1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trouble with models

  1. May 16, 2009 #1
    1. The problem statement, all variables and given/known data
    The transient behaviour of a capacitor has been studied by measuring the voltage from across the device as a function of time. The following data has been obtained.

    Time (s)/Voltage (v)
    0 / 10
    1 / 6.1
    2 / 3.7
    3 / 2.2
    4 / 1.4
    5 / 0.8
    6 / 0.5
    7 / 0.3
    8 / 0.2
    9 / 0.1
    10 /0.07
    12 /0.03

    2. Relevant equations

    The following relationship for the change in voltage, V, with time, t, has been proposed:

    DV/DT = -k fcn(v)

    Where k is a constant and fcn(v) could have one of the following forms:

    1. fcn(v) = c (where c is a constant)
    2. fcn(v) = V
    3. fcn(v) = V2

    (fcn is an abbreviation of the term "a function of")

    I am to derive three corresponding expressions for V as a function of time.

    3. The attempt at a solution

    I am having trouble with where to start. I believe the aim is to find which equation, using the data, creates a linear relationship. Any help on how to derive the equations would be much appreciated.
     
  2. jcsd
  3. May 16, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Surely the problem is NOT to "find which equation, using the data, creates a linear relationship" because that's trivial- (1) is the only constant derivative and so the only linear function. The problem simply asks you to "derive three corresponding expressions for V"

    In (1), since fcn(V)= c, a constant, your equation becomes dV/dt= kc. What do you get when you integrate that? Use three of the given values to determine k, c, and the constant of integration.

    In (2), since fcn(V)= v, your equation becomes dV/dt= kV. That can be written as dV/V= kdt and integrated.

    In (3), since fcn(V)= V2 (I assume that's what you mean), your equation becomes dV/dT= kV2. That can be written as dV/V2= kdt and integrated.
     
  4. May 16, 2009 #3
    For the equation

    dv/dt= kc

    With intergration I get
    V(t2)-V(t1)=kc[t2-t1]

    I am unsure of how to use the 3 values to determine k c and the constant could you give me some idea please?
     
  5. May 16, 2009 #4
    Firstly, if k is a constant and c is a constant then k x c = a constant - lets' call it m.

    So you could re-write your equation as

    V(t2) - V(t1)=m x (t2-t1)

    Suppose t2=4 sec and t1=3 sec - could you not refer to the time-voltage values list and find the corresponding voltages at those times?
    This would lead you to a value of the unknown m over the interval from t=3 to 4 sec.

    You could also find the value of m for other time spans - choose whatever you wish. The point is that if the (first) proposed function is to fit the given data, then m would be constant over the whole range of the data. If it doesn't turn out to be so, you would discard this solution.

    This would seem to be the point of the exercise. Which proposed function best fits the data?
     
  6. May 16, 2009 #5
    Ok thanks for all that. I understand getting the model which fits best, however I am struggling with the integration parts, here is my attempts at the other 2.

    dv/v=kdt

    becomes

    ln (vt2)/(vt1) = k[t2-t1]

    and for

    dv/v2=kdt

    becomes
    k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]

    Is that correct, my next step would be to graph them
     
  7. May 16, 2009 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Looks correct, except that in the first post you had "-k".

    Also, the expression from integrating dv/v2 could be simplified.
     
  8. May 16, 2009 #7
    By simplify do you mean saying,

    t2 would be anytime t
    and t1 would be the initial voltage ie t1=0

    So simplifying the expressions would give:

    for ln (vt2)/(vt1) = -k[t2-t1]

    Becomes ln V(t)/V=Kt
    making that ln V(t)=ln V(-kt)

    and for k[t2-t1]= [(v(t2)-1)/-1] - [(v(t1)-1)/-1]

    becomes -kt= [(v(t)-1)/-1] - [(v-1)/-1]

    I am unsure of how to make v(t) the subject of the last one?
     
  9. May 16, 2009 #8
    Could it be

    V(t)-1= v-1 -kt
     
  10. May 16, 2009 #9
    Perhaps,

    [tex]V(t)^{-1}[/tex]=[tex]V_{0}^{-1}-{kt}[/tex]

    Where [tex]V_{0}[/tex] is the value at time t=0

    You might also write the equation in the form V(t) = ... etc.
     
  11. May 16, 2009 #10

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, for starters. Though I think you meant t1 would be the initial time, and v1 is the initial voltage at t=0.

    Okay so far.
    (You do mean ln(V(t) / V), right?)

    Not quite. Note that

    ln(V2/V1) = ln(V2) - ln(V1), not ln(V2)/ln(V1)​

    Use another simplification I had in mind, namely

    x-1 / (-1) = -1/x
     
  12. May 16, 2009 #11
    Firstly thanks for all your replies you are very generous with your time.

    ln(vt2) - ln(vt1) = -k(t2-t1)

    Becomes

    ln v(t) - ln v0 = -kt

    which then becomes

    ln v(t) = -kt + ln v0

    and I plot the ln v against time.

    and the last equation becomes
    1/v(t) = -kt + 1/v0

    and I plot the inverse of voltage against time!
     
  13. May 16, 2009 #12

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yup, that's the idea. :smile:

    p.s. You're welcome.
     
  14. May 17, 2009 #13
    Thank you to everyone who replied to this thread. This is my first experience with the Physics Forum and its just great! I will endeavor to be helpful to others on the site. You guys have given me a huge helping hand so thanks very much!

    Cheers
    Matthew
     
  15. May 28, 2009 #14
    Just an update!

    I got an A for my assignment, thanks again for your help.

    Regards
    Matthew
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trouble with models
  1. Integration trouble (Replies: 1)

  2. Trouble with tables (Replies: 7)

  3. Integration troubles (Replies: 4)

  4. Proof Trouble. (Replies: 3)

Loading...