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.

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)

After some more critial thinking on this problem, it occurred to me that I ought to derive the basis for the forumulas given above, that is, exactly how do we know what will be the total resistance when two (or more) resistors are connected in parallel. My analysis:

If a voltage, E, is applied across two resistors, A and B, connected in parallel and forming an equivalent resistance R, then the current that will flow in each resistor will be

I_{a} = E/A

I_{b} = E/B

The total current flowing in the combined resistance is then:

I_{total} = I_{a} + I_{b}

The voltage across the two actual resistors is the same, and this is the same volage across the equivalent resistance of the two in parallel. Now we express the relationship of the three currents in terms of voltage and resistance by using the Ohm's Law relationship I = E/R:

E/R = (E/A) + (E/B)

Now we multiply both sides of the equation by R/E to eliminate the common factors:

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

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

And that easily factors to:

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

Copyright © 2018 by James W. Hebert. Unauthorized reproduction prohibited!

Author: James W. Hebert

This article first appeared January 21, 2018.

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Author: James W. Hebert