# resistors power

I am trying to make a 10 $\mathrm{\Omega}$ 2 W resistor.

I was wondering if I can achieve it connecting either four parallel pairs together serially using 2.5 $\mathrm{\Omega}$ resistors or connecting 2 quad sets in parallel serially.

I'm using 1/4 W resistors.

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Can you please make a schematic of what connections you are meaning. – Kortuk May 29 '12 at 15:23

At that power and voltage you've got plenty of possibilities: 8 $\times$ 1.25$\Omega$ in series, or 8 $\times$ 80$\Omega$ parallel, or 2 parallel $\times$ 4 in series $\times$ 5$\Omega$ in parallel, series, or 4 parallel $\times$ 2 in series $\times$ 20$\Omega$.

It doesn't really matter what you do. Since in all solutions the current through the different branches is the same, the power will be equally distributed over all resistors.

Why are the resistor values different for every solution, and how do you find these values? I first look how many branches in parallel I have. If that's 2 then each branch should be 20$\Omega$, because placing two equal resistors in parallel halves their value. Then I see how many resistors each branch has. If that's 4 then each resistor should be 20$\Omega$ / 4 = 5$\Omega$.
The 2.5$\Omega$ you mention fits no solution.

I mentioned voltage. In some applications the series configuration is better than the parallel because a high voltage may thus be divided over the resistor values. Say you have 230V, which is too much for an 0603 resistor. Then you can't use eight 1M$\Omega$ in parallel, you have to place 15.6k$\Omega$s in series so that each resistor only gets 230V/8 = 29V.

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 Small addition: it's better to leave some headroom to the power rating, so maybe 2.x W would be more appropriate. – clabacchio♦ Jun 4 '12 at 8:43