PWM or Variable Resistor
One thing not explored in the video is how the load responds to the pulses of voltage and current produced by PWM. Depending on the type of load the result may not be as efficient nor as linear as expected.    

In the case of a DC motor, PWM frequency is an important variable which can have a dramatic effect. At very low PWM frequencies the motor will speed up and slow down as it is turned on and off. As the frequency is raised the motor's speed becomes more even, but it will still produce noticeable vibration and noise right through the audio range. Using an ultrasonic frequency (above 20KHz) eliminates the audible noise, but increases losses in the switching transistor and driver circuit.

Another effect of PWM is an increase in rms current and voltage relative to their average values. This causes extra loss in the resistive components of the load. 

Consider this simplified equivalent circuit of a 12V DC motor, being powered from a 6V supply to get half speed:-







simulate this circuit &ndash; Schematic created using CircuitLab

R1 represents the resistance of the motor's armature windings and brushes, and V2 is the voltage the motor generates while spinning at half speed. R1 drops the difference between the supply voltage and generator voltage, 6V-5V = 1V. Ohm's Law tells us that current = voltage / resistance, so the motor must be drawing 1V/1&ohm; = 1A. Power = Volts*Amps, so 1 Watt of power is lost in the motor.   

Now consider what happens if  instead of using a smooth DC supply we apply 12V with PWM. This rapidly pulses the motor on and off, lowering the average voltage to get the same effect as 6V DC. Logically you might think that applying 12V with a 50% PWM ratio would do it. However what actually happens is that during the 'on' part of the PWM cycle the motor receives the full 12V, so current jumps to (12V-5V)/1&ohm; = 7A. Since this current flows 50% of the time the average current is 7A*50% = 3.5A, not the 1A we wanted. This extra current makes the motor speed up. 

To get half speed we must lower the PWM ratio until the motor draws an average current of 1A while generating 5V. This requires a PWM ratio of 1A/7A = 14.3%. During the PWM 'on' time R1 is dropping 7V and passing 7A, so it wastes 7*7 = 49W of power. During the PWM 'off' time the current is 0A so there is no power loss. The average power loss is now 49W*14.3% = 7W, much higher than the 1W we were expecting!

Luckily, most motors also have considerable inductance, which resists current change. If the PWM frequency is made high enough then the current variations will be smoothed out and the PWM ratio to motor speed curve will become linear (ie. 50% PWM = 50% speed).

Due to this effect, to properly compare PWM vs. resistive speed control you must measure the motor speed, not just assume that it will be proportional to average voltage. Trying different PWM frequencies will show you the effect it has on efficiency and speed control linearity.