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Good @ Maths?

This is a discussion on Good @ Maths? within the Off Topic forums, part of the Entertainment category; Originally posted by Hostage+--><div class='quotetop'>QUOTE(Hostage)</div> <!--QuoteBegin-TunaFishyMe uhhh... arent you just plugging in the numbers? 4= 6(1-e^(-t/25^-6)(52)) 0.6 = 1-e^(-t/25^-6)(52)) 0.3=e^(-t/25^-6)(52)) ...
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  1. #16
    Senior Member festo's Avatar
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    Originally posted by Hostage+--><div class='quotetop'>QUOTE(Hostage)</div>
    <!--QuoteBegin-TunaFishyMe
    uhhh...
    arent you just plugging in the numbers?

    4= 6(1-e^(-t/25^-6)(52))
    0.6 = 1-e^(-t/25^-6)(52))
    0.3=e^(-t/25^-6)(52))
    ln 0.3 = (-t/25^-6)(52))

    (1.2)(52)(25^-6) = t
    t = 2.5 * 10^-7


    Btw, are you use the value of C is not 25*(10^-6)?

    Can you explain the math reasoning behind this? It was the only part I didnt understand..[/b]
    Havent checked the answer yet too busy on another assignment, although all it basically is, is working with exponential, indices and pluggin in the numbers

  2. #17
    TunaFishyMe
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    Uhh, thats natural logarithms.

    Its used for like working in different bases. Many things use log to measure.
    e is euler's constant. Which is like 2.something something sometihng.
    if i had x = e^10. I can just say ln x = 10.

    uhhh...the answer is wrong then. YOu have to change the constant of C. I just took it as soeminthg^-6. Not as something * 10^-6

  3. #18
    Senior Member SlmShady's Avatar
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    Default Re: Good @ Maths?

    Originally posted by Festo
    The instanteous current i amps flowing in a capacitor at a time t secs is given by:
    I= I(1-e^-t/CR), where I = 6A, R = 52Kohms, C= 25^-6 F
    Determine the time t when the instanteous current I is 4A
    thats physics fag...btw we did that a whiel ago but i forgot it, u have to take ln (natural log) to get rid of the e and work with the exponent.

    and ln(e) is 1.

  4. #19
    Senior Member devilscommandments's Avatar
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    ok lets show everyone how smart we are cuz we copy formulas from text books and solve them in a calculator...

  5. #20
    TunaFishyMe
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    Default Re: Good @ Maths?

    Originally posted by SlmShady+--><div class='quotetop'>QUOTE(SlmShady)</div>
    <!--QuoteBegin-Festo
    The instanteous current i amps flowing in a capacitor at a time t secs is given by:
    I= I(1-e^-t/CR), where I = 6A, R = 52Kohms, C= 25^-6 F
    Determine the time t when the instanteous current I is 4A
    thats physics fag...btw we did that a whiel ago but i forgot it, u have to take ln (natural log) to get rid of the e and work with the exponent.

    and ln(e) is 1.[/b]
    ln (e) is error.
    ln e1 is 1 xD!!

  6. #21
    Vlein
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    674764746ex(888***-t-ii)/poop.8745^45)[p-09]

  7. #22
    Senior Member festo's Avatar
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    i = I(1-e^-t/CR) Plug in all the shit...

    4 = 6(1-e^-t/25E-6*52000) Multiply out the brackets
    4 = 6 - 6e^-t/25E-6*52000) Take the 6 over
    -2 = -6e^-t/25E-6*52000) Divide by 6
    -1/3 = -e^t/25E-6*52000) Take log of each side and *-1 to make time positive
    -ln(1/3) = t/25E-6*52000
    1.098612289 = t /1.3 Times by 1.3 to get t
    1.428595975 = t

    Think thats correct plugged it back into original formula to get 4A

  8. #23
    TunaFishyMe
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    looks right

  9. #24
    cccch
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    stuff like this isnt really hard if you know how to do it. Just work them down. If festo wasnt drunk all the time he coulda had this done already.

  10. #25
    Senior Member C4pt4iN's Avatar
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    e = 2,7182

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