I have a strain gauge mounted in a voltage divider as in the image below:
simulate this circuit – Schematic created using CircuitLab
where \$R_{g}\$ is the resistance of the strain gauge when it is not strained (120 \$\Omega\$), \$R\$ is a 1200 \$\Omega\$ resistor, \$V_{app}\$ is the voltage applied on the voltage divider (1.1 V) and \$V_{meas}\$ is the voltage measured with a lock-in amplifier.
When the strain gauge is not strained, \$V_{meas}= V_{1} = \frac{R_{g}}{R+R_{g}}*V_{app} = 0.1 V\$
How to calculate \$V_{meas}\$ when the strain gauge is strained?
This is a shortened version of my question. The original formulation is hidden behind the spoiler.
The strain across the strain gauge is \$x = \frac{1}{GF} * \frac{\Delta R_{g}}{Rg}\$ (eq. 2) where GF is the gauge factor. I am not certain how to express \$x\$ as a function of \$V_{meas}\$. So far, I used the following method: when the strain gauge is strained, eq. 1 becomes: \$V_{meas} = V_{2} = \frac{R_{g}+\Delta R_{g}}{R+R_{g}+\Delta R_{g}}*V_{app}\$. From there I get \$\frac{\Delta R_{g}}{R} = - \frac{R}{\Delta R_{g}} * \frac{V_{2}}{V_{2}-V_{app}} - 1\$ (eq. 3). By inserting eq. 3 in eq. 2, I have \$x = - \frac{1}{GF} * (\frac{R}{\Delta R_{g}} * \frac{V_{2}}{V_{2}-V_{app}} + 1)\$ (eq. 4). Using eq. 1, I get \$\frac{R}{R_{g}} = \frac{V_{app}-V_{1}}{V_{1}}\$ and inserting this equation in eq. 4: \$x = - \frac{1}{GF} * (\frac{V_{app}-V_{1}}{V_{1}} * \frac{V_{2}}{V_{2}-V_{app}} + 1)\$. Knowing the values of \$V_{app}\$ and \$V_{1}\$, I obtain: \$x = - \frac{1}{GF} * (10*\frac{V_{2}}{V_{2}-1.1} + 1)\$ where \$V_{2}\$ is the voltage measured with the lock-in amplifier when the strain gauge is strained. Could someone confirm these equations? Thank you!
