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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