Think about the losses involved with using a pot. Suppose you have a 1HP/750W motor that runs on 100VDC (because why not). This comes out to 7.5A of current. You want to run it at 50%? Suppose then you need to drop it to 50VDC. Suppose for the sake of this example, the effective resistance of the motor is 13.33Ohm (this is a really gross approximation). Your potentiometer would have to be about 13.33 Ohm too. The input power is 375W. Power to the motor is 187.5W. Power burnt in your potentiometer is *also* 187.5W. This is a pretty poor efficiency - 50%.

Now, we go to PWM. We use FETs or IGBTs to switch the connection of the motor in such a way that it's either on 100%, or 0% (off).

Suppose we only use one FET with an RDSon of 0.1Ohm.

100V input, 13.3333Ohm of the motor, 0.1Ohm of the fet. Power delivered to the motor is 738.9W, power lost in the fet? 0.075W. Our delivered efficiency while on is 99.99%.

Now we PWM the thing, 50% of the time, the motor is on. 50% of the time, it's off.

Let's call the time when it's on "t1" which is t1 = 0.5 T (T is the period of the frequency)

The time t2 = (1-0.5)T (the off time)

This means, that we can average the "power" delivery:

Pavg = 738.9W * t1/T + 0*t2/T

Pavg = 738.9 * t1/T

Since duty cycle is what is recovered from t1/T

D = t1/T

Pavg = 738.9W*D

For 50% on? Pavg = 370W

25%? 184.7W

The value add? We're not wasting power with the potentiometer. This is _very_ efficient.