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

Aug 24, 2014

I think you mean $\Omega$ and not $W$. The answer is $500 \Omega$. If you actually mean $W$, the question should be resistors consuming $1000$ $W$. BTW, resistors don't normally consume $1000$ $W$. We call resistors that consume $1000$ $W$ a heater.

Resistors in parallel use the formula:

$\frac{1}{{R}_{T}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}}$
$\frac{1}{{R}_{T}} = \frac{1}{1000} + \frac{1}{1000}$
$\frac{1}{{R}_{T}} = \frac{2}{1000} = \frac{1}{500}$
${R}_{T} = 500 \Omega$

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 the same across the resistors in parallel, so ${V}_{T} = V$. So, summing up the wattage, we get:
$2000 = \frac{{\left({V}_{T}\right)}^{2}}{{R}_{T}} = \frac{{V}^{2}}{{R}_{T}}$
$2 \frac{{V}^{2}}{R} = \frac{{V}^{2}}{{R}_{T}}$
${R}_{T} = \frac{1}{2} R$

Since we are not provided with any voltage or resistance, we simply have to leave the answer as ${R}_{T} = \frac{1}{2} R$, where $R$ is the resistance of 1 resistor. This is the same relation as calculating 2 - $1000 \Omega$ resistors in parallel.