 0 real power. i.e. the motor does no work. This is overly ideal, as the windings have some resistance AND the friction of the bearings etc - require some work, but a free running Induction Motor then has very bad Power Factor ( PF = |Real Power|/Apparent Power) .

Running a motor under no load - - it is debatable if it is bad for the motor, short term no, long term depends on the motor. It is however bad for your electrical supply(if it is a large motor) - as it needs to supply the current - and the power system will experience higher losses overall (I^2R), and not deliver any power. ( Like the mailman driving by your house - but you get no mail )

As the motor does more work the PF increases to about 0.95 - under the absolute best conditions.

For a DC machine (or 3 PH PM Motor - BLDC) - the current is directly proportional to the torque ( think actual work done by the motor) as mentioned above. It behaves differently in many ways - so the IND Motor case is quite different than the OP.

 

[psparky]

Hello PSparky - the case you describe is typical for an induciton motor - important to note this is an AC machine - so we need to think of V and A in vector form. I know complicated, but necessary. If the V and I waveforms are 90Deg out of phase (as would happen with a pure inductor or capacitor) - you can have full current and full voltage BUT 0 real power. i.e. the motor does no work. This is overly ideal, as the windings have some resistance AND the friction of the bearings etc - require some work, but a free running Induction Motor then has very bad Power Factor ( PF = |Real Power|/Apparent Power) .

Not complicated...I'm all over vectors like white on rice. Full comprehension of power factor as well. We are talking about a JWL + R situation...I got it.

Running a motor under no load - - it is debatable if it is bad for the motor, short term no, long term depends on the motor. It is however bad for your electrical supply(if it is a large motor) - as it needs to supply the current - and the power system will experience higher losses overall (I^2R), and not deliver any power. ( Like the mailman driving by your house - but you get no mail )

Gotcha...I guess it would be pretty silly to burn that much power for no reason.

As the motor does more work the PF increases to about 0.95 - under the absolute best conditions.

Copy.

For a DC machine (or 3 PH PM Motor - BLDC) - the current is directly proportional to the torque ( think actual work done by the motor) as mentioned above. It behaves differently in many ways - so the IND Motor case is quite different than the OP.

I don't know squat about DC motors...almost never work with them in industry...at least not yet~~!~!

 

[I_am_learning]

Does it pull 124 amps with no load on it and just free spinning? Is this bad for a motor?

It won't pull full_load current at no_load. Only a fraction of it.
Is this bad? Probably not, at least not for the windings. But since free spinning speed is greater than Full_load speed, bearing might not appreciate it. :)

How about a full load that actually is more than the motor can handle and slows down the normal RPM...does it increase the amperage then?...I'm assuming it does. I'm assuming it again feels less of a magnetic field (resistance) and increases amps accordingly...obviously bad for motor...or then again the "overloads" in the motor will pop when the current reaches the bad news state.

Straighten me out.

All most every motor has short-time over load capacity. So, if you put more load than full_load (i.e. more load than recommended by manufacturer), as you have suggested, it will consume more current. And obviously, its bad for the motor to consume higher than rated current for long duration of time. If there are protection systems, then they will pop out. :)

 

[psparky]

Good answers people.

Oddly enough...motor windings have less resistance with more load...and more resistance with less load!

I remember that because I got the test question wrong in sophomore year!

It's tricky because it's the exact opposite of what you might think. But when you think about it...it's the only way it can work...V=IR...or in this case...
V=I*(JWL + R)...which will automatically shift the current out of phase with the voltage.

 

[FOIWATER]

it won't hurt the motor to draw it's no-load running amperage. No

Just try and read the last post I posted, it is simple.. motor's draw only one current based on supply voltage, because they have a fixed resistance at a fixed temperature.

The motor motors and generates at the same time, because as it starts to move it satisfies the need for induction to take place (relative motion between a conductor (armature coil) and a magnetic field (motor field)). This generation is counter emf and is relative to the amperage of the motor (effecting the strength of the field) and the motion of the armature coil.

A voltage is induced in the armature which, according to lenz' law, BUCKS the source voltage and subtracts from it yielding a total LOWER average armature voltage.

The motor therefore draws Less current as it speeds up.

It therefore draws max current when it is not moving, ie locked rotor, stall, start-up...whatever

As long as it is not moving it draws ONE value of current for a fixed input voltage. It doesn't care if it's couple to a 16 cylinder diesel engine or a paper pinwheel, it will exhibit one value of torque.

pssparky they will be designed to be able to handle no-load current.. it is stall current which is threatening. which the motor exhibits at start-up as well as stall. this is the purpose of soft-starters on motors such as autotransformer starting, resistor starting, et cetera.

 

[FOIWATER]

Good answers people.

Oddly enough...motor windings have less resistance with more load...and more resistance with less load!

I remember that because I got the test question wrong in sophomore year!

It's tricky because it's the exact opposite of what you might think. But when you think about it...it's the only way it can work...V=IR...or in this case...
V=I*(JWL + R)...which will automatically shift the current out of phase with the voltage.

nope... this is wrong.

The resistance does not change with load... you need to research counter emf.

The counter emf is dependent on speed... as such the motor draws different currents based on different speeds depending on how the fields are connected (series or shunt).

Has nothing to do with the load.

Connecting a load slows the motor down, which slows the relative motion between the armature coil and the motor field, which decreases the counter emf induced in the armature, which decreases the buck on source voltage, which causes the motor to SEEM to try harder... but it's not trying harder. the counter emf just decreased under load because the motor moves SLOWER under load. resistance is never changing, unless temperature is changing.

PS:

Let me try to explain counter emf for you in such a way that it makes sense.

Counter emf is generator action in a motor... all DC motors such as engine cranks, experience this phenomenon.

More CEMF is developed as the motor speeds up, therefore it experiences ZERO CEMF at start-up.

The function of the CEMF is to BUCK the source voltage.

at start-up,the motor receives the full 24 volts... as it accelerates... it receives LESS of this supply voltage because it is developing more and more counter emf.

remember your left and right hand rules for motors and generators respectively? FORCE FIELD CURRENT ---> THUMB INDEX MIDDLE respectively.

therefore the armature current when motoring is opposite the armature current caused by CEMF (or, generation while motoring).

It's sort of a touchy subject when you hear it for the first time but, it is CRUCIAL to understand motor behavior. motor resistance does not change though, it has all to do with this CEMF phenomenon.

 

[psparky]

nope... this is wrong.

The resistance does not change with load... you need to research counter emf.

The counter emf is dependent on speed... as such the motor draws different currents based on different speeds depending on how the fields are connected (series or shunt).

Has nothing to do with the load.

Connecting a load slows the motor down, which slows the relative motion between the armature coil and the motor field, which decreases the counter emf induced in the armature, which decreases the buck on source voltage, which causes the motor to SEEM to try harder... but it's not trying harder. the counter emf just decreased under load because the motor moves SLOWER under load. resistance is never changing, unless temperature is changing.

PS:

Let me try to explain counter emf for you in such a way that it makes sense.

Counter emf is generator action in a motor... all DC motors such as engine cranks, experience this phenomenon.

More CEMF is developed as the motor speeds up, therefore it experiences ZERO CEMF at start-up.

The function of the CEMF is to BUCK the source voltage.

at start-up,the motor receives the full 24 volts... as it accelerates... it receives LESS of this supply voltage because it is developing more and more counter emf.

remember your left and right hand rules for motors and generators respectively? FORCE FIELD CURRENT ---> THUMB INDEX MIDDLE respectively.

therefore the armature current when motoring is opposite the armature current caused by CEMF (or, generation while motoring).

It's sort of a touchy subject when you hear it for the first time but, it is CRUCIAL to understand motor behavior. motor resistance does not change though, it has all to do with this CEMF phenomenon.

I was refferring to AC motors...I assume you are refferring to DC motors?

I will act accoringly either way.

 

[FOIWATER]

counter emf pertains to all electric motors.

resistance will not change in AC or DC motors, all dependent on CEMF

 

[FOIWATER]

In fact, AC motors will draw up to 6 times full load amperage at starting.

They are where soft starting becomes crucial in industry

 

[psparky]

In fact, AC motors will draw up to 6 times full load amperage at starting.

They are where soft starting becomes crucial in industry

Very aware of this as breakers I size are typically way, way larger than full load amps. A G.E. or square D chart is great for this...or if there actually is a MOCP number available I will use that.

I'll take a look later when I have ample time and try to digest what you said...thanks.

 

[psparky]

Wait...it just clicked...I think I know what you mean.

In a normal magetic field...the current is driving the magnetic field.

Now as soon as the magnetic field changes speed due to exterior forces or perhaps start up or whatever...now all the sudden the current has a counter effect on it from the change in the existing magnetic field...now the magnetic field is now driving or resisting the current that was originally driving it...

And in the case of start up...there is no back EMF...no rotating magnetic field...nothing...just bare wire...hence full load locked rotor current on start up...for a few cycles anyway.

In the ball park?

 

[FOIWATER]

let me try my best.

You have a DC shunt wound motor, very basic..

assume the shunt and armature are fed by two separate sources.

you can think of your shunt and armature as two separate coils in very close proximity physically.

Your armature has a very low resistance about half an ohm. Your shunt has a little higher resistance, between 1 and 10 ohms, depending on the motor.

you connect your motor to a power source.

the motor accepts current through its winding, both the field and the armature based on their resistance. At this point, the value for accepted current is at a maximum value, supply voltage/ coil resistance = current drawn.

The current through the coil develops a magnetic field around the coil.

We now, due to lorentz' force law, have the basic necessities for motoring. we have a current carrying conductor places in a magnetic field, it will now experience torque.

As the conductor begins to move now however, we have the conditions met for GENERATION... based on Faraday induction law.. which states that relative motion between a conductor and a magnetic field will induce a voltage in said conductor. The conductor my friend, is the armature.. and the voltage is the counter electromotive force. Its value is dependent on the value of field flux, and the speed of the motor rotor (dphi/dt).

The counter electromotive force that is induced, causes a current to flow in the armature, which due to lenz' law, OPPOSES the current direction which induced the voltage, that being the SOURCE VOLTAGE.

Therefore, the faster we spin the rotor, the more counter electromotive force is induced in the motors armature. and therefore, the more it opposes the source voltage, IE: less voltage!

Now look at ohms law...

V = IR

It has been said here that for the current to change under load (speed), the resistance MUST be changing. try my friend to imagine that the VOLTAGE is changing with speed (load), not the resistance. The voltage changes because of CEMF---> and this new value of voltage yields a new value for current drawn. for a fixed resistance and supply voltage.

higher speed = more counter emf induced = more bucking cemf = less current drawn.

So, at high speeds... the motor draws little current.

Now, add a load, and people say, "now the motor draws more current, it is trying harder to turn the load!" ... no.

It draws more current, because it slowed down, and as it slowed down, the armature coil was cut by the magnetic lines of flux of the field less over time, and hence a lower cemf was induced in the armature, meaning MORE of the source voltage was received by the motor, meaning MORE current was drawn.

This is as good as I can explain it, sorry