growth & decay calculation

Discussion in 'Homework Help' started by spark8217, May 31, 2013.

  1. spark8217

    Thread Starter Member

    Aug 29, 2011
    64
    2
    Hi
    will someone please check the following calculation

    IL=e^-t/τ
    If t=2τ, then
    IL=e^2t/τ
    IL=e^-2
    I STEADY = 107/66*10^-3=0.001621 Amps
    IL= 0.001621*0.632(63.2%)
    IL= e(66000)(1.39393934*10^-6)/(0.046)
    IL=0.001621(e^(-2)) = 2.193784941*10^-4

    cheers
    spark8217
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    Sloppiness, sloppiness everywhere.

    Whatever IL is, it apparently is a pure number. It can't be a current because the right hand side is dimensionless.

    Also, note that your equation is actually

    IL=[e^(-t)]/τ

    Because unary negation is highest priority followed by exponentiation followed by multiplication/division. You need to get in the habit of being explicit with your expressions when you write them in text (and it's not the easiest thing to do) because if you don't then you will almost certainly mess things up when writing code, whether in a program or in a simulator or something else.

    How on earth do you get that result?

    Assuming you are talking about the original equation and not the one directly before it. And note that IL is still dimensionless.

    This is sure out of the blue. Where do these numbers come from? And how do the units magically appear at the end?

    More magic! Now there's a new factor that comes out of nowhere. And where does the 0.632 come from? It's certainly not from the e^(-2)

    At this point I'm totally lost and can't even begin to guess where any of these numbers come from or what they mean. Your first expression yields IL=5.44. You second expression for IL differs by more than four orders of magnitude. Do you really need someone to point out that they don't agree?
     
  3. spark8217

    Thread Starter Member

    Aug 29, 2011
    64
    2
    Hi
    thanks for the feedback after looking back can only agree,will re -work

    cheers
    spark 8217
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,743
    4,795
    Sounds good.

    Re-work, re-post, and I'll re-rip. :D

    That's how we learn.

    Something that might help.

    Your starting equation.

    <br />
I_L=e^{\frac{t}{\tau}}<br />

    As noted, if I_L is a current, this equation is wrong because the right-hand side is dimensionless. If the scaling current is "1" (hopefully meaning "1A"), then that needs to be included. This can be done a couple of ways. You can explicitly include it:

    <br />
I_L=1A \cdot e^{\frac{t}{\tau}}<br />

    Or you can introduce a scaling parameter:

    <br />
I_L=I_0 e^{\frac{t}{\tau}}<br />
I_0 = 1A<br />

    When you do use numbers, always, always, always track whatever units are associated with those numbers. The units are a fundamental part of the quantity. Most errors you will make will mess up the units, allowing you to spot and fix most mistakes almost immediately instead of three of four pages of algebra later (trust me, the half page of algebra that seems long now will sometimes pale in comparison to what you may, at times, have to wade through in the "real world"). Also, mistakes that go uncaught lead to errors in the final system and while a doctor that makes a mistake can kill someone, engineers that make mistakes can kill people in job lots. Finally, always, always, always ask if your answers make sense. If they don't, then hammer away at it until either you find the mistake or you understand why the answer is correct and that it does make sense.
     
    screen1988 and spark8217 like this.
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