# How to set power when given fixed supply voltage?

WARNING: noob question!

If given a supply of voltage V, how do I "set" to have a load that dissipates P amount of heat? I think i have a mental block here:

Since R is the only variable I can play with, I thought a higher resistance would give more heat since $P = I^2 \cdot R$. But, then wouldn't higher resistance also reduce the current that passes since open circuit = infinite resistance?!! Also, $P = \dfrac{V^2}{R}$ !?

PS: I am a software programmer who is tasked to reverse engineer an electrical design software that was written with excel formulas and VBA. I need to know some of this to know if the formulas are correct in the first place.

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When finished - run your conclusions past an experienced electronics designer. Or else! :-). –  Russell McMahon Oct 11 '11 at 10:24
The software is only used to pick up electrical components with the right ratings and costs at the qoutation stage. Hence, no worries. Will not exploded in my face! =) –  Jake Oct 13 '11 at 5:31
This will explode in someone's face if the software is wrong, but its component selection is trusted and enacted. –  Kaz Feb 16 at 4:15

You're almost there:

$P = V^2/R$

so

$R = V^2/P$

so for a given V and P you just choose R to give the required power.

E.g. if you have 12 V and you want to dissipate 60W then plug those numbers in and get $R = 2.4 \Omega$.

A possibly simpler way to look at this is to break it down into two steps:

(1) for a given V and P you can calculate the required current, $I = P / V$

(2) when you know the current, I, you can calculate the required load resistance as $R = V / I$

So using the above example, you'd get:

$I = P / V = 60 / 12 = 5 A$

$R = V / I = 12 / 5 = 2.4 \Omega$

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