Calculating R with 2 resistors parallel

Instead of thinking of these things as "resistors", try thinking of them as "conductors". After all, that's what they do: conduct.

A resistor with resistance \$R\$ is a conductor with conductance \$S=\dfrac{1}{R}\$.

When you provide multiple conductors connecting one point to another, the conductances simply add. What could be more intuitive? When you provide an additional path for current to flow, more total current flows.

Conductors \$S_1\$ and \$S_2\$ in parallel have a total conductance of:

\$S = S_1 + S_2\$

If you want to express \$S_1\$ and \$S_2\$ as resistances \$R_1\$ and \$R_2\$, you get:

\$S = \dfrac{1}{R_1} + \dfrac{1}{R_2}\$

And, if you want to express the total conductance S as a resistance R:

\$\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}\$

\$R = \dfrac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}}\$

Which is the usual expression for the total resistance of two resistors in parallel.

Trivia: the unit of conductance (i.e. inverse ohms) is sometimes called the "mho" ('Ohm' backwards), and is written with an upside-down Omega symbol: ℧. The official SI name for this unit is siemens ("S").

Of course there is, but it does not look pretty. Make the divisors equal, and add the terms. for three resistors you get

\$ \dfrac{R2 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R3}{R1 \cdot R2 \cdot R3} + \dfrac{R1 \cdot R2}{R1 \cdot R2 \cdot R3}\$

\$ = \dfrac{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)}{R1 \cdot R2 \cdot R3} \$

now do the \$ \dfrac{1}{n} \$ and you get:

\$ \dfrac{R1 \cdot R2 \cdot R3}{(R2 \cdot R3) + (R1 \cdot R3) + (R1 \cdot R2)} \$

The top line is easy, it is the product (multiplication) of all resistors. The bottom line is the sum of the products of all leave-one-out combinations. For two that reduces to the pretty formula:

\$ \dfrac{(R1 \cdot R2)}{(R1+R2)}\$