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Re: Solve Simple Relationship


From: Thomas D. Dean
Subject: Re: Solve Simple Relationship
Date: Fri, 27 Aug 2021 09:33:38 -0700
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Thunderbird/78.11.0

On 8/27/21 9:04 AM, Doug Stewart wrote:


On Fri, Aug 27, 2021 at 11:13 AM Thomas D. Dean <tomdean@wavecable.com <mailto:tomdean@wavecable.com>> wrote:

    On 8/27/21 8:09 AM, Thomas D. Dean wrote:
     > I have a voltage divider I want to calibrate. I tried this by
    exchanging
     > the resistors and measuring the resultant rfesistance.
     >
     > 1.6516 = 3.3 * R1 / (R1 + R2)
     > 1.6484 = 3.3 * R2 / (R1 + R2)
     >
     > R1 is around 1005 ohms
     > R2 is around 1003 ohms
     >
     > I think this is simple.  But evidently, it is beyond me.
     >
     > Tom Dean

    And, worse, I can not clearly state the problem:  rfesistance = voltage.



I think that you are making it too hard.

Choose r1 say 1000
  now calculate r2 from your first eq.
I get 998.06


I want to calibrate the voltage divider. Two equations, two unknowns. Should have a solution.

The idea is to have a known value for R1 and R2 so I can measure Vs,
(= 3.3?) in the future.

Maybe, iterate R1, calculate R2 with the first and see if it satisfies the 2nd equation.

I wrote a function
function [y] = eq_r(x)
 y(1) = 1.6516 - 3.3 * x(1) / (x(1) + x(2))
 y(2) = 1.6484 - 3.3 * x(2) / (x(1) + x(2))
endfunction

fsolve returns values that do not satisfy the equations. Very dependent on X0 supplied.

Tom Dean



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