cal and electrical circuits

Discussion in 'Homework Help' started by pie, Jul 30, 2008.

  1. pie

    Thread Starter New Member

    Jul 30, 2008
    3
    0
    Can someone help me understand a concept? I did the problem, but I don't understand something.

    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]
    [​IMG]

    My book replaced the + with just -. When I graph them, they look different, but both have a limit of 5 as t approaches infinity. The red line, where I use + is wrong according to the book. I don't know why they change the + to just -, does anyone know? Can the equation ever have the + instead of -? If so, what would that mean? Would that be a spark? Would the direction of the flow change? What does the red graph mean compared to the green graph?
     
  2. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    I got 5-e^(-3t).

    In your 7th line, you use +/-. The left hand side of the equation decides the value is absolute, so you don't need to do that. On the 9th line, you converted e^(-3t-3C) to e^(-3t)e^(3C); It should be e^(-3t)e^(-3C).
     
  3. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    I think it (the green) means you initially had a 12V potential applied for about 95.9 msec. Then the 60V was applied. Ignore t<0. You might want to wait for a second opinion.
     
    Last edited: Jul 30, 2008
  4. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    According to your problem statement, you only need to find the limiting value of the current. If this is the case, then the differential equation is a red herring. The current at t=infinity is simply V/R, or 5 amps. Perhaps you have to prove this by evaluating the differential equation at t=infinity.
     
    Last edited: Jul 31, 2008
  5. pie

    Thread Starter New Member

    Jul 30, 2008
    3
    0
    Oh I see that. Thank you. But I guess since I change that into a constant, it didn't matter in the end.

    Yes I only need to find the limit. It's 5 as t approaches infinity. And I understand the green graph at at a practical matter. But in theory or in some strange chase, is the red graph ever possible? Even if you ignore all rules of physics, what would cause the red graph?
     
  6. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Where did you come up with the idea that absolute value was required in the evaluation of the integral? I don't claim to be a math wizard. I just looked it up, and the resources I have said nothing about absolute value - just like your book.
     
  7. blazedaces

    Active Member

    Jul 24, 2008
    130
    0
    Integral (1/x) = ln(abs(x))

    Now, to answer the OP, as said somewhere else, you only need the limiting value of the current, that is when t -> Infinity so the +/- 0 doesn't really make a difference, it still = 5 (as you can see on the graph).

    I don't know why the book chooses the positive though. I think they just didn't care at that point...

    -blazed
     
  8. pie

    Thread Starter New Member

    Jul 30, 2008
    3
    0
    Yes I know the final answer. But I'm trying to understand the concept behind this. If the constant, A, was another sign, that would change the sign the book used. So when I plot them, one comes from y1(0)=infinity and the other if y2(0)=0. And y2(0)=0 looks right. When the time is zero, how can you have any current? But would there ever be a case where at 0 time, you'll have a real high amount, like in the case of y1?
     
  9. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    pie,

    What is it that you don't understand. You have never explained what it is.

    Your switching back and forth using T,t and I,i is confusing.

    You don't have to worry about 15-3i being negative. At t=0, i is zero, so 15-3i is positive. The highest value i attains is 5, so 15-3i is never negative.

    You don't have to sweat whether in ln u, u is positive or negative. Example: Integral of 1/x from -5 to -2 = ln |-2| - ln |-5| = ln 2 - ln 5 = -.91629 . The absolute value automatically takes care of the sign.

    What concept? What are A and B supposed to be? The solution required two integrations, and the constants can be added together. ln(15-3i) = -3(t+c) . Since the initial conditions are t=0 and i=0, e^-3c = 15 . Back substituting gives 15-15e^-3t = 3i ===> 5-5e^-3t = i . Do any of your plots look like that equation? If there were a initial current, then the value of the constant c would be different and the equation would change.

    Ratch
     
    Last edited: Aug 7, 2008
Loading...