# 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$\Omega$.
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.