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Resistors In Parallel

by James W. Hebert

The total resistance R of several resistors, A, B, C and so on, connected in parallel can be found from:

R = 1 / [(1/A) + (1/B) + (1/C) + …]

This same relationship can perhaps be more clearly described as:

1/R = 1/A + 1/B + 1/C + …

It is also frequently mentioned that for the specific case of only two resistors the resistance can be more easily found from:

R = AB/(A+B)

I have been reading that for years in reference books, but never have derived it myself. Now is the time to figure this out. It takes a bit of alegraba to get from the first form to the second, which I will demonstrate. Back to the initial form and for just two resistors, A and B:

1/R = 1/A + 1/B

We have a fractional expression with different denominators, A, B, and R. The least common denominator is ABR; multiply both sides of the equation by it:

ABR(1/R) = ABR(1/A) + ABR(1/B)]

Next we reduce that by dividing out the common factors:

AB = RB + RA

On the right side, factor out the R:

AB = R(B + A)

R is now isolated as a variable; divide both sides by (B + A) and flip the relationship:

R = AB/(B+A)

I think I first saw this relationship when I was about ten-years-old and just learning about electricity. More than fifty years later, my curiosity about how to derive this finally overcame my inertia, and I solved it myself.

Finding What Resistor to Parallel to Make a New Value

If a value of resistance, R, is desired, but on hand we only have a resistor, B, of higher resistance, it is possible to create the desired resistance by placing a second resistor, A, in parallel. To find the value of A in terms of R and B we can manipulate the equation we derived earlier:

AB = RB + RA

We need to isolate the term A:

AB - RA = RB

Factoring:

A(B-R) = RB

A = RB/(B-R)

Copyright © 2018 by James W. Hebert. Unauthorized reproduction prohibited!
Author: James W. Hebert
This article first appeared January 21, 2018.

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