How to set power when given fixed supply voltage?

Question

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}\$ !?

Thanks in advance!

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.

Answer 1

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\$

Answer 2

Yes, given a constant-voltage supply, less resistance means more current through the load resistor and more power dissipated in the resistor. You might think of an idealized voltage source as being capable of supplying infinite power (fixed voltage x infinite current), and load resistance is the extent to which the load "resists" drawing infinite current.

In reality the voltage source can only supply current up to some limit. A bench supply will often have both voltage and current limit, where it will continue to output the set voltage but limit current to the set limit. In other types of supply such as an on-board switching regulator you might see the output voltage fall to something less than the expected output voltage if the current goes above its capability. In some designs a fuse might blow if you draw too much current from the device.