# What is the series equivalent of two 1000 W resistors in series?

Aug 24, 2014

I think you mean $\Omega$ and not $W$. The answer is $2000 \Omega$.

Resistors in series are simply summed (added) up.

If you actually mean $W$, then we need a different calculation:

$P = V I = V \frac{V}{R} = \frac{{V}^{2}}{R}$
$1000 = \frac{{V}^{2}}{R}$
This is for 1 resistor.

Recall that the voltage is split between the resistors, so total voltage is double, ${V}_{T} = 2 V$. So, summing up the wattage, we get:
$2000 = \frac{{\left({V}_{T}\right)}^{2}}{{R}_{T}} = \frac{{\left(2 V\right)}^{2}}{{R}_{T}}$
$2 \frac{{V}^{2}}{R} = \frac{4 {V}^{2}}{{R}_{T}}$
${R}_{T} = 2 R$

Since we are not provided with any voltage or resistance, we simply have to leave the answer as ${R}_{T} = 2 R$, where $R$ is the resistance of 1 resistor.