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## Re: Solve Simple Relationship

 From: Doug Stewart Subject: Re: Solve Simple Relationship Date: Fri, 27 Aug 2021 12:50:01 -0400

On Fri, Aug 27, 2021 at 12:34 PM Thomas D. Dean <tomdean@wavecable.com> wrote:
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

I believe there is an infinite set of solutions.
if you set r1=100 you can calculate r2
and if you set r1=1000 you can calculate r2
etc.

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DAS 