Electronics, Wheatstone bridge, with a twist.

[brokenbarbie]

Homework Statement

The bridge circuit below is used to monitor temperature using a thermometer. The thermometer RT has resistance of 100 Ω at 0°C and a linear coefficient with temperature of 0.005 Ω/°C. What is the sensitivity of the bridge (ΔVOUT/ΔT) around 100 °C for ΔT = 1°C? How does it compare with sensitivity around – 100 °C?

Homework Equations

(ΔVOUT/ΔT) = Sensitivity

R_T=(1+aΔT)

The Attempt at a Solution

I attempted to calculate output voltage,

First I calculated R_T=100* (1 + 0.005 * 100) = 150Ω which is the resistor of the transducer

Then I calculated v₁ and v₂

v₁ = 10*(200/200+200) = 5V
v₂ = 10*(150/200+150) = 4.29V

hence Δv=0.71V

and sensitivity = 0.71/1 = 0.71I'm not sure what I've done makes any sense, I followed an example for a wheatsone bridge, any help will be appreciated greatly, thanks =D.

 

[gneill]

Hi brokenbarbie, Welcome to Physics Forums.

You have the right idea about determining Vout by first finding your v1 and v2. But things go off the rails from there

The first step should be to find an expression for Vout rather than a particular value at a particular temperature. Just leave R_T as a symbol and find the expression for Vout. You can plug in an expression for R_T later.

Then you want to head towards finding ##\frac{d}{dT}V_{out}##, either by way of calculus, or I suppose you could "brute force" it by plugging in temperatures +/- the ΔT and determining the corresponding ΔVout.

 

[Evaaeroqueen]

Hello gneill, would you kindly show me how to do the remainder of the question. If it's not too much trouble could you post the full solution?

Thanks in advance

 

[gneill]

Hi Evaaeroqueen, Welcome to Physics Forums.

Hello gneill, would you kindly show me how to do the remainder of the question. If it's not too much trouble could you post the full solution?

Thanks in advance

No, that would be against forum rules; We can't jut do your homework for you. We can provide hints, corrections, and so forth, but you must show some effort and do the bulk of the work.

Cheers.

 

[Nickie]

Your approach is correct, but there are a few things that can be improved in your solution.

Firstly, when calculating the resistor value for the thermometer, you should use the given linear coefficient of 0.005 Ω/°C. This means that at 100°C, the resistance of the thermometer will be 100*(1+0.005*100) = 150Ω.

Secondly, when calculating v1 and v2, you should use the resistance value of the thermometer at the given temperature (100°C). This means that v1 = 10*(150/150+200) = 5V and v2 = 10*(150/150+150) = 5V. This results in Δv = 5V-5V = 0V.

Finally, the sensitivity should be calculated as the change in output voltage (Δv) divided by the change in temperature (ΔT). In this case, the change in temperature is given as 1°C, so the sensitivity would be 0V/1°C = 0V/°C.

To compare the sensitivity at -100°C, you can follow the same steps as above, but use the resistance value of the thermometer at -100°C (100*(1+0.005*(-100)) = 50Ω). This will result in a sensitivity of 0V/1°C = 0V/°C, which is the same as at 100°C.

Overall, your approach was correct, but there were a few minor errors in your calculations. Keep in mind that the sensitivity of a Wheatstone bridge is dependent on the ratio of resistances, so it will remain the same regardless of the actual resistance values (as long as they are proportional).