How do I learn RPL and solve this problem with it?

[Dieter]

(09-26-2017 01:02 PM)David Hayden Wrote:  I'd solve it like so:
(...)

Great, finally a possible method that is better than "brute force". ;-)

I tried a similar, but somewhat different approach. Here is how it works:

Let T be the target resistance.
Now do the following for all resistors > T and ≤ 2T:

1. Pick the first/next value for R1 from the E12 series
2. Calculate the value of R2, which is the resistor parallel to R1 that yields a total resistance of T
3. Take the two closest resistors R2a and R2b and determine the total resistance R1||R2a and R1||R2b
   Or simply consider only the resistor R2c that is closest to R2.
4. If the tolerance is not yet met, repeat with next R1
5. Else display R1 and R2a or R2b and quit
   

Instead of quitting as soon as a combination within the allowed tolerance is found you can also run through all possible R1 (i.e. all resistors > T and ≤ 2T) and return the combination with the overall lowest deviation. But due to the individual tolerances of the two resistors this may not make much sense. On the other hand this returns the best possible combination even if it is not within tolerance.

Here is an example:

Try to find a combination for a target resistance of T = 350 Ω

The first larger R1 in the E12 series is 390 Ω
R2 = 350*390 / (390–350) = 3412,5 Ω
So R2a = 3300 Ω and R2b = 3900 Ω
390*3300 / (390+3300) = 348,78 Ω which is T – 0,35%
390*3900 / (390+3900) = 354,55 Ω which is T + 1,30%
The first value (3300 Ω) is within tolerance, so we already got a solution.
Otherwise repeat these steps with R1 = 470, 560 and 680 Ω.
Higher R1 values are not required, as 820 Ω already is > 2*350 Ω.

This way a combination of two parallel resistors can be found. Here the best result was 348,78 Ω. In a final step, the gap to 350 Ω can be closed with another 1,2 Ω resistor in series. But since we are already within the tolerance of the individual resistors this is not really required.

What do you think?

Dieter