0
\$\begingroup\$

I have this wheatstone bridge: ![enter image description here

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?

\$\endgroup\$
0

2 Answers 2

1
\$\begingroup\$

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:

schematic

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:

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

\$\endgroup\$
1
\$\begingroup\$

Consider this equivalent circuit:

schematic

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\$).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.