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Two variable resistors (RS & RE) and one static resistor (RC) are connected to a high impedance input analog input pin of a micro-controller. The static resistor is connected to ground and the two variable resistors are connected to digital outputs of the micro controller where they can be driven to +V or to ground. By driving RE or RS either high or low in any configuration and sampling the voltage from the resulting voltage divider is it possible to deduce the value of RS, and if so what formula would describe it? The value of the variable resistors is to be considered the same between successive scans.

schematic

simulate this circuit – Schematic created using CircuitLab

Not required for the question but RS is assumed to be between 1kΩ and 100kΩ and RE is assumed to between 200Ω and 20kΩ. RC can be any constant.

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  • \$\begingroup\$ I'm voting to close this question as off-topic because it is an unattempted homework problem. Write a system of equations and solve it. \$\endgroup\$ – Chris Stratton Aug 14 '18 at 3:02
  • \$\begingroup\$ Yes, you can solve this using two of these states. In the state a, the total parallel resistance of RS and RE in parallel can be calculated, as voltage Va will be determined by the current flowing through RC, which has a known resistance. Following this, in state b, you can use this relationship to provide the necessary information to solve for both values. Like the man said, homework is fine but you have to show some work. Hint. You need some combination of ohm's law and series/parallel resistance. \$\endgroup\$ – K H Aug 14 '18 at 3:14
  • \$\begingroup\$ It's been a while since I had homework. This side project to read pressure data from a resistive sensor matrix. I distilled down to its most fundamental part. Rs is a single sensor that is actively being sensed. Re is the current leakage path thru all of the sensors on inactive rows on the current column and Rc is the pulldown resistors at the controller. I thought it was solvable but I was banging my head on the systems of equations and started doubting my self as its been some time. V Rc/(Rc + Rse) = Va, V Rsc/(Rsc + Re) = Vb. I'll take another stab at it tomorrow knowing its possible. \$\endgroup\$ – Nick Ryan Aug 14 '18 at 4:21
  • \$\begingroup\$ Try Norton equivalent with Pots \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 14 '18 at 4:32
  • \$\begingroup\$ why do you have Rc ? \$\endgroup\$ – jsotola Aug 14 '18 at 13:38
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Setup system of equations

\$ \begin{cases} { V }_{ in } \dfrac{ { R }_{ c } }{ { R }_{ c } + { R }_{ s } || { R }_{ e } } = { V }_{ a } \\ \\ { V }_{ in } \dfrac{ { R }_{ s } || { R }_{ c } }{ { R }_{ s } || { R }_{ c } + { R }_{ e } } = { V }_{ b } \end{cases} \$

Solve for Rs

\$ { R }_{ s } = { R }_{ c } \dfrac{ \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ a } - { V }_{ b } } \$

Also can be solved for Re

\$ { R }_{ e } = { R }_{ c } \dfrac{ \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ b } } \$

It can also be solved for any two of the three circuits.

\$ ({ V }_{ a }, { V }_{ b }) : { R }_{ s } = \dfrac{ { R }_{ c } \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ a } - { V }_{ b } }, { R }_{ e } = \dfrac{ { R }_{ c } \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ b } } \$

\$ ({ V }_{ a }, { V }_{ c }) : { R }_{ s } = \dfrac{ { R }_{ c } \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ c } } ,{ R }_{ e } = \dfrac{ { R }_{ c } \left( { V }_{ in } - { V }_{ a } \right) }{ { V }_{ a } - { V }_{ c } } \$

\$ ({ V }_{ b }, { V }_{ c }) : { R }_{ s } = \dfrac{ - { R }_{ c } \left( { V }_{ b } + { V }_{ c } - { V }_{ in } \right) }{ { V }_{ c } } , { R }_{ e } = \dfrac{ - { R }_{ c } \left( { V }_{ b } + { V }_{ c } - { V }_{ in } \right) }{ { V }_{ b } } \$

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