# Wheatstone bridge, only one R value known

• Samuelriesterer
In summary, the conversation discusses a Wheatstone Bridge circuit used in sensor and measurement circuits. The circuit consists of a sensor, an adjustable resistor, and a voltmeter. The resistance of the sensor changes with temperature or light intensity. The resistance of the thermistor in ice is given by R = 100 + 0.04T, and in order to balance the bridge, Ra must be adjusted such that the voltmeter reads zero. The relation between Rs, Ra, and R1 when the meter reads zero is Ra = R1^2/Rs. When the thermistor is placed on a counter at 20°C with a voltage of 9V, the meter reading can be determined using Rs = 100 +
Samuelriesterer
Problem statement, work done, and equations:

I cannot figure how to get the other resistor values, seems like too few knowns

A Wheatstone Bridge is often used in sensor circuits and other measurement circuits. The basic circuit is drawn below:

In the circuit, Rs is the sensor and might be a thermistor for measuring temperature or a photoresistor for measuring light intensity. These change resistance when the temperature or light intensity changes.

Ra is an adjustable resistor or potentiometer (pot for short). It will have a knob on it to turn for adjusting the resistance.

Real thermistors have a resistance which changes nonlinearly with temperature. For this question, we will assume the resistance is linear over the temperature range of interest.

(a)With the thermistor placed in ice, Ra is adjusted so that the volt meter in the center of the drawing reads zero. Figure out the relation among Rs, Ra and R1 when the meter reads zero. Assume the meter has a resistance of 10^13Ω (essentially infinite).

##\frac{Ra}{R1} = \frac{R1}{Rs} → Ra = \frac{R1^2}{Rs}##

(b) With Ra adjusted as above, the thermistor is then placed on a counter where the temperature is 20°C. With V = 9V, determine the meter reading. The resistance of the thermistor is given by R = 100 + 0.04T where T is the temperature in degrees Celsius and the resistance is in Ohms.

##Rs=100+0.04(20) = 100.8 ohms##

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Samuelriesterer said:
I cannot figure how to get the other resistor values, seems like too few knowns
Samuelriesterer said:
thermistor placed in ice
Samuelriesterer said:
R = 100 + 0.04T
Samuelriesterer said:
Ra=R1 2 Rs \frac{Ra}{R1} = \frac{R1}{Rs} → Ra = \frac{R1^2}{Rs}
Looks like everything you need.

I can't get Ra because I don't have the value of R1.

Samuelriesterer said:
I can't get Ra because I don't have the value of R1.
The resistance ratios in (a) can give you R1 in ohms.
EDIT: my apologies, an algebraic error led me to the wrong conclusion

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I must be missing something because I only see that the thermistor is Rs and when in ice T = 0 Celsius so the resistance is 100 ohms. It does not give me Ra so it could arbitrarily be anything and still hold to the ratios.

What other equations can you write? Like Rs/(Rs + R1) - R1/(Ra + R1) = 0, and what can you do with them?

I know I can solve this with a system of equations with another equation if I had one but I don't. The equation you gave is essentially the same as mine.

No current through voltmeter.

I know the circuit is balanced, that is what gave me the ratios in (a). I can't find any other relevant equations.

I don't think the problem is solvable without knowing R1.

Is the sum of the parallel circuits equal to each other. That is, R1 || Rs = Ra || R1?

I think you can make a practical assumption about the choice of resistance values when the circuit is initially zeroed. Suppose you want the possible negative and positive excursions from balance to have equal range? What potential would you assign to the nodes where the voltmeter connects?

I suppose we could have a voltage through the meter if it had "infinite resistance" which would make the current essentially zero. In any case, wouldn't the potential through the meter just be equal and opposite? Or are we looking at actual arbitrary values?

Samuelriesterer said:
I suppose we could have a voltage through the meter if it had "infinite resistance" which would make the current essentially zero. In any case, wouldn't the potential through the meter just be equal and opposite? Or are we looking at actual arbitrary values?
I don't know what you mean by the phrase "potential through the meter". Potential difference is measured between two points (such as across a component) while current flows through a component or wire.

Assuming a very high impedance voltmeter, you can remove the voltmeter from the circuit without disturbing the potential difference between the nodes or the potentials at those nodes with respect to some reference.

What I was hinting at was that there is a logical choice for the potentials at those two nodes for the balanced bridge condition.

Isn't the logical choice for the potentials at the two nodes equal to each other? I don't know if anybody is seeing anything here but everyone is only telling me what I already know. I still can't seem to find a concrete value for R1 or Ra. I need another equation so I can solve the system.

Samuelriesterer said:
(a)With the thermistor placed in ice, Ra is adjusted so that the volt meter in the center of the drawing reads zero. Figure out the relation among Rs, Ra and R1 when the meter reads zero. Assume the meter has a resistance of 10^13Ω (essentially infinite).

##\frac{Ra}{R1} = \frac{R1}{Rs} → Ra = \frac{R1^2}{Rs}##

(b) With Ra adjusted as above, the thermistor is then placed on a counter where the temperature is 20°C. With V = 9V, determine the meter reading. The resistance of the thermistor is given by R = 100 + 0.04T where T is the temperature in degrees Celsius and the resistance is in Ohms.

##Rs=100+0.04(20) = 100.8 ohms##

You need to give the reading of the voltmeter (the potential difference between its terminal) when Rs = 100.8 Ω and Ra is the same as it was set when the bridge was balanced at T=0 °C. That means, Rs=100 Ω, ##Ra = \frac{R1^2}{100}##. R1 is not given. Write the potential difference in terms of R1. The voltmeter has almost infinite resistance, no current flows through it, You know that the emf of the battery is 9 V.

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Investigate the circuit when it is not balanced - the voltage between the 2 points is not zero, but no current is flowing from the one arm to the other through the voltmeter.

Samuelriesterer said:
Isn't the logical choice for the potentials at the two nodes equal to each other? I don't know if anybody is seeing anything here but everyone is only telling me what I already know. I still can't seem to find a concrete value for R1 or Ra. I need another equation so I can solve the system.
That they are equal to each other is a given: the bridge is purposely balanced at that time by adjusting the pot.

But what would be the logical choice for the potential at those nodes, say with respect to the battery's (-) terminal? How can you achieve that given the known value of the strain gauge resistance?

OK so:

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

Then solve for R1?

Samuelriesterer said:
OK so:

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

Then solve for R1?
Sure, if you already have values for both Ra and Rs.

But you seem to be avoiding putting a number to the potential at the nodes, and I can't figure out why that is. What's a logical value for it when the bridge is calibrated? One third of V? One tenth? Three quarters? Would there be any advantages to one of those choices? How about another choice?

Well the logical choice would be 1/2 V?

Samuelriesterer said:
Well the logical choice would be 1/2 V?
Yes. Can you think of a reason? (Even if you can't, it's a natural choice by esthetics and symmetry, right?).

What does the choice of node mean for R1?

gneill said:
... What's a logical value for it when the bridge is calibrated? One third of V? One tenth? Three quarters? Would there be any advantages to one of those choices? How about another choice?
Sure, one can predict roughly what Rs will be and hence choose R1 to match, giving you the ideal 1:1 ratio. But,
1- the formula for calculating the sensor's resistance is stated to be an approximation
2- Even if you knew it precisely, you would not have two R1s that exactly matched it. (You could make these adjustable and adjust them to match and be equal, but then why on Earth show them as fixed and Ra as variable when it too would be =R1?)
So assuming original balance was at 1:1 (the logical choice IMO) must be at least slightly in error, even if that were what had been attempted.

For my money, R1 must be known to make this a sensible and soluble question.

Edit - PS the reason for 1:1 is to maximise sensitivity.

Samuelriesterer said:
Isn't the logical choice for the potentials at the two nodes equal to each other? I don't know if anybody is seeing anything here but everyone is only telling me what I already know. I still can't seem to find a concrete value for R1 or Ra. I need another equation so I can solve the system.
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.

Merlin3189 said:
For my money, R1 must be known to make this a sensible and soluble question.

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 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.

Samuelriesterer said:
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 ...

Absolutely. Agree with you from #1.

And I'll go along with rudeman
rude man said:
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!

Samuelriesterer said:
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 ...

Samuelriesterer said:
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.

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.

## 1. What is a Wheatstone bridge?

A Wheatstone bridge is an electrical circuit used to measure an unknown resistance by comparing it with a known resistance.

## 2. How does a Wheatstone bridge work?

A Wheatstone bridge works by balancing the ratio of two known resistances (R1 and R2) to the unknown resistance (Rx). When the bridge is balanced, there is no current flowing through the galvanometer, indicating that the ratio of R1 to R2 is equal to the ratio of Rx to the fourth resistor (R4).

## 3. What is the purpose of using a Wheatstone bridge with only one known resistance?

The purpose of using a Wheatstone bridge with only one known resistance is to measure the value of the unknown resistance. By balancing the bridge, the unknown resistance can be calculated using the known resistance and the other two resistances in the bridge.

## 4. What are the advantages of using a Wheatstone bridge?

One advantage of using a Wheatstone bridge is its high accuracy in measuring unknown resistances. It is also a simple and inexpensive circuit to construct. Additionally, it can be used to measure a wide range of resistances.

## 5. Are there any limitations to using a Wheatstone bridge?

One limitation of using a Wheatstone bridge is that it requires a known resistance to be present in the circuit. Additionally, any small changes in temperature or the quality of the components can affect the accuracy of the measurements. It also requires a power source to operate, which may not be suitable for certain applications.

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