Wheatstone bridge, only one R value known

[gneill]

I think that we can agree that Rs is approximately 100Ω. It may vary by some amount, if for no other reason than manufacturing tolerance for a given run of parts.

So choosing R1 to be 100Ω is a reasonable start, and places the bridge's node potentials near half the available potential. Variable resistor Ra balances the bridge to take care of these tolerances. The balanced bridge should still be fairly close to V/2, leaving the most leeway for excursions in either direction when the strain gauge resistance varies in operation.

 

[rude man]

OK so:

$V_s (\frac{R_s}{R_s +R_1} - \frac{R_1}{R_a +R_1})=0$

Then solve for R1?

No. The bridge voltage is zero only when Rs is at zero C temperature ...

 

[Merlin3189]

Absolutely. Agree with you from #1.

And I'll go along with rudeman

Tell you what, since they didn't give you R1, pick any value you want and calculate the bridge voltage! They have to give you an A.

Work it out for R1=100.
Then work it out for another value of R1. You will get a different answer, proving that the question is lacking the value for R1.

Assuming that the intention was for R1 to be around about 100, you will get what the answer was probably intended to be, had the question been set properly!

 

[rude man]

That is what I have been thinking all along. But it sounds like some of these other people see something that I don't. 1/2 V seems like a logical by symmetry but can't R1 be arbitrarily anything here?

I think they're seeing ghosts ...

 

[NascentOxygen]

but can't R1 be arbitrarily anything here?

In the design, R1 can be used to limit the power to the sensor. The sensor is a resistor, after all, so is subject to self-heating and heat can influence its reading.

 

[Samuelriesterer]

Just an update to those who were wondering. I asked my teacher and he said to just pick a value for R1 because yes, there are infinite possibilities.