# Wheatstone bridge, only one R value known

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1. Mar 8, 2015

### 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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2. Mar 8, 2015

### Bystander

Looks like everything you need.

3. Mar 8, 2015

### Samuelriesterer

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

4. Mar 8, 2015

### Staff: Mentor

The resistance ratios in (a) can give you R1 in ohms.
EDIT: my apologies, an algebraic error led me to the wrong conclusion

Last edited: Mar 10, 2015
5. Mar 8, 2015

### Samuelriesterer

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.

6. Mar 9, 2015

### Bystander

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

7. Mar 9, 2015

### Samuelriesterer

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.

8. Mar 9, 2015

### andrevdh

No current through voltmeter.

9. Mar 9, 2015

### Samuelriesterer

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

10. Mar 9, 2015

### rude man

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

11. Mar 9, 2015

### Samuelriesterer

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

12. Mar 9, 2015

### Staff: Mentor

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?

13. Mar 9, 2015

### Samuelriesterer

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?

14. Mar 9, 2015

### Staff: Mentor

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.

15. Mar 9, 2015

### Samuelriesterer

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.

16. Mar 10, 2015

### ehild

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.

Last edited: Mar 10, 2015
17. Mar 10, 2015

### andrevdh

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.

18. Mar 10, 2015

### Staff: Mentor

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?

19. Mar 10, 2015

### Samuelriesterer

OK so:

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

Then solve for R1?

20. Mar 10, 2015

### Staff: Mentor

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?