# Power Ratings—Given a power value and nothing else, how can resistor values be calculated?

I am going through "Circuit Analysis for Dummies". On pg 18, it says, "If you calculate the power dissipated as 0.1 watts, then a 0.25-watt resistor can handle this amount of power. A 0.125-watt resistor should be able to handle that amount as well, but when it comes to power ratings, err on the larger side."

The only information they had provided prior to this are these two well-known equations, $$P=i^2*R$$ and $$p = v^2/R$$. Given P = 0.1, how do I get the resistance without the voltage or current? I've gone far back into the book to ascertain that I'm not really missing other values...

## 2 Answers

You're mixing up the power rating of a resistor and its resistance. Note that it said a 0.25-watt resistor, not a 0.25-ohm resistor.

The power rating of a resistor is the amount of power that resistor can safely dissipate before it's at risk of failing. It's exclusively a property of a real resistor; an ideal resistor would be able to dissipate an arbitrarily large amount of power without failing. Meanwhile, a real resistor is made of real materials that have constraints on the kinds of conditions they can be put through. For example, maybe the contacts of the resistor are vulnerable to oxidation at a high enough temperature, or maybe the plastic film casing of the resistor melts if enough power is dissipated. The important thing to remember is that the power rating of a resistor has nothing to do with its resistance.

For example, you can get a 100-ohm resistor with a power rating of 100 watts, like this one by ARCOL, and you can get a 100-ohm resistor with a power rating of 2 watts, like this one by Yageo. It's entirely determined by the engineering of the resistor.

It's usual to work out the resistance you need from the requirements of the circuit that you are using it in, that is from considerations other than power. For example if the makers tell you that an LED can take a current of 5 mA, and will then have a pd of 1.8 V across it, what resistance would you need to put in series with it in order to run it at 5 mA off a 9 V battery? The answer is 1.44 k ohm, so you'd probably choose the easily available 1.5 k ohm.

Now you'd work out the power dissipated in the resistor, perhaps using $$P=I^2 R$$. It comes to 0.04 W, so you'd be fine with a 0.125 W resistor!