I'm missing something fundamental in my understanding of the torque of a motor.
Consider the following analogy:
I have a 10kg dumbell on the ground and a 1kg dumbell next to it. Let's say that I lift the 10kg dumbell off the ground with a certain force 'F'.
Now let's say I use the same force 'F' to lift the 1kg dumbbell. Naturally, I would lift the 1kg dumbell much faster, right?
Now coming to electric motors, let's say I attach a 10kg load to the motor shaft. The motor would produce a torque T and rotate at a certain rpm. Now, I put 1kg load on the shaft.
As per the dumbell analogy, wouldn't the 1kg load cause an increase in speed because it's lighter than 10kg?
But we know that a motor's torque keeps changing with the load. So for a 1kg load, does the motor produce just enough torque required to rotate a 1kg load?
If yes, how would the motor know what the load is so as to produce the right amount of torque?
If no, them wouldn't the motor produce the same amount of torque for all loads and only the rpm keeps changing?
Best Answer
Assuming we are talking about a shunt wound or permanent magnet brushed DC or BLDC motor, torque is proportional to current, so the more torque load you put on the motor the more current it will draw to match it. However the current causes a voltage drop in the winding resistance, which reduces rpm. How much the rpm drops depends on the resistance. A large motor with thick low resistance wire could lift a heavy weight without its rpm dropping significantly, while a small motor with thinner wire might drop down to stall speed and not draw enough current to lift the heavy weight at all.
To understand the relationships between torque, current, rpm and voltage, consider this simplified model of a PMDC motor:-
simulate this circuit – Schematic created using CircuitLab
When the motor spins it generates a voltage proportional to its speed. This is a result of Faraday's Law, which says that moving a wire through a magnetic field induces a voltage in it proportional to its velocity, and occurs whether it is driven by an external force or by its own power. Depending on how it is constructed (number of turns, size, stator or armature slots, magnet poles etc.) each motor has a characteristic 'Kv' (Velocity constant), commonly expressed in rad/s/Volt or rpm/Volt.
Output torque is proportional to current according the Lorentz force law, which says that the magnetic force on a wire is proportional to current x magnetic field strength (which is constant in a permanent magnet or shunt wound DC motor). For a particular motor this relationship is a constant, called Kt (Torque constant), often expressed in N.m/A.
By symmetry, Kt is the inverse of Kv. So if you know one then you know the other. If Kv = 10.5 rad/s/V (100 rpm/V) then Kt must be 0.095 N.m/A.
For this simplified model we will ignore any internal losses due to magnetic hysteresis, eddy currents, friction and air drag. If the motor has no load then torque and current are zero, and no voltage is dropped across Rm. That means the generator voltage must equal the supply voltage, and the motor must be spinning at 100 rpm/V * 10V = 1000 rpm.
Now put an external torque load on the motor. It will slow down and produce less voltage, causing voltage to be dropped across Rm. By Ohms Law a current must then flow, producing torque. Speed continues to drop and current rises until output torque matches the externally applied torque.
If the torque load is high enough to stop the motor spinning then there is no generator voltage and the full 10V is dropped across Rm, causing a current of 10V/1Ω = 10A. With Kt = 0.095 N.m/A and current = 10A, the 'stall' torque must be 0.95 N.m.
Now imagine an otherwise identical motor which has thicker wire, reducing Rm to 0.1Ω. The same 10A current (producing the same 0.95 N.m torque) now only drops 1V across Rm, so the generator must be producing 9V. This motor must be spinning at 9*100 = 900 rpm (only 10% less than the no-load speed) while lifting the weight that stalled the other motor.