On Mon, Feb 26, 2001 at 12:40:15PM +0000, Mike Harrison wrote:
> Any maths wizards out there....?
> Given a conventional resistance bridge :=20
>=20
> Vin
> +----+
> R1  R2
> +----|--- Vout (differential)
> |    +---
> R3  Rx
> +----+
> GND
>=20
> What is the formula to determine Rx (Or, better,  the ratio Rx/R3)
> given the ratio Vout/Vin. R1=3DR2 if that makes anything easier.
>=20
> My algebra isn't quite up to juggling it round - I got as far as the
> following before my brain started to hurt :=20
>=20
> Rx/(Rx+R2) =3D V-(R3/(R1+R3)
> where V is the ratio Vout/Vin
>=20
This is what I've got using the FREE symbolic algebra package "maxima"
(available at http://www.ma.utexas.edu/users/wfs/maxima.html )
The "C3" line contains the answer...

wzab@myhost:~$ maxima
GCL (GNU Common Lisp)  Version(2.3) pi=B1 wrz 22 14:49:21 CEST 2000
Licensed under GNU Library General Public License
Contains Enhancements by W. Schelter
Maxima 5.4 pi=B1 wrz 22 14:48:33 CEST 2000 (with enhancements by W. Schelte=
r).
Licensed under the GNU Public License (see file COPYING)
(C1)  V1 : Vin*R3/(R1+R3);

                                    R3 Vin
(D1)                                -------
                                    R3 + R1
(C2) V2 : Vin*Rx/(R2+Rx);

                                    Rx Vin
(D2)                                -------
                                    Rx + R2
(C3) solve(Vout=3DV1-V2,Rx);

                           (R2 R3 + R1 R2) Vout - R2 R3 Vin
(D3)               [Rx =3D - --------------------------------]
                               (R3 + R1) Vout + R1 Vin
(C4)=20
--=20
                                HTH
                              Wojciech M. Zabolotny
        http://www.ise.pw.edu.pl/~wzab  <--> wzab@ise.pw.edu.pl

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