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Wheatstone bridge problem unable to understand

  1. Nov 5, 2014 #1
    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 found out Vout expression but don't know how to differentiate it. Can someone help please?

    Vout = (10*Rt/Rt+100) - 5

     

    Attached Files:

  2. jcsd
  3. Nov 5, 2014 #2

    gneill

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    Staff: Mentor

    Before getting to the differentiation, note that the temperature coefficient has units Ω/°C. So you should re-think your expression for Rt in your Relevant equations. Make sure that the units are consistent throughout.

    Your expression for Vout is good as far as it goes (although it could use another pair of parentheses to make the denominator unambiguous). You should be able to substitute in your expression for RT to give you an expression for Vout(T). Then you can find dVout/dT or use a "brute force" method by calculating Vout when small changes in temperature occur around a given temperature.
     
  4. Nov 5, 2014 #3
    Sorry yeah Rt should be RT=R*(1+aΔT). So after I sub in for RT it will be give me equation in R and T. I am not sure how to differentitate R and T or do I have to make a differential equation and integrate instead? Please reply
     
  5. Nov 5, 2014 #4

    gneill

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    Check the units. The "1" in the parentheses is unitless. What are the units of aΔT? Can you add a unitless number to a number with units?
    You want to find how Vout changes with a change in T. That is, dVout/dT.
     
  6. Nov 5, 2014 #5
    I am using RT=R*(1+aΔT) as its the equation for accuracy of linear fit. if I remove 1 from the equation the Rt value might be wrong, don't you think...
     
  7. Nov 5, 2014 #6

    gneill

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    You could leave the equation in that form, but then you'd have to determine the required value for the constant a such that units would work out. You cannot sum values with different units.

    Note that there are are other algebraic forms for a straight line.
     
  8. Nov 5, 2014 #7
    I took Rt as R*aΔT and then replaced the equation so it became

    Vout = 10*R*aΔT/(R*aΔT+100) -5

    then I differentiated with respect to T to get

    dv/dT = 10*R*a/(R*a+100) -5

    I am not sure if that was correct...I know ΔT can be written as dT itself but can't figure how to use it. Any suggestions please..
     
  9. Nov 5, 2014 #8

    gneill

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    The ΔT in the Rt equation is the change in temperature from some reference temperature. In this case you're given the reference resistance ##R_o = 100 \Omega## at reference temperature ##T_o = 0C##. So in fact

    ##\Delta T = (T - T_o)##

    and since ##T_o = 0C##, ##\Delta T = T## for the Rt expression.

    An appropriate form for Rt would be the slope intercept form: y = mx + b.
     
  10. Nov 5, 2014 #9
    How about I use ΔR/R0 = aΔT and then find the ΔR. After that I can add the ΔR to R0 to get Rt. That should be correct right?
    and for y = mx+b can't we just do Rt = aT+1 assuming a as the gradient and 1 as the intercept.
     
  11. Nov 5, 2014 #10

    gneill

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    aΔT has units of Ohms. ΔR/R0 would be unitless. So no, that can't work. How would you find the ΔR?
    aT has units of Ohms. 1 is unitless. The intercept occurs when T = 0C, and it's equal to R0.

    I think you're struggling to retain the R(1 + aT) form when it's not required. You have a reference resistance and temperature of 100 Ω at 0C. You have a temperature coefficient of 0.005 Ω/°C . Multiplying that coefficient by a temperature change will yield a change in resistance in Ω.

    So moving right along, Rt(T) = aT + R0; slope intercept form where a takes on the role of slope and R0 the intercept.
     
  12. Nov 5, 2014 #11
    Okay so if I put that Rt into the Vout equation I get

    Vout = 10(aT+R0)/(aT+R0+100) - 5

    Now differentiating this with respect to T I get

    dv/dT = 10(a+R0)/(a+r0+100) - 5

    Is that it? I just need to sub in R0 and a values to solve this question?
     

    Attached Files:

  13. Nov 5, 2014 #12

    gneill

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    Sure. You could replace Ro with its numerical value at this point to get rid of the extra constant.
    Hmm. The T variable should appear in the derivative. And surely the constant 5 must disappear since the derivative of a constant is zero... You might want to verify your work on the derivative. If in doubt, post your steps.

    Once you've got the expression for dV/dt then you can plug in values of T to obtain dV/dt at those temperatures.
     
  14. Nov 5, 2014 #13
    So Vout = 10(0.005T+100)/(0.005T+200) -5

    dv/dT = 0.05T+1000/0.005T+200

    is that correct? If I differentitate T then it will totally disappear. But shouldn't 1000 and 200 disappear as they are constants as well?
     
  15. Nov 5, 2014 #14

    gneill

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    Your derivative is not correct. How do you find the derivative of a quotient? (that is, suppose f and g are both functions of x, then how do you find the derivative of f/g ?)
     
  16. Nov 5, 2014 #15
    Okay I used quotient rule. so I did

    vu'-uv'/v^2

    where v is 0.005T +200 and v' is 0.005
    U is 0.025T and U' is 0.025

    dv/dT = 0.025 - 1.25*10-4 T/(0.005T+200)

    Should I just sub in the T values to get dv/dt?
     
  17. Nov 5, 2014 #16

    gneill

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    ...or with parentheses to make the order of operations clear: (vu'-uv')/v^2
    Okay, not sure where your u comes from. The expression or Vout that you had before was:

    Vout = 10(aT+R0)/(aT+R0+100) - 5

    Ignoring the 5 because it'll disappear under differentiation, and subbing in 100 for R0 gives:

    f(T) = 10(0.005T + 100)/(0.005T + 200)

    Surely u is 10(0.005T + 100) = 0.05T + 1000
    Of course. You're finding the slope of the resistance curve at the given temperature.
     
  18. Nov 5, 2014 #17
    I multiplyed with 5 to make it one fraction and then I took u and v. If I differentiate 5 and then take u and v won't it be some sort of double differentiation?

    If I take u as 0.05T + 1000 then I get

    dv/dT = 5.05-2.5*10^-4T/(0.005T+200)

    Is this Dv/dT correct or the previous one?
     
  19. Nov 5, 2014 #18

    gneill

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    I don't understand multiplying by 5... If you wanted to make one fraction out of the lot you would have to multiply the 5 by the denominator of the other term, then combine the resulting numerators. But that's added work for nothing. The 5 is a constant and will disappear under differentiation, so you can ignore it altogether.

    d/dx (f(x) + const) = d/dx f(x)

    Neither. What happened to squaring the denominator? You should end up with a T2 term in the denominator.
     
  20. Nov 5, 2014 #19
    Okay at the denominator I had v^2 which is (0.005T+200)^2 which gives me a quadratic equation 2.5*10^-5 T^2 + 2T +40000 so

    dv/dT = 15/(2.5*10^-5 T^2 + 2T +40000)

    Is that correct?
     
  21. Nov 5, 2014 #20

    gneill

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    Getting there. The numerator looks a bit off. Let's start by clearing the decimals from the original function. If you multiply the numerator and denominator by 200 you'll have:

    ##10 \frac{200(0.005 T + 100)}{200(0.005 T + 200)} = 10 \frac{T + 20000}{T + 40000} ##

    You can keep the 10 separate, since it will just multiply the result. That is, d/dx 10f(x) = 10 d/dx f(x).

    Now your u and v comprise easy to handle values. You can leave the denominator of the derivative, v2, as ##(T + 40000)^2##.
     
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