New Method for Calculating Parallel Resistors

[paulfr]

I teach HS Physics part of which is Electric Circuits,
I have discovered a rule / technique for parallel resistors that I never encountered
in all my 30+ years in electronics engineering, nor in any textbook on Circuits.
It is what I call " The N + 1 Rule "

We all know the Reciprocal Rule
1 / RT = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

AND
we know that for 2 resistors, this becomes the Product over the Sum of the 2 R's

BUT
The N+1 Rule is this
1/ Find N = the ratio of the two R's
2/ Add 1 to it to get N + 1
3/ Divide the largest R by N+1

E.g.
4 and 20 ohms
N = 20/4 = 5
N+1 = 6
RT = Rtotal = 20/6 or 10/3
Check
Product = 80
Sum= 24
RT = 80/24 = 10/3

It is quite useful when the numbers are large and thus the Product is very large.
No need to remember it and then do long division.
e.g.
300 and 50 becomes much easier and thus faster with N+1 than with Product-Sum.
300/50 = 6 ==> Rtotal = 300/7
Check with Product Sum Rule
300 (50) / 300 + 50 = 15000/350 = 300/7

It works even when N is not an integer.
e.g.
500 and 300 ohms
500/300 + 1 = 5/3 + 1 = 8/3
Rtotal = 500 / (8/3) = 1500/8
Check with Product Sum Rule
500(300) / (500 + 300) = 150000 / 800 = 1500/8

Have any of you ever seen this

Just curious and wondering why it is not in all the textbooks on Circuits.

Comments solicited

[Kalvin]

Neat trick, but I would argue that the reciprocal rule is a) easier to remember and b) more universal applicable c) carries the physical meaning of the resistors in parallel. People are using calculators (numeric or symbolic) anyway which make this a bit outdated trick. And in real life (tm) one can either guesstimate the result of the parallel combination for quick ball park computations, or if more precision is needed one will reach for the calculator.

[KaneTW]

Your trick follows from simple algebra for the 2-resistor case, and not entirely correct (it's not the largest of the two, but the dividend of N)

1/R = 1/R1 + 1/R2
R = 1/(1/R1 + 1/R2)
R = R1/(1 + R1/R2)

or R = R2/(1+R2/R1), same principle.

[magic]

Why, that's a perfectly sensible technique, easy to remember and carries the physical meaning of paralleling resistors

I mean, if R2 is N times smaller than R1, it conducts N times more and therefore the network conducts N+1 times more than R1 alone. By Ohm's law, that's N+1 times less resistance.

Can even trivially add a different R3, M times smaller than R1, then it is R1/(N+M+1).

You can also produce this formula by taking the well known 1/(1/R1+1/R2+1/R3) and multiplying by R1/R1.

[Siwastaja]

Seems handy and practical whenever you have the ratio N figured out already - this would be in a resistor divider circuit, where you design for a certain N, and need to calculate the parallel combination to figure out the output impedance.

[Kalvin]

I mean, if R2 is N times smaller than R1, it conducts N times more and therefore the network conducts N+1 times more than R1 alone. By Ohm's law, that's N+1 times less resistance.

Yes, you are totally right! 

[RoGeorge]

Indeed, the 'change of variables' method (https://en.wikipedia.org/wiki/Change_of_variables) is a nice and well known math trick.  Sometimes it's very helpful for symbolic evaluations too, not only for numerical evaluations.

Now, should this be put in a physics textbook?  No, because:
- it is offtopic, it is math, not physics
- it will only add confusion for the student
- yet another "random" thing to be memorized for the exam, with no correspondence to the physics phenomena
- for real life values, like e.g. 4.7 and 5.6, it doesn't help

In a class about the physics of the harmonic oscillator, it would be very confusing to present an algorithm for fast numerical approximation of cos(x), wouldn't it?

[vk6zgo]

I mean, if R2 is N times smaller than R1, it conducts N times more and therefore the network conducts N+1 times more than R1 alone. By Ohm's law, that's N+1 times less resistance.

Yes, you are totally right! 

I use it all the time for applications where the ratio is easy to determine by observation.

You can, of course use a similar process for any pair of resistors in parallel.

Determine the LCM of the values of the resistors, so, for instance, calling the LCM  Rm, if LCM = 3 R1, & also= 5 R2, then Rt = Rm/8.

[Brumby]

I have to agree with these (which make the exam point irrelevant):

- it is offtopic, it is math, not physics
- it will only add confusion for the student
- for real life values, like e.g. 4.7 and 5.6, it doesn't help

I'll also add that where you have more than 2 resistors in parallel, it gets very awkward, very quickly. 

The standard reciprocal of the sum of reciprocals is a very simple formula and works for any number of parallel resistors (and series capacitors).

[magic]

where you have more than 2 resistors in parallel, it gets very awkward, very quickly

It scales linearly with the number of resistors

[T3sl4co1l]

I do that implicitly when doing quick estimates mentally or on paper.

Reduce the problem to its simplest form.  If we have values of 4700 and 10000, obviously don't do long division by 14700.  That'll take forever!

First notice that 4700 is about half 10000.  The approximate problem reduces to 1 || 2.  We note the denominator is 3 and the numerator is 2, and we're done.  We can go slightly further and write 3 || 6 = 2, all whole numbers.  Multiply back up to the original values (i.e., by 10000/6) and you have your answer, 3.33k.

4700 is not actually half, but a few percent low (about 6%), so our result will be not quite as many percent low as well (about 4%?).  This gets into the range of typical resistor tolerance, so we might leave it as good enough, or further approximate it, or fully calculate it, or choose 4.99k resistors because hell, my time is literally worth a great many times the resistors themselves!

I feel, in general, ratios are not taught nearly as well as they deserve to be.  Grade school application as I recall seemed mostly to waste time evaluating factored expressions that could more easily be solved throwing everything together.  I don't think you can motivate it in the fully abstract.  The greatest power comes with units, dimensional analysis.  You can avoid many algebra errors this way, by associating a unit to each variable in your expression and follow the unit parity through your work -- as soon as two different units are being added, you know exactly which terms encountered an error!


Ratios also help with some algebraic and higher analyses, where a messy equation (typical electronic example, an RLC filter with piles of physical variables per term) can be greatly simplified, with judicious choice of ratios stuffed away into auxiliary expressions (e.g., ω_0, ζ).

The highest level of this may be Feynman integration, where the limits of integration are pulled through the ratios, giving a dimensionless inner expression, which may be no easier to integrate, but only needs to be solved once and we can let a computer think about that instead.

This is probably harder to justify at an early school level, because only technical fields will use much of it (these exact topics are usually covered in physics courses), but I think a good argument can still be made for a lighter-weight version for general use.

Tim

[Brumby]

where you have more than 2 resistors in parallel, it gets very awkward, very quickly

It scales linearly with the number of resistors

How good is your mental arithmetic?

That's the only reason for trying this sort of "trick", IMHO.

Reduce the problem to its simplest form.  If we have values of 4700 and 10000, obviously don't do long division by 14700.  That'll take forever!

First notice that 4700 is about half 10000.  The approximate problem reduces to 1 || 2.  We note the denominator is 3 and the numerator is 2, and we're done.  We can go slightly further and write 3 || 6 = 2, all whole numbers.  Multiply back up to the original values (i.e., by 10000/6) and you have your answer, 3.33k.

This is the sort of thing I would do for rough calculation, understanding there is an error margin by doing so.  If I wanted to be more precise, then I'd break out a calculator