Q:

How to determine watts and ohms for circuit board resistors?

I am about to embark on my first soldering job to fix a module that is broken inside my truck (compass/temperature display).  It is a well known design deficiency in this particular vehicle brand as the failure has happened a lot.  In most cases, one of the two 51 ohm resistors overheats and for reasons explained in this article, which also, to a fair degree outlines the fix.
What confuses me is how he came up with the idea to replace a single 51 ohm resistor with three 1W 150 ohm resistors (the formula for calculating net resistance):

A little research told me that the resistors have values of 51 Ohms. Figuring in the 12V power supply, that means they are probably dissipating about 3 Watts each. That equates to a lot of heat in such a small package. No wonder they keep failing. Who thought this design was a good idea? I decided to replace the single resistor with three 1W 150 Ohm resistors in parallel, and extend them off the board and out into the air so they can cool by convection. In my case, it was only one of the two resistors that kept breaking loose. Some people only have trouble with one of them. Others have trouble with both of them. I only replaced the one that has repeatedly failed. If the other one ever fails I will replace it too.

I went to Radio Shack to get the supplies necessary but they didn't have 1W 150 Ohm resistors, only 1/2 W 150 ohm so I got those thinking that in such small wattages (?), it may be okay.
I am looking for an explanation how to, in such parameters (12V DC, small electricity usage, etc), you would calculate what wattage and ohmage is necessary for your resistor.  I haven't done any electronics but have lots of experience with residential electrical work and how electric load is balanced and how circuits should be structured.

A:

The equation is basic Ohms Law with parts replaced for power calculation.
The formula is:
$$
P=\frac{V^2}{R}
$$
As it's got 12V across the resistor, and the resistor measures 51Ω, that becomes:
$$
P=\frac{12^2}{51} = 2.82W
$$
So you need at least 3W to be safe.
Placing resistors in parallel results in lower resistance.  The long form of the formula for that is:
$$
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}
$$ 
So taking three 150Ω resistors in parallel gives you:
$$
\frac{1}{R_T} = \frac{1}{150} + \frac{1}{150} + \frac{1}{150}
$$
$$
\frac{1}{R_T} = 0.02
$$
$$
R_T = \frac{1}{0.02} = 50\Omega
$$
If all the resistors are the same value then it is a simple as dividing the value by the number of resistors, and the power ratings just add together, 20 three 0.5W resistors would give 1.5W of power.  Not enough for your application.
If you cannot get 1W resistors then you would be better with six 0.5W resistors of double the value - so 300Ω each.

