Now, I'm just looking for a wattmeter, that isn't dependant on Amps,
but resistance and voltage.

I was thinking, that it would find the amps, with I=RV (R=resistance
and V=Voltage) and then finding the the watts with P=VI, from the
measured voltage and the calculated amps. BTW, this is for AC power,
variable voltage, amps, and power.

Is there a cheap solution for this (DIY, premade for >$75USD)?

Since it is AC power, P=VI isn't total correct.  S (apparent power) =
IV and P=VI cos(phase angle between V and I) or P=VIPf where Pf is the
power factor and equal to the cosine of the angle between V and I.  P
is always less than S and as a general rule the are pretty close in
value.  If you are looking to ignore this difference and are not
looking for real time numbers (which I assume since you are looking to
calculate P), all you need is a digital multimeter, then measure V and
R seperatly and P~V/(R^2).  Note: measure R without power source
disconnected otherwise you would be measureing your Resistance in
parrellel with the power source's resistance thus lowering the
resistance and increasing your power by a factor of a square.

One side note:  I=V/R not RV.

I hope this helps

Kmclean, BSEE, EIT
Electrical Engineer

Well, see the problem is that I need realtime Wattage output, and in
the applications that I plan to use it in has a constant change in
voltage and resistance. So, the multimeter idea is not going to work.

And thanks for the part on I, I meant to write it as I=V/R.

how much power are you expecting?

Is your load purely resistive?  Or is it inductive/capacitive?

If it's purely resistive, your apparent power (calculated) and active
power (measured on the wattmeter) will be fairly equal.

If your load is inductive/capacitive, they'll hardly ever be equal -
I*V most likely won't equal your measured P.

If you're unsure about your load being resistive or
inductive/capacitive, if it's more like a light bulb or toaster, it's
mainly resistive.  If it's more like a drill, motor, or microwave -
it's inductive/capacitive.

kmclean - is that right?  My background is mechanical, but I'm
beginning a MS program in Electrical Engr...

Sorry about a delay in comments on my part. I live in Florida and had
to rush to Oklahoma to deal with some family issues. Now here we go.
The load is inductive I believe (I'll be using this to measure the
power going to speakers, drills, and things like that). I'm expecting
a power input that can range anywhere from 50W to 600W. Resistance
would be something like 1-7000 ohms.

Hope that answers some questions.