Connect with us

A little help with some basic calculus

Discussion in 'Electronic Basics' started by Steven O., Feb 6, 2005.

Scroll to continue with content
  1. Steven O.

    Steven O. Guest

    Okay, this is a homework problem, but I'm middle-aged and already have
    my B.A., and just returning to school to pick up a little more
    knowledge. Besides, you can clearly see (below) that I'm giving this
    my best shot. So, someone please help me out here. For a basic
    electronics class, we are given that the temperature coefficient for
    the resistance of a material is given by:

    a (for alpha) = (1/R)(dR/dT), and we are asked to show that:

    R2 = {[ 1 + a(T1 - Ts) ] / [1 + a(T2 - Ts)]} R2

    where Ts is the "reference temperature".

    But here is how the math works out for me:

    a = (1/R)(dR/dT)

    dR/R = a dT Take indefinite integral of both sides....

    ln R = a T + Ts, where Ts is said reference temperature

    Assume R1 corresponds to T1, and R2 to T2, then....

    ln R1 = a T1 + Ts, and ln R2 = a T2 + Ts

    ln R1 - ln R2 = a (T1 - T2), so Ts is already gone.

    ln (R1/R2) = a (T1 - T2), exponentiate both sides...

    R1/R2 = exp (a [T1 - T2])

    exp x is approximately 1 + x, so we have,

    R1 = R2 { (1 + a [T1 - T2]) }, which is not the professor's desired
    result. Did the professor screw up, or have I forgotten some basic
    calculus or algebra in twenty years since college? Thanks in advance
    for all replies....

    Steve O.

    "Spying On The College Of Your Choice" -- How to pick the college that is the Best Match for a high school student's needs.
  2. Andrew Holme

    Andrew Holme Guest

    This is wrong. The constant of integration isn't Ts.
    Your maths looks fine to me, apart from that constant of integration - which
    cancels-out anyway

    Maybe your teacher did it this way :

    [ ln R ] Limits Rs -> R = [ aT ] Limits Ts -> T

    ln (R1/Rs) = a[T1-Ts]
    ln (R2/Rs) = a[T2-Ts]

    R1/Rs = exp( a[T1-Ts] )
    R2/Rs = exp( a[T1-Ts] )

    If we apply the approximation now ...

    R1/Rs ~ 1 + a[T1-Ts] eqn 1
    R2/Rs ~ 1 + a[T1-Ts] eqn 2

    Divide eqn 1 / eqn 2

    R1/R2 = { 1 + a[T1-Ts] } / { 1 + a[T2-Ts] }
  3. If I follow the intent, I get:

    dR/R = a dT
    ln R + C1 = a*T + C2
    ln R = a*T + C2 - C1
    ln R = a*T + C

    solve C for initial conditions, which is Rref and Tref

    C = ln Rref - a*Tref

    Plugging back in, we get:

    ln R = a*T + ln Rref - a*Tref
    ln R = a*T - a*Tref + ln Rref
    ln R = a*(T - Tref) + ln Rref
    ln R - ln Rref = a*(T-Tref)
    ln (R/Rref) = a*(T-Tref)
    R/Rref = e^(a*(T-Tref))
    R/Rref = e^(a*T)/e^(a*Tref)


    R/Rref = A*e^(a*T), where A = e^(-a*Tref)

    I think that puts me in rough agreement with you, to this point. Substituting
    1+x for e^x:

    R/Rref = (1+a*(T-Tref))
    R = RRef * (1+a*(T-Tref))

    I suppose if there were two different R's, say R1 and R2, then their ratio would

    R1/R2 = [RRef * (1+a*(T1-Tref))] / [RRef * (1+a*(T2-Tref))]
    R1/R2 = [(1+a*(T1-Tref))] / [(1+a*(T2-Tref))]

    That would look about like your teacher's ratio, wouldn't it?

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day