Re: Solve Simple Relationship

On Fri, Aug 27, 2021 at 12:34 PM Thomas D. Dean <[email protected]>
wrote:

> On 8/27/21 9:04 AM, Doug Stewart wrote:
> >
> >
> > On Fri, Aug 27, 2021 at 11:13 AM Thomas D. Dean <[email protected]
> > <mailto:[email protected]>> 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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