A Stupid Ohm's Law Problem

[bob p]

Ok, I confess that I'm scratching my head trying to solve a simple Ohm's Law problem. I'm going to hide behind the excuse that its 1 a.m. now and I'm getting sleepy...

I have a 12-ohm, 300-watt power resistor and a 22-ohm, 400-watt power resistor. I'd like to wire them in parallel to obtain a 7.76-ohm dummy load. (not quite 8 ohms, but close enough). I know that the power will not be distributed evenly across the two different impedances, so that the answer isn't going to be 600-watts.

What's the new power rating of the resistor circuit once they're wired in parallel?

(If somebody tells me that the answer is 665 Watts I'll be pleasantly surprised.)

[Sock Puppet]

I get:

Max V across 300 = sqr(300X12) = 60V
Max V across 400 = sqr(400X22) =93.8V

New power rating=V^2/R = 60^2/7.76 = 464W

It's early here...

S.

[bob p]

Thanks. I was solving for current in the first step, and I guess I was barking up the wrong tree.

Interestingly, your approach tells me that I might be better off creating the dummy load by using a 24-ohm, 200-watt resistor than the 22-ohm, 400-watt resistor in parallel with the 12-ohm, 300-watt resistor. Is that right? It would seem that after switching the resistors, the total power rating of the parallel load stays about the same at 450W vs. 464W (maximum voltage remains determined by the lower impedance resistor), while the new parallel Z comes out to exactly 8 ohms.

Math:

Max V across 300 = sqr(300X12) = 60V
Max V across 200 = sqr(200X24) =69.28V

New power rating=V^2/R = 60^2/8 = 450W

Its funny how the math works -- the results aren't quite what I would have expected them to be, as it doesn't really matter if the resistor with the higher Z is rated at 200W or 400W.

argh -- 02:30 and time for bed!