I am trying to write a general expression for the equivalent resistance of $n$ parallel resistors. Of course the well-known formula is

$$ R_{eq} = \left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \cdots + \dfrac{1}{R_n}\right)^{-1}. $$ In the case of two parallel resistors, this can be simplified to a product-over-sum expression:

$$ R_{eq} = \dfrac{R_1R_2}{R_1+R_2}. $$

When more than 2 resistors are in parallel, it's easier to just use the first equation, but for fun I decided to try to derive a product-over-sum expression for $n$ resistors. This is quite simple to derive but I'm having trouble with how I would express the sum in the denominator. In English, the denominator turns out to be "the sum of all possible combinations of $n - 1$ resistances"; for example, if we have 4 parallel resistors, their equivalent resistance is

$$ R_{eq} = \dfrac{R_1R_2R_3R_4}{R_1R_2R_3 + R_1R_2R_4 + R_1R_3R_4 + R_2R_3R_4}. $$

The numerator is easy to express using product notation, but like I said I don't know how to express the denominator in the general case of $n$ resistors. I expect it would look something like this:

$$ R_{eq} = \dfrac{\prod\limits_{i=1}^{n}R_i}{\sum\limits_{i=1}^n(\text{something...})}. $$

Also it would be interesting to generalize the denominator to not just take all possible combinations of $n-1$ quantities, but all possible combinations of any number $k$ of quantities.


1 Answer 1


Quite trivial if you have seen some conditions/inequalities below the $\sum$ or the $\prod$ symbols

The expression in the denominator you are looking for is $$\sum_{i=1}^n\prod_{\substack{j=1 \\ j\neq i}}^n R_j$$

For the generalization to the 'sum taken $k$ at a time', let us find the sum with respect to the set $S={x_1,x_2,...,x_n}$

We write $$\sum_{\substack{{T\subseteq S} ,\\ {|T|=k}}}\prod_{x\in T} x$$

which would translate to english as 'summing over all k sized subsets of S and multiplying the elements inside them' which is nothing but the sum of the products of any k elements of S

  • 2
    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – CrSb0001
    Oct 7, 2023 at 19:58
  • $\begingroup$ I’ve seen the inequality/condition below the sum/product symbols before, but I forgot about that notation and didn’t think to use it. Very helpful, thanks! $\endgroup$
    – William
    Oct 7, 2023 at 23:30

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