How to rewrite $\frac{1}{\sum_{j=1}^{N}\frac{1}{R_j}}$?

Question

This is a very simple question but I don't seem to find a solution online. I would like to rewrite this sum which appears very often in series and parallel electrical circuits.

$$R_{\text{Eq}} = \frac{1}{\sum_{j=1}^{N}\frac{1}{R_j}}$$

In the case of two resistances, we have:

$$R_{\text{Eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{1}{\frac{R_1 + R_2}{R_1R_2}} = \frac{R_1R_2}{R_1+R_2}$$

In the case of three, we have:

$$R_{\text{Eq}} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}} = \frac{1}{\frac{R_1R_2 + R_1R_3 + R_2R_3}{R_1R_2R_3}} = \frac{R_1R_2R_3}{R_1R_2+R_1R_3+R_1R_2}$$

And so on...

With this in mind, I would like to write something like:

$$R_{\text{Eq}} = \prod_{j=1}^{N} R_j \sum_{j=1}^{N} \frac{1}{\cdots}$$

But I don't have enough knowledge in math to write those combinatory terms in the denominator.

Thanks for reading!

Answer 1

The equivalent resistance of $n$ resistors in parallel, with resistor $i$ having resistance $R_i>0$, is given by

$${1 \over \sum_{i=1}^n {1 \over R_i}}={\prod_{j=1}^n R_j \over \sum_{i=1}^n {1 \over R_i} \prod_{j=1}^n R_j}={\prod_{j=1}^n R_j \over \sum_{i=1}^n \prod_{j\neq i}R_j},$$

or if you wish to expand it out for $n\geq 3$,

$${R_1\cdot...\cdot R_n \over (R_2\cdot...\cdot R_n)+(R_1\cdot R_3\cdot...\cdot R_n)+...+(R_1\cdot...\cdot R_{n-1})}.$$

Answer 2

If you're really interested in the math of it, then go for it, but I prefer to rewrite this as

$$\frac{1}{R_{\text{Eq}}} = \sum_{j=1}^{N} \frac{1}{R_j}$$

and just remember that "conductances add in parallel", where conductance is the reciprocal of resistance.