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Wheatstone bridge

When determining the millivolt reading I'm confused on the order in which you subtract V(2-4) and V(1-3). Would it be: V(1-3) - V(2-4) or V(2-4) - V(1-3)

I've seen it done both ways, so how would you know?

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As drawn you don't know the proper order of subtraction. You would just determine the voltage "difference" across the mV symbol. You could then enter that value at the mV symbol and place the + and - polarity marks as appropriate.

If the mV symbol had been initially drawn with polarity you would then know to subtract the labeled - side from the labeled + side.

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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:

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

Using your given values we get:

$$\text{V}_\text{A}-\text{V}_\text{B}=10\cdot\left\{\frac{120}{480+120}-\frac{\text{R}_4}{480+\text{R}_4}\right\}=\frac{4800}{480+\text{R}_4}-8\tag8$$

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Polarity is usually calculated as Larger polarity - Smaller polarity. It's just to avoid negatives. You put polarities on after calculation.

This is represented by V(1-3) - V(2-4) or V(2-4) - V(1-3).

If V(1-3) > V(2-4), you use V(1-3) - V(2-4) with positive polarity on V(1-3) and negative polarity on V(2-4).

If V(1-3) < V(2-4), you use V(2-4) - V(1-3) with positive V(2-4) and negative V(1-3).

If V(1-3) = V(2-4), voltage is 0.

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The order you do the subtraction will just change the polarity of the result, so you can select the order to give the result you require.

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