ISC PV Cell Question (Isolating a variable)

Discussion in 'Homework Help' started by Ryaner, Jul 5, 2013.

  1. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    Hey I am wondering how I isolate the Isc variable here.

    Question
    I presume I use the Voc formula, just wondering how I isolate Isc. I believe the answer should be 4 and 2.
     
  2. WBahn

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    Mar 31, 2012
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    The Voc equation should be

    <br />
Voc\;=\;0.0257V \ln \left( \frac{I_{SC}}I_0} + 1 \right)<br />

    The units are important (and I'm having to guess that it is volts and not millivolts or something else, which I shouldn't have to do).

    If you have

    a = ln(b)

    How do you solve for 'b'?
     
  3. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    That V you added is not in the data provided, I write the questions word for word. I am presuming it is a constant. I already made a attempt of breaking the ln(isc/io) into log Isc - log Io and working from there.
     
  4. WBahn

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    I'm not surprised that it wasn't in the data provided. Which means YOU have to guess. If you guess wrong, then bad things happen. In the real world, "bad things" can include hundreds of people dying when something you designed fails. If YOU develop the habit of religiously using and tracking units, then you will spot when a vendor doesn't and can take a step back and find a way to determine the units that should be used. Otherwise you will just proceed to blindly crank your way to financial ruin or a lengthy prison stay for criminal negligence.

    Don't break the ln() up that way. Answer my question.
     
  5. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    WBahn I think I will wait for the help of someone a little more friendly, I don't find anything you are saying remotely helpful.

    Regards.
     
  6. WBahn

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    That's your choice. I don't know what is so unfriendly about asking you to solve for 'b' in that expression and doing so would show you the path to find the answer you are looking for in just two or three more simple steps (this one is the hardest).
     
  7. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    I obviously don't know the maths required to get the answer for this problem, otherwise I wouldn't be here. You then proceed to ask me the maths required for the problem.

    If you have a solution, please post it.
     
    Last edited: Jul 6, 2013
  8. WBahn

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    We won't do your homework for you, which is what you clearly want. If you don't have the math to work this problem, then you are taking a class you shouldn't be in. You either need to drop this course and take a math course that is going to fill in your gaps or you need to spend some time filling them in yourself.

    One thing I have seen is that many students are, unfortunately, only taught how to do logarithms and exponentials by being told which button to press on a calculator. If that is the case here, it is not your fault that your skills are weak, but it is still your responsibility to correct them. I put together a mathematics review a number of years ago that might be helpful, but I have to find the link to it. I'll try to post that later today.
     
  9. WBahn

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  10. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    I looked at your page, nothing in the log section that I did not see before. Only log I can see that could apply was the one you told me was wrong!!

    log(x/y) = log(x) - log(y) seems like the most relevant equation to use. If not why don't we use it ? Does this not apply because of the +1 ??
     
  11. WBahn

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    What is the relationship between the logarithm function and the exponential function?

    If log10(y) = 3, what is y?
     
  12. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    Log10(y) = 3
    y = 10^3

    Log10(y)=3
    10(y)=10^3
    y=(10^3)/10
    y=100

    Right or wrong ? If it is wrong please explain where I went wrong please. Posting links doesn't really help as I am more than capable of reading a book.

    Also this is not homework. I finished my degree already.
     
    Last edited: Jul 7, 2013
  13. WBahn

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    Both. You gave TWO different answers (and apparently didn't even realize it) and one of them is right and one of them is wrong.

    Now that you are finally posting some work, it is actually possible to do so.

    Log10(y)=3
    10(y)=10^3

    Where do you get this from?

    What does log10(y) = 3 mean?

    From the reading:

    <br />
z = x^y<br />

    is equivalent to

     <br />
y = \log_x(z)<br />

    So do some simple manipulation and exponentiate both sides of the last equation using x as the base:

     <br />
y = \log_x(z)<br />
\ <br />
x^y = x^{\log_x(z)}<br />
\ <br />
z = x^{\log_x(z)}<br />

    This result says that if you rase the base to the logarithm, in that base, of anything you get the anything.

    Therefore

    <br />
\log_{10}(y)=3<br />
\ <br />
10^{\log_{10}(y)}=10^3<br />
\ <br />
y=10^3=1000<br />

    Reading and comprehending are clearly two different things.

    And this is supposed to be reassuring?
     
  14. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    First one was the solution for Log10(y) = 3, Second one was the solution for Log(10y) = 3. Figured I would do both.

    So were does this leave me in relation to isolating the Isc variable.

    The log poses the issue for me here. Why does it not make sense to do the following, Log((Isc/Io)+1) into (Log(Isc) - Log(Io) +1). Does the 1 not make the use of that formula possible ? If you could explain it like you did above that would be great.
     
    Last edited: Jul 7, 2013
  15. WBahn

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    Log10(y) is not the same as log(10y); log10() is a function that returns the base-10 logarithm of the argument, while log() is a function that either returns the base-10 logarithm or the base-e logarithm, depending on which convention is used by the person that wrote it.

    But I pretty much see what you were trying the do.

    It actually leaves you very close to the answer, if you will just now do what I asked originally. Trust me that this is leading you someplace.

    If you have

    a = ln(b)

    How do you solve for 'b'?

    Which is why I am trying to walk you through it.

    For two reasons: (1) It doesn't lead you to an answer, at least not very cleanly, and (2) it's wrong.

    You are trying to claim that

    log(a+b) = log(a) + b

    Don't think so? You simply have a=Isc/Io and b=1.
     
  16. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    A = Ln(B)
    Ln(B)=A
    E^Ln(B)=e^A
    B = e^A
    e^A= B

    This is probably wrong, I will make another attempt at it today.
     
  17. WBahn

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    Nope, this is just fine.

    Now let's complicate it just a tiny bit each time. Solve each of the following for 'A'.

    <br />
V_1 = V_0 \ln(A)<br />

    <br />
V_1 = V_0 \ln(CA)<br />

    <br />
V_1 = V_0 \ln(CA+B)<br />
     
  18. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    First Solution

    V1 = VoLn(A)
    VoLn(A) = V1
    VoLn(A)/Vo = V1/Vo
    Ln(A) = V1/Vo
    e^ln(A) = e^V1/Vo
    A = e^V1/Vo

    Second Solution

    Working on it atm
     
  19. WBahn

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    You basically have it correct, but only because I know what you mean and not by what you wrote. What you wrote is:

    A = (e^V1)/Vo

    because exponentiation has higher precedence than division. What you needed to write was:

    A = e^(V1/Vo)

    With a bit of practice you should be able to go from

    ln(A) = V1/Vo

    to

    A = e^(V1/Vo)

    directly. But, in the meantime, by all means list the intermediate step(s).

    For the next one, don't make it harder than it is. Use the exact same steps that you use above to get rid of the ln() function before you even try to do anything with what is inside the ln() as the argument.
     
  20. Ryaner

    Thread Starter New Member

    Dec 14, 2011
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    Second Attempt.

    V1 = VoLn(CA)
    VoLn(CA) = V1
    VoLn(CA)/Vo = V1/Vo
    Ln(CA)=V1/Vo
    e^ln(CA) = e^(V1/Vo)
    CA = e^(V1/Vo)
    A = (e^(V1/Vo))/C

    Not sure if it is correct, I looked at what you said in relation to CA. If that is the class is the method not the same with the added part of isolating the variable I am after.

    Sorry for the delay in getting this attempt to you, I've been very busy with work.

    V1 = VoLn(CA+B)
    VoLn(CA+B) = V1
    VoLn(CA+B)/Vo = V1/Vo
    Ln(CA+B)=V1/Vo
    e^ln(CA+B) = e^V1/Vo
    CA+B = e^V1/Vo
    CA = (e^V1/Vo)-B
    A = ((e^V1/Vo)-B)/C


    Again if I am leaving the isolating of whats inside the ln until I have removed it I can't see the method changing.
     
    Last edited: Jul 14, 2013
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