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!
Best Answer
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.