- #1
archaic
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I did this for fun, if you want to do it in an unconventional way, this would be useful in programming.
For ##n## resistors in parallel:
$$R_{total}=\frac{\prod_{i=0}^{n-1} R_{i}}{\sum_{i=0}^{n-1}(\prod_{j=0}^{n-2} R_{i+j \mod n})}$$
For ##n## resistors in parallel:
$$R_{total}=\frac{\prod_{i=0}^{n-1} R_{i}}{\sum_{i=0}^{n-1}(\prod_{j=0}^{n-2} R_{i+j \mod n})}$$
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