I am afraid 1.2Ohm will not cut it. :?
At 9.6V (let's call it 10V just for simpler math) and 5Amp current
load is 10/5=2Ohm. With stall current of 10A, load resistance is
down to 1Ohm.
Power used by the tool (DC motor) is 10*10/2=50Watt continuous
and 100W stall. If you put 1.2Ohm in series current will be limited
to about 5Amp all the time. At 5A current through 1.2 Ohm resistor,
you get 6V dropon resistor and only 4 (or 3.6V) on motor.
The resistor would burn some 6*6/1.2=30Watt (and that's 650% over
it's rating) while motor would get only ca. 20 Watt.
Compared to 50Watt he has now, that's dramatic power loss and
huge waste of batteries. Besides 30Watt is lot's of heat, I have
soldering irons as low as 8 and 15Watt ;-)
Resistor that would be used for this application must be VERY low.
I would use 0.005 Ohm / 2Watt. They are available through regular supply
channels. At only 5mOhm and at full 10Amp stall current,
voltage drop is only 0.05Volt and heat disipation is just 0.5Watt.
Regards,