A motor rated for 3S (11.1 V) has an internal resistance of 0.12 ohm. The maximum current is 22 A.
11.1 V / 0.12 ohm = 92.5 A
Doesn't this mean that by supplying a 11.1 V three phase current, the motor will burn up instantly? How does an electronic speed control (ESC) prevent the current from exceeding 22 A?
It doesn't, the motor itself does. Once the rotor starts spinning, the motor produces a voltage that opposes the flow of current; this is commonly called "back EMF (electromotive force)".
The motor's speed increases until the back EMF reduces the current flow to the level needed to account for the actual physical load on the motor (plus losses).
The heavy current you calculate is drawn only for an instant, just as the rotor starts spinning. If the rotor is prevented from spinning, then that current will be drawn indefinitely, and yes, it can destroy the motor.
Don't forget about the inductance and back EMF. If you were to put 11.1V DC across the winding you would wind up with 92.5A of current in that phase, but the impedance to an AC signal is higher. Once the motor starts turning it generates an internal voltage, the back EMF that fights the drive voltage. In many drives the current is controlled via current feedback from each phase, so that the drive current can't exceed the maximum. Other schemes have over current protection provide by a comparator on a sense resistor at the bottom of the 3 phase bridge.
An inductor may have a DC resistance of 0.1 ohms and if you supplied it with 10 volt DC it would dissipate 1000 watts and probably near enough instantly fry and smoke.
The thing about a motor is that it either commutates (dc motors) or it is fed from AC (which is another way of commutation) . The point is that the voltage to the "active" coil part of a motor reverses polarity often enough to prevent the DC situation causing burn-out.
