# What is the equivalent resistance of three resistances of 12 Ω each connected in parallel?

Apr 24, 2015

For the total resistance when the resistors are in parallel to each other, we use:

$\frac{1}{{R}_{T}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \ldots + \frac{1}{{R}_{n}}$

The situation you describe seems to be this:

So there are 3 resistors, meaning we will use:

$\frac{1}{{R}_{T}} = \frac{1}{{R}_{1}} + \frac{1}{{R}_{2}} + \frac{1}{{R}_{3}}$

All resistors have a resistance of $12 \Omega$:

$\frac{1}{{R}_{T}} = \frac{1}{12} + \frac{1}{12} + \frac{1}{12}$

Total up the Right Hand Side:

$\frac{1}{{R}_{T}} = \frac{3}{12}$

At this point you cross multiply :

$3 {R}_{T} = 12$

Then simply solve it:

${R}_{T} = \frac{12}{3}$

${R}_{T} = 4 \Omega$