I think this can be better explained with motor curves. This curve shows the values when **effective voltage (PWM and battery voltage) is fixed** and you vary the load on the motor.

**Current is always proportional to torque**, with no exceptions. It is also proportional to [effective voltage - back EMF], and back EMF is proportional to the actual speed. This is why in the graph, as RPM goes up, current/torque goes down. For a higher KV motor, the slopes in the graph would be different. All of the slopes would be steeper, and the speed curve would have a higher y-intercept (free speed) and lower x-intercept (stall torque). Because the amps/torque curve is steeper, **higher KV motors require more current for the same load**.

Now suppose we double the battery voltage, doubling the effective voltage. All the slopes in the curve stay the same, but the axis values are doubled. For example, the 12000 rpm point becomes 24000, and the 0.6 oz in point becomes 1.2 oz in. Because current is proportional to torque, the **stall current doubles when the voltage is doubled**. If you double the input voltage, your motor will also be capable of twice the torque and therefore twice the current. So if something goes wrong in the ESC, it will cause much more damage to the motor at a higher voltage.

Here is where it gets confusing. If we overlay the original curves on top of the new curves (double voltage) and pick a point on the x (torque) axis, we see that with double the voltage, the motor is spinning faster when delivering the same torque/drawing the same current. This makes sense from a power perspective, we have double the electrical power, so we should have more mechanical power (torque * speed) too. Now, as a thought exercise, suppose we change the gearing (which preserves mechanical power) so that the speed is halved, and the torque is doubled. But the load is still the same, so the system doesn’t need twice the torque that we just gained. So the operating point on the motor curve moves to the left, because the motor is now only delivering half the torque is used to before the gearing change. We didn’t change the mechanical power, but we halved the current, so **by increasing the voltage, less current is required to produce the same power**. Moving left on the curves also tends to mean better efficiency, so increasing the voltage also helps with the [mechanical power / electrical power] ratio.

Hope this wasn’t too confusing.