# Electronics, Wheatstone bridge, with a twist.

1. Nov 7, 2013

### brokenbarbie

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

(ΔVOUT/ΔT) = Sensitivity

RT=(1+aΔT)

3. The attempt at a solution

I attempted to calculate output voltage,

First I calculated RT=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 = 0.71

I'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.

Last edited: Nov 7, 2013
2. Nov 7, 2013

### Staff: Mentor

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 RT as a symbol and find the expression for Vout. You can plug in an expression for RT 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.

3. Nov 12, 2013

### 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?

4. Nov 12, 2013

### Staff: Mentor

Hi Evaaeroqueen, Welcome to Physics Forums.

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.