The figure below shows two elements connected in *parallel*.

- One element has resistance R
_{1} - The other element has resistance R
_{2} - The equivalent parallel resistance between A and B is:

$$\frac{1}{{\text{R}}_{\text{A}\text{B}}}=\frac{1}{{\text{R}}_{1}}+\frac{1}{{\text{R}}_{2}}$$

This is a different-looking rule than for series resistances. We say that parallel resistances are "inversely additive." To calculate the equivalent resistance for elements in parallel, you would first take the inverse of each individual resistance, add those up, and take the inverse of the sum.

Here is an example of how this works. Let's plug some numbers into the parallel elements figure.

How can we calculate the equivalent parallel resistance here? Let's dive in and use our equation directly:

$$\begin{array}{l}\frac{\text{1}}{{\text{R}}_{\text{AB}}}\text{=}\frac{\text{1}}{\text{10}}\text{+}\frac{\text{1}}{\text{20}}\\ \text{=}\frac{\text{2}}{\text{20}}\text{+}\frac{\text{1}}{\text{20}}\\ \text{=}\frac{\text{3}}{\text{20}}\end{array}$$

Since $\frac{1}{{\text{R}}_{\text{AB}}}=\frac{3}{20}$ we would have an equivalent parallel resistance in this system of ${\text{R}}_{\text{AB}}\text{=}\text{20}/\text{3\Omega}$ .

Here is another example that you can try yourself. If R_{1} = 5 Ω and R_{2} = 7 Ω in the parallel system figure above, show that the equivalent parallel resistance is 35/12 Ω.