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Dimensional Analysis

  1. May 2, 2009 #1
    [tex]

    V(t)=IR(1-e^{t/RC}) \rightarrow ln|IR - V(t)| = ln|IR|-t/RC

    [/tex]

    The first equation below has units of voltage.

    the logarithm has units of voltage, but what does ln|IR| and ln|V| physically mean?


    The purpose of this question is that if you plot the natural logarithm of the voltage versus time, it should yield a straight line with slope -1/RC and intercept ln|V_c|.

    From which you can determine the capacitance given a known resistance.
     
  2. jcsd
  3. May 2, 2009 #2

    tiny-tim

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    Hi Nusc! :smile:

    yes, you're right …

    technically the equation is dimensionally wrong …

    anything inside a ln() has to be dimensionless, so it should really be ln(|(IR - V)/IR|) = -t/RC,

    or ln(|(IR - V)/V0|) = ln(|IR|/V0|) - t/RC, where V0 is some voltage chosen at random just to make the dimensions right. :rolleyes:

    In practice, we choose V0 = 1 in whatever units we're using, but without bothering to mention it. :wink:
     
  4. May 2, 2009 #3
    I don't like the idea of introducing an arbitrary voltage.

    What if your actually measuring the voltage of the capacitor? The resistance is due to the voltmeter. I need to know what, in your case, V_0 physically is and where does it arise? The voltage of what?
     
    Last edited: May 2, 2009
  5. May 3, 2009 #4

    tiny-tim

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    Hi Nusc! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    ok, then use the first formula, ln(|(IR - V)/IR|) = -t/RC. :smile:
    V0 is arbitrary … it can be anything

    in practice, it's 1 volt (so as not to change the maths!) :wink:
     
  6. May 3, 2009 #5
    You said in practice, can you convince me from an experimental point of view? I just don't like introducing quantities such as V0 or IR without any justification.

    I_c(t) + I_R(t) = I

    i=dQ/dt implies I_c(t) = dQ/dt = d/dt C*V(t) = C d/dt V(t)

    V_R(t) = I_R(t) * R


    The above equations imply

    C d/dt V(t)+V(t)/R = I

    Then we get V(t) = IR(1-e^(-t/RC))

    This is understood mathematically but from an experimental point of view I'm not sure whether this is safe. This is important to me.

    Thanks
     
  7. May 3, 2009 #6

    tiny-tim

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    I don't follow you …

    the second equation obviously is a solution of the first, and there's nothing dimensionally suspect about it …

    what worries you about it? :confused:
     
  8. May 3, 2009 #7
    Sorry, so that second equation you mentioned:
    V(t) = IR(1-e^(-t/RC)) is a solution of the first C d/dt V(t)+V(t)/R = I, yes and there is nothing dimensionally suspect about it. What worries me is introducing V0 (or arbitrarily dividing by IR in ln(|(IR - V)/IR|) = -t/RC.)
     
  9. May 3, 2009 #8

    diazona

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    The division by IR is not arbitrary at all - that's what you get when you solve the differential equation [tex]C \frac{\mathrm{d}V}{\mathrm{d}t} + \frac{V}{R} = I[/tex]. As you know (I suppose), whenever you do an integral, you wind up with an "arbitrary" constant that is in fact determined by the boundary terms (for example, that [tex]V(0) = 0[/tex] or whatever it is). Here IR is that constant.
     
  10. May 4, 2009 #9

    tiny-tim

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    Can't improve on diazona's :smile: reply!

    (and if using, or imagining, V0 worries you , then … don't use it! :wink:)​
     
  11. May 4, 2009 #10
    If you don't use it then the log of that function doesn't physically make sense, I just needed a justification for V0.
     
  12. May 4, 2009 #11

    tiny-tim

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    I meant use ln(|(IR - V)/IR|) = -t/RC instead.

    The mathematical justification for V0 is that it works.

    There's no physical meaning to it … as i said, V0 is arbitrary. :wink:
     
  13. May 4, 2009 #12
    Let's prove it then:
    [tex]
    c\frac{d}{dt} V(t)+\frac{V(t)}{R}&=&I,
    [/tex]
    [tex]
    I &=& e^{\int^{t}_{t_{0}} \frac{1}{RC} dt} =e^{ \frac{t}{RC}}
    [/tex]
    [tex]
    \frac{d}{dt}(V(t)e^{\frac{t}{RC}}) &=& \frac{I}{C}e^{\frac{t}{RC}
    [/tex]


    [tex]
    V(t)e^{\frac{t}{RC}}|^{t}_{t0} &=& \frac{I}{c} \int_{t0}^{t} e^{\frac{t'}{RC}} dt'
    [/tex]
    [tex]
    V(t)e^{\frac{t}{RC}} - V(t0) &=& \frac{IRC}{C} e^{\frac{t'}{RC}}|^t_{t0}
    [/tex]
    [tex]
    V(t)e^{\frac{t}{RC}} - V(t0) &=& IRe^{\frac{t}{RC}}-IRe^{\frac{t0}{RC}}
    [/tex]

    Taking t0 to be 0
    [tex]
    V(t) - e^{\frac{-t}{RC}}V(0) &=& IR( 1 - e^{\frac{-t}{RC}})
    [/tex]
    [tex]
    IR-V(t) &=& e^{\frac{-t}{RC}}(1-V(0))
    [/tex]
    [tex]
    ln|IR-V(t)| &=& -\frac{t}{RC}(1-V(0))
    [/tex]

    I don't see where the division of V0 you speak of comes from the differential equation and if you set V0 to be 1 in this case the RHS gives zero. Please prove it to me and don't just say it's true. The RHS has limits so it's not arbitrary.
     
    Last edited: May 4, 2009
  14. May 4, 2009 #13

    tiny-tim

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    Hi Nusc! :smile:

    Sorry, I don't understand that at all.

    You seem to be introducing V0 for no reason … you don't need it unitl the line in which "ln" first appears. :wink:

    (are you confusing V0 with V(t0)?)
     
  15. May 4, 2009 #14

    diazona

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    I can't follow your solution, Nusc, but here's mine:

    First, this is the differential equation we want to solve.
    [tex]C\frac{\mathrm{d}}{\mathrm{d}t} V(t)+\frac{V(t)}{R} = I[/tex]
    Now, there are many different ways to solve this equation. One is to change variables by [tex]V(t) = u(t) + IR[/tex] and rewrite the equation as
    [tex]C\frac{\mathrm{d}}{\mathrm{d}t} u(t)+\frac{u(t) + IR}{R} = I[/tex]
    which, by simple algebra, becomes
    [tex]\frac{\mathrm{d}}{\mathrm{d}t} u(t) = -\frac{u(t)}{RC}[/tex]
    and then
    [tex]\frac{\mathrm{d}u}{u} = -\frac{\mathrm{d}t}{RC}[/tex]
    Now we can integrate on both sides,
    [tex]\int_{t_0}^{t} \frac{\mathrm{d}u}{u} = -\int_{t_0}^{t}\frac{\mathrm{d}t}{RC}[/tex]
    When you do the integral on the left, you get
    [tex]\int_{t_0}^{t} \frac{\mathrm{d}u}{u} = \mathrm{ln}\frac{u(t)}{u(t_0)}[/tex]
    and the one on the right gives
    [tex]-\int_{t_0}^{t}\frac{\mathrm{d}t}{RC} = -\frac{t}{RC}[/tex]
    Equating the two, we get
    [tex]\mathrm{ln}\frac{u(t)}{u(t_0)} = -\frac{t}{RC}[/tex]
    and exponentiating:
    [tex]\frac{u(t)}{u(t_0)} = e^{-\frac{t}{RC}}[/tex]
    and then multiplying by [tex]u(t_0)[/tex]:
    [tex]u(t) = u(t_0) e^{-\frac{t}{RC}}[/tex]
    Now we can substitute back in [tex]u(t) = V(t) - IR[/tex], the inverse of the previous substitution:
    [tex]V(t) - IR = (V(t_0) - IR) e^{-\frac{t}{RC}}[/tex]
    and then
    [tex]V(t) = V(t_0)e^{-\frac{t}{RC}} + IR\left(1 - e^{-\frac{t}{RC}}\right) [/tex]
    If [tex]V(t_0) = 0[/tex], the first term is just zero and we're left with
    [tex]V(t) = IR\left(1 - e^{-\frac{t}{RC}}\right) [/tex]

    The need for an arbitrary constant [tex]V_0[/tex] comes in when you do the logarithmic integral. In elementary calculus they tell us that
    [tex]\int_{t_0}^{t} \frac{\mathrm{d}V}{V} = \mathrm{ln}V(t) - \mathrm{ln}V(t_0)[/tex]
    But then again, in elementary calculus, all variables represent pure numbers - no units. Unfortunately, this doesn't make sense if V has units, because you can't take the logarithm of a unit. For that reason, we introduce a constant [tex]V_0[/tex] with the same units as [tex]V(t)[/tex] and say that
    [tex]\int_{t_0}^{t} \frac{\mathrm{d}V}{V} = \mathrm{ln}\frac{V(t)}{V_0} - \mathrm{ln}\frac{V(t_0)}{V_0}[/tex]
    Why is this valid? Well, as you know, it's a property of logarithms that
    [tex]\mathrm{ln}a - \mathrm{ln}b = \mathrm{ln}\frac{a}{b}[/tex]
    So in the first case
    [tex]\mathrm{ln}V(t) - \mathrm{ln}V(t_0) = \mathrm{ln}\frac{V(t)}{V(t_0)}[/tex]
    and in the second case
    [tex]\mathrm{ln}\frac{V(t)}{V_0} - \mathrm{ln}\frac{V(t_0)}{V_0} = \mathrm{ln}\frac{V(t)/V_0}{V(t_0)/V_0} = \mathrm{ln}\frac{V(t)}{V(t_0)}[/tex]
    They're equal to the same thing and, thus (transitive property of equality) equal to each other.

    P.S. Thanks for the words of support tiny-tim ;-)
     
  16. May 4, 2009 #15
    The I i defined above is just the integration factor.

    I am not introducing V0 for no reason, V(t0) is just V(0) for t0 = 0.
     
  17. May 4, 2009 #16
    I got the same result in a different way. So I'll just quote what I don't understand.
    You mention logarithmic integral, but when I talk about the logs in first few posts, I just took the log of the solution for V(t). There's no reason to integrate again.

    Recall,

    students are measuring the voltage of a charging capacitor. The resistance comes from the voltmeter 10M ohms.
    So the idea is that using the equation for a log, students get a linear fit and they can determine the capacitance. The log solution I wrote at the begining works, but the physical meaning of logarithm with units inside does not make sense physically.

    Because what your saying when you choose V0 = 1 V means that the capacitor is initially charged at 1 volt. This is not true. The capacitors are discharged when you short the circuit.

    And please show me explicitly how you get from:
    [tex]
    V(t) = V(t_0)e^{-\frac{t}{RC}} + IR\left(1 - e^{-\frac{t}{RC}}\right)
    [/tex]
    to
    [tex]
    ln(|(IR - V)/V(t_0|) = -t/RC
    [/tex]
     
    Last edited: May 4, 2009
  18. May 5, 2009 #17
    Any ideas?
     
  19. May 6, 2009 #18

    tiny-tim

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    No, V0 is not V(0), V0 is arbitrary.

    In V(t) = IR(1 - e-t/RC), you have already chosen the constant of integration so that V(0) = 0.

    You then rewrite that as IR - V(t) = IRe-t/RC,

    and in order to take logs, you then have to divide by an arbitrary V0 (which we always take to be 1 volt, because dividing by 1 is so easy :smile:),

    to get ln((IR - V(t))/V0) = ln(IR/V0) - t/RC

    (or, of course, you can just write ln((IR - V(t))/IR) = -t/RC)
     
  20. May 6, 2009 #19
    How sad.


    Thanks!
     
  21. May 6, 2009 #20

    diazona

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    um... does that mean all is clear now?
     
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