I think that power supply DC current must be higher than 1.25A of DC current. (15W is pretty small amount of power)
Electromotive force depend mostly on current so you will be reducing the current in the circuit to reduce FFB. I can not help you more because I do not have the PSU or motor specs. You can anyways calculate the motor equivalent resistance to figure out what the additional resistor should be if you want to reduce current (FFB) e.g. to 80% of original. (and open your physics book if you are rusty on basic circuit theory
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15W isn't much, no, but when it comes to motors you apply the equation F=BIL, Meaning the force (F) equals the product of the magnetic flux of the permanent magnets (B), the current in the coil (I) and the length of the coil (L), so if the magnets were strong enough and the coil was long enough the motor could theoretically work on milliamps. Also I'm pretty sure that 15W would quite easily kill a person.
Also electromotive force doesn't depend on current at all, it's the other way around. Electromotive force, EMF, is what most people call voltage. When you pass a voltage through a conductor, current is the result; you don't pass current through a conductor to get a voltage because current doesn't exist without the voltage in the first place.
To put it more simply, if you have a 12V 1.25A regulated power supply, you won't always have 12V at 1.25A from the supply. You always, always get 12V because that's how transformers work (the ratio of primary to secondary coils in the transformer steps down the 110/240V from your wall to a definite, unchanging voltage), but the current depends on what the 12V is flowing through. Ohm's law states that V = IR (I is current), which can be written as I = V/R. That means that 12V through 10R (10 ohms) will give you 1.2A, 12V through 20R will give you 0.6A, and so on.
So you see, increasing the resistance with resistors will reduce the current because the same amount of voltage is going through a higher resistance, which means less current is being produced.
I don't know what kind of resistance you'll want to add, but I'd estimate the motor is approximately 10-15 ohms or so; if you apply Ohm's law, 12V divided by a range of 10-15 ohms gives you 1.2 to 0.8A. I'd be surprised if it's as low as 10 ohms because running at a steady 1.2A on a 1.25A power supply is very close to the limit. If I were you I'd basically assume some values for the force of the motor, the length of the coil and the strength of the magnets and use 1A for the current in the F = BIL equation, so you could say F is 100, B and L together equal 100, because 100(F) = 100(BL) x 1(I). Then use V = IR to drop I from 1 to, say, 0.7, and that should in theory, reduce F (the thing you have an issue with) to 70% of the original. V, B and L are absolute constants, R is what you want to change, I and F are what will change as a result. You don't need to be incredibly accurate but I would guess the resistor will be something like 5 ohms or so, maybe less. Then you need to consider the power dissipation in the resistor, the equation for that is P = VI. If you have 12V at 1A, you need a 12W rated resistor at least, but I'd get one higher than that to be safe.
I hope this helps you or anyone in some way, I've spent about an hour typing this post trying to make it make some sense. In case you're wondering where I'm getting my information I'm an electronic engineering student and I deal with stuff like this all the time. I think I'd been drinking when I posted my last reply so I wasn't in the mood to really think about it properly...
Edit: As an aside, in case you're wondering why the power supply is 12V and not lower, like say 5V, it's because it's much easier to get 1A (for example) from 12V; if R = V/I then 12V/1A is 12R, 5V/1A is 5R. It's just much easier to increase the voltage than it is to decrease the resistance, especially when you're running off of at least 110V in the first place.
