(Almost) Wheatstone bridge

Question

I have this wheatstone bridge:

Rmeasure is the impedance of my voltmeter between A and B. My goal is to have the equation of R1/R2. Because Umeasure changes if \$R_2\$ or \$R_1\$ changes.

the values of \$R_3\$ and \$R_4\$ are the same and known. E is known as well as Rmeasure. Only \$R_1\$ and \$R_2\$ are unknowmn. I started with a Millman on point B

$$V_B = \frac{ \frac{V_A}{R_m} + \frac{V_D}{R_1} + \frac{V_C}{R_2}}{ \frac{1}{R_m} + \frac{1}{R_1} + \frac{1}{R_2}}$$

and then I get lost in my equations when I try to isolate $$\frac{R_1}{R_2}$$

Could you help me?

Answer 1

In theory, we have \$\text{R}_\text{measure}\to\infty\$, when that is the case we have the answer I wrote. You can use the same approach when that is not the case.

Well, we have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

The input resistance can be found:

$$\text{R}_\text{in}=\frac{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\tag1$$

Now, the input current is given by:

$$\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_\text{in}}=\text{V}_\text{in}\cdot\frac{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag2$$

Now, the current \$\text{I}_1\$ is given by:

$$\text{I}_1=\frac{\text{R}_3+\text{R}_4}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_1+\text{R}_2}\tag3$$

Now, the current \$\text{I}_2\$ is given by:

$$\text{I}_2=\frac{\text{R}_1+\text{R}_2}{\text{R}_1+\text{R}_2+\text{R}_3+\text{R}_4}\cdot\text{I}_\text{in}=\frac{\text{V}_\text{in}}{\text{R}_3+\text{R}_4}\tag4$$

So, we get:

• $$\text{V}_\text{A}=\text{I}_1\cdot\text{R}_2=\frac{\text{V}_\text{in}\text{R}_2}{\text{R}_1+\text{R}_2}\tag5$$
• $$\text{V}_\text{B}=\text{I}_2\cdot\text{R}_4=\frac{\text{V}_\text{in}\text{R}_4}{\text{R}_3+\text{R}_4}\tag6$$

So, we also get:

$$\text{V}_\text{A}-\text{V}_\text{B}=\text{V}_\text{in}\cdot\left\{\frac{\text{R}_2}{\text{R}_1+\text{R}_2}-\frac{\text{R}_4}{\text{R}_3+\text{R}_4}\right\}\tag7$$

Answer 2

Consider this equivalent circuit:

^{simulate this circuit – Schematic created using CircuitLab}

From the left to the right you have this:

$$E_A = E \frac{R_4}{R_3 + R_4}$$

$$R_A = \frac{R_3 R_4}{R_3 + R_4}$$

You also know \$R_M\$ and measured the voltage (so you know the current). So you know everything until here (current and voltages): you know the total voltage at RB + EB and the current.

The problem is that you don't have a single pair \$R_B\$ and \$E_B\$ that gives the known results so far. These two values depend on the values of \$R_1\$ and \$R_2\$, and not only on their ratio:

$$R_B = \frac{R_1 R_2}{R_1 + R_2} = R_1 \left(\frac{R_2}{R_1 + R_2}\right)$$

$$E_B = E \left(\frac{R_2}{R_1 + R_2}\right)$$

\$E_B\$ and \$R_B\$ share a common term which can be determined by the ratio between the resistors. The voltages and current at these elements are known, consequently, each ratio results in a different value of \$R_1\$ (and also \$R_2\$).