Resistors – What Limits the Heating Power Output of a Resistor Heater?

First concentrate on each of your 12 ohm resistors. Their max rated dissipation is 10 watts. The quickest way to find volts and current from power and resistance is max_voltage = sqrt(max_power * resistance) (or you could simply calculate current and hence power for each of several applied voltages). In this case, sqrt(10*12) is roughly 11v.

As you have a 12v supply, you need 2 in series, where the 24ohm total will give you 0.5A, and so 3 watts per resistor. With a 10w rated resistor, if you feel it's getting too hot at 3w, that may mean it's not heatsinked sufficiently, or your expectations need recalibrating. A 10w resistor would need to be big to be air-cooled, or to get very hot. If it looks like it can be bolted down, do bolt it to a heatsink. If it's ceramic with no flat surfaces, then it's supposed to run very hot in air.

Two parallel strings of two series resistors will give you a total of 12 watts. Is that enough for your application?

[hooligan alert] In my world (turn it up until it blows up, then back off a bit), 12v on a 12ohm resistor will give you 12 watts, which a 10w resistor will probably survive for a while, as long as you keep it cool. Four such resistors in parallel would give you 48 watts, enough? [/hooligan alert]

You were generally correct in your figurings BUT calculated for new power loss of 50 Watts total. The heating pad would get 1/3 of that or about 16 Watts and the resistor about 34 Watts. As Rseries is added the current decreases (so total power drops) AND the portion available to the eatong pad drops - so power is related to I^2.

Throughout "=" === "~=" :-)

Assume 150 W , 120V heating pad. Power = V x I so I = Power/V
I = 150/120 ~= 1.25A = 1250 mA.

How big is existing R?
Power = V^2/R so R = V^2/Power
So Rpad = 120^2/150 = 96 Ohms.

For 50W in pad you want new I to be smaller by a facti\or of 3^0.5 as power is to current squared.

So new Ipad = 1250mA/3^0.5 = 1250/1.732 = 720 mA.

As I = V/R you want total new R to be 1.732 x as large as before so added series R is 0.732 X existing R Rseries = 0.732 x 96 ~= 70 Ohms.

Sanity check.
I = V/R = 120/(96+70) = 720 mA
Power = I^2 x R = (0.72)^2 x 96 = 50 Watts in heating pad. Power in 70 Ohm resistor = 36 Watts.

_____________________

Diode + Resistor. As Spehro say - adding a series diode is an easy way to halve the power to about 75 Watts. You can then if wanted add a smaller series R to lower the power further.
From above (75/50)^0.5 = 1.22
Extra series R = Rpad x 0.22
= 96 R *0.22 = 21 Ohms. (eg 20, 22, ...)
Power in 20 Ohms = 20/96 x 50W = 10W.
Rate resistors at ~= 50% of max rated power for good lifetimes.
2 x 10 Ohm 10W resistors on series or 2 x ~= 40R 10W resistors on parallel would be "about right".