In my (extremely rudimentary) understanding, the amount of current flowing in a circuit is determined by a) its resistance, and b) the voltage of the power source (voltage from beginning to end), which forces the charge to flow through.

Why then, do people talk about a device "drawing" extra current when e.g. a motor encounters a heavy force? If anything, I would expect this to increase the resistance in the circuit, and thus decrease the current that flows through. What say does a load in the circuit have in how much charge is forced through? How can it draw more out?

Alternately: where is my understanding of these interactions flawed? :)

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    \$\begingroup\$ I suspect they themselves don't quite know what they mean by 'drawing' current. However, a "load" is essentially a device to which power is delivered. Thus, increasing the load on, e.g, a motor, requires the motor to deliver more power and, assuming the voltage to the motor is (more or less) constant, this means an increase in current through the motor (electrical power is the product of voltage and current). \$\endgroup\$ Jan 9, 2014 at 2:53
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    \$\begingroup\$ It might be better to say a device or load "accepts" or "allows" current to flow, rather than saying it "draws" current. A common misconception by beginners seems to be that a power supply will force its rated current through the load - this is incorrect, only as much current as the load will accept will flow. \$\endgroup\$ Jan 9, 2014 at 3:07
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    \$\begingroup\$ If you are asking about motors rather than the term "draw", the resistance does not change under load but something called the back-EMF does. The back EMF of a freely spinning motor will cause the current to be at its minimum. The back-EMF disappears when the motor is stalled and you are left with the resistance of the windings which is relatively small. \$\endgroup\$ Jan 9, 2014 at 3:25
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    \$\begingroup\$ Also note that if you plug up a vacuum cleaner, the motor pitch increases. This is not, however, because the motor is working harder. It is working less hard! It is not able to move air, and so it has nothing to do. Being more lightly loaded it does what lightly loaded motors do: it speeds up. \$\endgroup\$
    – Kaz
    Jan 9, 2014 at 5:56
  • \$\begingroup\$ possible duplicate of Do electrical devices "take what they need" \$\endgroup\$
    – Phil Frost
    Jan 9, 2014 at 16:06

7 Answers 7


Think of it as "drawing" extra breath whilst jogging as opposed to walking.

A circuit under normal conditions will appear as a certain impedance. For instance, a DC motor operating without a mechanical load will spin at a rate determined by the number of its windings, contacts, permanent magnets etc. As a load is applied to the shaft, the rotor decelerates, reducing the impedance of the windings being contacted. Simplistically, the impedance is determined by the speed (frequency) at which it spins. As the windings are inductive, the reduction in angular frequency reduces the impedance. As a result, the current increases, so it "draws more breath", so to speak.

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    \$\begingroup\$ Ah, excellent! So for a person who has not yet really studied AC, I can for now think of it (very approximately) as the resistance in the main circuit actually being reduced when the motor is working against something? All is right in the world again. Thanks! \$\endgroup\$ Jan 10, 2014 at 21:52
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    \$\begingroup\$ Yes that is correct. You will find during your studies, that the impedance (complex resistance) of a winding is dependent on the AC frequency. In a motor there are many other contributing variables but the essence is that XL = 2*pifL. So the complex impedance reduces as frequency reduces as suggested in the inductor reactance expression. The DC (real) resistance does not change, but the complex impedance is the vector sum of R and XL so it is higher as AC is applied, or in the case of a DC motor, is generated by switching windings through brushes and contacts. \$\endgroup\$
    – Martin
    Jan 10, 2014 at 22:38
  • \$\begingroup\$ So in case of dc with resistance the current or power drawn will be constant as the resistance is constant. and the drawn power or current depends entirely on resistance right? \$\endgroup\$
    – Alex
    Dec 10, 2023 at 7:47

By "drawing extra current", most people mean exactly that, as for "being under [extra] load", it means a similar thing, i.e. the device has to deliver more power to it's load.

If you think about it, if the current (and therefore power assuming constant voltage) in a motor decreased under heavier load, then a fundamental law of physics would be broken - more power going out than in. There is much detail that could be gone into here, but essentially Counter Electromotive Force (or back EMF) is the reason motors and other similar things work as they do (see Lenz's Law also). A rough analogy is maybe that you can think of it as the motor drive actually being physically connected to the load (similarly with a transformer)

It's quite easy and informative to do a quick test with a (preferably smallish) motor, half decent (or bench) power supply, and multimeter (or scope with current probe)
Set the circuit up to measure the current through the motor, then observe current variations from start up on up to full speed unloaded, then apply a small load and gradually increase the load until the motor stalls (if safe for the motor and yourself - check datasheet, in any case make it briefly. Most half decent datasheets will give data on winding resistance, stall current, unloaded current, graphs, etc)


When Engineers talk they want you to know that they are very smart and have their own language just to prove it... Kinda just like "COP TALK" ie: "the perp exited the vehicle and entered the establishment at which time he secured the purchase of a coca-cola product just prior to re-entering his vehicle ...."

For electrical or electronic purposes:

"Drawing" means consuming, pulling or using some resourse....Just as you may "DRAW" the milk up the straw from your cup, which is the source, in electronics you can "DRAW" more current from the source. Used almost exclusively in the term "Drawing Current"

"Current" is a term we use to visualize/represent/measure electricity running through a wire.

When people say a device is "drawing current" they simply mean that the device is pulling or using power from the power supply. In electronics we are always obsessively worried about how much juice our devices are requiring or currently using. Maybe we are worried because we are using a battery and we are worried we may kill the battery if we "DRAW TOO MUCH CURRENT" Or perhaps we may have a problem with a device. Maybe it doesnt turn, or light up, or make the correct sounds.... BUT!!, it does seem to be "Drawing a lot of current" Thats valuable information to the troubleshooter and it sure does make you sound smart to people that dont know what you are talking about.

A "LOAD" is simply a device, or part of a circuit that is using power from some electrical source. When looking at an electrical circuit that operates something such as a washing machine, air conditioner or hard drive in a computer, we say that component represents or is in fact "THE LOAD" of the circuit.

BUT!!!!! The amount of LOAD can change depending on what the device is doing at any specific time.

Alternatively, the "LOAD" may also be referring to the total amount of power that is being pulled from the power source. So, we may also say, "The air conditioner is putting a heavy LOAD on our circuit or collectively the air conditioner, washing machine and the computer is too much LOAD for a single household circuit.

When a device is under load that means it is doing work and it needs more power from the power source or, "Power Supply". You may/will see the term, "It is under heavy load" Obviously, HEAVY LOAD gives the idea that the device is working very hard, perhaps near its maximum capability. A light load assumes the opposite and also obvious.

So the correct question is not really why does a device under load draw more power, it is really more of an understanding of what the term load actually means in this complex Engineers vocabulary. The fact that a device is "Under load" actually directly means that the device is simply using power or current from the supply. So, for any specific circuit, the heavier the Load becomes, the more current or power will be taken or drawn from the power supply. A circuit that is NOT under load is not drawing any power or current at all.

Please note that "Power" and "Current" have been used in my explanation just as a generic way to describe electrical energy being used by a device or circuit. Technically, "Current" and "Power" although they are both directly related, are 2 different measurements and have 2 different values for any given circuit.

I hope this helps and hits the heart of your question..

Thanks for reading,

Keith Danhardt

SORRY! I only saw the top portion of your question about drawing currents and loads... Throwing in the motor thing shows me that you have a higher level of understanding than I was actually writing for.

Now the motor thing does throw in a bit of a monkey wrench into things simply because of the incredibly interesting things that actually go on when a motor is running.

It seems pretty obvious that any device that has a heavier load will require more power to operate. For example a motor that is currently lifting 10 lbs should pretty much draw double the power when the weight changes to 20lbs. Not exactly though, due to many reasons like friction and other factors, but suffice to say that logic would dictate that a machine that does double the amt of work should use double the energy (everything else being equal). So in a way, "LOAD" may be fairly described as the amount of work a machine is doing. So, the heavier the lifting in our example, the heavier the LOAD, the more power is required. Pretty straight forward.

So looking at the motor thing strictly as a DC Ohms law thing, and considering your level of understanding, there should not be any question why a heavier load will increase the current in a circuit. . When the load gets greater, effectively the resistance of the load decreases. So if the applied voltage stays the same but the load resistance goes down, obviously the current must go up. Simple ohms law. Only problem is, the numbers dont work.

Looking at this from a straight Resistance, Voltage, Current relationship, the motor thing doesnt seem to make electronic sense. The numbers dont calculate out the way you thought. And this is the exact reason that I chose not to pick AC theory or communications as my major field of study. When you get into these theories, things start to appear to break the old OHMS law thing. Notice I said appears. When you finally sit down and do 4 pages of mathmatical equations, all based directly on and in accordance with ohms law, it all works out and proves out to be exactly what they said should happen even though it doesnt seem to make sense at first glance..

What is actually going on in a motor when it is running is a complex series of interracting events that are all affecting the current flow in their own way. Along with friction, heating of the windings, and some other minor things taking place, there is something called counter EMF. This is the most influential factor believe it or not.

When you are running an electric motor(lets please just stick with a dc motor for our purposes. My brain is already starting to hurt just thinking of trying to explain the AC motor.), theoretically the only power that is consumed is the loss in the friction of the bearings and the coil windings. Otherwise an electric motor would "theorectically" draw no power. Because of the design of the electric motor, it actually generates its own electricity. .......in a way....... Just as a transformer or electric generator work , the electric motor also employs the idea that a charging coil of wires will actually contain energy in the magnetic field that surrounds itself when a electric current is passed through it. When this field collapses, it induces a voltage on the surrounding coil of wires 100% equal and opposite of the current that was used to charge the coil initially. (minus the losses in the coil.) This is called the counter EMF. In a Transformer or generator device, the resulting electrical current produced is sent to its load or power supply to be used as needed. But in the electric motor, this reverse current flows back into its own power supply having the effect of seemingly replacing the current that was originally pulled from it. Now add in the heating of the wires, the effect of the permanant magnets that are also part of the DC motor, and other factors and for me at least, it becomes mathmatically imposible to calculate.... Well, not really, but...Get a Watt meter and measure the actual power. Much easier..Do the math one time in your life to prove out the theory but after that just trust the Watt meter. If you attempt too many of these type calculations in your life time your head will explode so please use extreme caution.

One thing missing from the above explanation is that although we were talking about a dc motor we are still dealing with an alternating current building and collapsing on a coil because as a DC motor rotates, it is constantly reversing the polarity of the charge in the coil wires effectively producing an AC voltage. This could probably take a lot bigger and better explanation but i got to cut it off somewhere.

Ok, so now to explain why the current increases when the motor is held back or even stopped while full power is still applied. Now that the motor is stopped lets say, the magnetic field around the coil never collapses. Without the motor turning you are simply running full voltage directly into a straight piece of wire. A long coiled wire perhaps, but still not much electrical resistance. So without the on and off switching from the rotation of the motor, the full voltage from the power supply is constantly applied to the motor coil. Then the coil starts to pull massive amounts of current from the supply and at the same time heating up the coil wires trying to expell this basically shorted energy. Hence! Current goes through the roof and you most likely destroy the coil windings. It is easy to look at it on the surface and say, since the motor is not turning it must be drawing less current/power... But as I hope I have explained, thats not really how it works..... Once again I hope my explanation has helped..


  • \$\begingroup\$ That was quite an answer! Thanks for your explanation. \$\endgroup\$ Jan 10, 2014 at 23:44
  • \$\begingroup\$ Ah, yes, this also explained to me why resisting the motor's motion might cause things to overheat. You're effectively short-circuiting it. Thanks again! \$\endgroup\$ Jan 10, 2014 at 23:46

At the end of the day your confusion stems from Ohms Law. Ohms law only applys to components that are ohmic (that is to say their resistance changes very little with respect to other variables, its a bit stupid, in that the definition basically means "A component that adheres to Ohm's law).

Fact is the majority of components are non-ohmic. This of course means that its resistance can change and thus the current changes. But since resistance is something we made up (there is nothing physical about resistance), the actual change is a change in current, which amounts to a change in Voltage/Current or Resistance.

Okay...so there are plenty of objects out there that are non-Ohmic. The simplest I can think of is the solenoid (or electromagnet). An "ideal" solenoid has an imaginary resistance and is very non-ohmic. It resists CHANGES in current. Given that these exists in all motors, you should be able to understand that motors are NON-OHMIC.

Thus current draw is not proportional to voltage on motors.

Another way to think about it is with pipes. A simple pipe will flow more water the more pressure you put behind it. The thicker the pipe, the more water flow per pressure.

However, you can also have things like cisterns in your plumbing. When you turn on the water pressure, lots of water flows. However, when the tank starts to fill, the water stops flowing. Can you talk to me about the resistance of the cistern?

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    \$\begingroup\$ I'm trying to get my head around "there is nothing physical about resistance". \$\endgroup\$ Jan 9, 2014 at 9:14
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    \$\begingroup\$ thinkgeek.com/product/60fd \$\endgroup\$ Jan 9, 2014 at 10:58
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    \$\begingroup\$ How does one "measure" resistance? It is merely the ratio of current to voltage. If there was no current, or voltage, you CANNOT measure resistance. For some things, this ratio is mostly constant, but for most its not true. \$\endgroup\$
    – Aron
    Jan 9, 2014 at 15:13
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    \$\begingroup\$ It's a physical property of materials, just like density or dielectric constant. \$\endgroup\$
    – pjc50
    Jan 9, 2014 at 17:35
  • \$\begingroup\$ @pjc50 please describe the how you define the physical property of resistance for a plasma. \$\endgroup\$
    – Aron
    Jan 10, 2014 at 1:56

1) We refer it as "drawing" current because, When a load is connected to the source,electrons tends to align themselves and move towards positive terminal of the source.For sake of measuring amount of electrons(charge) and also the time taken,we have quantity called current. And we assume the direction of current to be opposite to that of flow of electrons.So it is termed as 'drawing' current from source,which actually means source drawing electrons from load.

2)We cannot jump into a conclusion, that load draws more current directly. Consider a resistive load connected to a dc source, it draws a current of V/R. Suppose a short circuit happens the R gets dropped to zero.Then current is V/0 ie., infinity. This is undesirable fault in a circuit. And on OC,it draws zero current. So ,when a source is connected to a load, it draws optimal current.


The power delivered to a motor is V x I. If the voltage is constant and the motor demands more power (because its mechanical load increases), more current is taken from the voltage supply.


The electrical potential of a point or "node" is the extent to which it attracts electrons; if allowed to do so, electrons will flow from points of lower potential to points of higher potential, doing work in the process. It is possible to push electrons from points of higher potential to points of lower potential (that's what batteries and power supplies do), but doing so requires adding energy from somewhere else.

A device is said to "draw current from a node" if it allows electrons from other (typically lower-potential) nodes to flow to to that node. A device is said to "sink current from a node" if electrons from other nodes flow to that node, whether that other node is of lower potential (so it can simply let the electrons flow there) or the other node is at higher potential and the electrons need to be pushed. A device is said to "source current to a node" if electrons from that node to flow to other nodes (again without regard for relative potentials). I don't know of any term which would be like "source", but with the implication that electrons were simply being allowed to flow.


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