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.