How to interpret this circuit configuration? (Sensor bridge)

[Benighted]

So, I'm asked to demonstrate the formulas for finding out the voltage labeled here as ΔU, which I'm given in my project document:

I think I would be able to make the demonstrations without any outside help, if only I knew how I'm supposed to interpret this circuit branch in the middle. Is it connected in parallel to the branches with the resistors? Not only the two points seem to be disconnected, but even if they were connected, since that branch has no resistive elements on it, I would be inclined to believe no current flows through it, which obviously contradicts the assignment. What kind of circuit part is an interrupted conductive wire connected in parallel to two branches with resistors on them?

I've attempted to collapse the resistors into series, then parallel equivalent resistances, but couldn't find a way to include the required branch into the new configurations. Also tried to apply Kirchhoff's laws, but there are 6 currents involved and couldn't find any way to simplify them out of the resulting equations. The whole thing just has me very confused.

The formula to be proven for this circuit is: ΔU=U/4 *ΔR/R * 1/(1+0.5*(ΔR/R)).

 

[gneill]

Hi Benighted. Can you recognize two potential dividers in the circuit? Can you write expressions for the potentials at the two open terminals with respect to the top center node?

 

[Benighted]

Hmm. Potential divider. This means a node in the circuit? I think I see four: top, bottom, left, right. Let's call these points A, B, C, D. So, between points A and C, the voltage on the top left resistor is:
V₁=V_A-V_C
and between A and D,
V₂=V_A-V_D

 

[gneill]

No, a potential divider (perhaps more commonly called a voltage divider) is a particular arrangement of resistances and a voltage source. Do a web search on it.

 

[CWatters]

Not only the two points seem to be disconnected

Correct, they are not connected together.

but even if they were connected, since that branch has no resistive elements on it, I would be inclined to believe no current flows through it

The two points are not connected together so you are correct in assuming that no current flows between these two points.

which obviously contradicts the assignment.

This does not contradict the assignment. The assignment asks you to find the voltage between the two points not the current flowing between them.

 

[Benighted]

@gneill: I did, now I remember having once attended some sort of lesson on it.
Here's an updated labeled image of the same circuit:

In that case, the voltage U is divided at point A between R₁ and R₂ (R+ΔR), and at point B between R₃ and R₄. Is that correct?

In the Wikipedia page, there's an example of a voltage divider with the resistors in a series configuration, and the formula for that is given as
$V_out = \frac{R_2}{(R_1+R_2} V_in$
Is that the case for a parallel configuration as well?

@CWatters: Oh. I thought that, because U=I*R, if I=0, then U has to be 0 as well.

 

[CWatters]

The Wikipedia page and that equation is definitely applicable to this circuit.

Can I suggest you redraw the circuit with the resistors shown vertical rather than horizontal.

@CWatters: Oh. I thought that, because U=I*R, if I=0, then U has to be 0 as well.

Ohms law is used to calculate a voltage drop through a resistor. However there are plenty of other situations where you might have a voltage difference that you can't use Ohms law for. For example the terminals of a 6V battery have a voltage difference of 6V even when there is nothing connected to the battery.

 

[gneill]

In that case, the voltage U is divided at point A between R1 and R2 (R+ΔR), and at point B between R3 and R4. Is that correct?

Not quite. The potential U is across points A and B. C is in between, separated by resistors R1 and R3. Similarly, D is between A and B separated by R2 and R4.

As CWatters suggests, re-drawing the circuit might make it more obvious.