Exponential function and equations

Discussion in 'General Electronics Chat' started by vead, Sep 5, 2016.

  1. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
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    Hello
    Exponential function that take argument any return value of e raised to power x
    F(t) =e^t
    F(0)=1,f(1)=2.718, f(2)=7.38,f(3)=20....etc
    It can be seen function is raising exponentials. I understand exponentials function. But I am having problem how they relates to electronics. How voltage and current signal represent as exponential function. So I trt to understand, look below example charging current flowing in series RL circuit as exponentials function over time
    IMG_20160905_164528.jpg
     
  2. Papabravo

    Expert

    Feb 24, 2006
    11,083
    2,159
    I'm not sure what your question is. An exponential function with a negative exponent will asymptotically approach 0 as time goes to ∞. If you subtract an exponential function with a negative exponent from a constant, like 1, then the result will approach 1 as time goes to ∞. An exponential function with a positive exponent represents a system that is unstable and has the potential to destroy itself.
     
  3. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
    11
    Have you seen the formula in last line? There is formula for current, that flow in RL series circuit. It is exponentials function. I don't understand that formula. I can solve questions but I don't understand concept behind the formula
     
  4. Papabravo

    Expert

    Feb 24, 2006
    11,083
    2,159
    At t=0 the value of the exponential function is 1, and 1 -1 = 0 so that is the value of the current at t=0. At some large value of t the value of the exponential is very nearly equal to 0, and so the value of 1 - ε, where ε is a very small number, is just a little less than 1 which is the current at that very large value of t.
     
  5. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
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    M
    How to derive formula for current that is flowing in circuit
    _20160905_205735.JPG
     
  6. wayneh

    Expert

    Sep 9, 2010
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    This is trivial if you know calculus and differential equations. Do you?
     
  7. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
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    My attempt
    _20160905_220912.JPG
     
  8. wayneh

    Expert

    Sep 9, 2010
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    You're close. Can you solve: e - IR - L•dI/dt = 0 for I = ƒ(t)?
    Consider the boundary conditions when the switch opens or closes, when t=0 or t=infinity

    Hint: Separate the dI and I, and dt terms so you can integrate them separately.
     
  9. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
    11
    What do you mean by e, I think voltage correct. Please confirm
    _20160905_231342.JPG
    My mistake small i and capital I are same things
     
    Last edited: Sep 5, 2016
  10. wayneh

    Expert

    Sep 9, 2010
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    Yes, e or emf. Note my previous hint.

    dI/dt = (V - IR)/L = (V/R - I)/(L/R)
    dI /(I - V/R) = - dt/(L/R)
    Integrate
     
  11. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
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    By taking integration answer is - (R /L) *t
    _20160906_011601.JPG
     
    Last edited: Sep 5, 2016
  12. wayneh

    Expert

    Sep 9, 2010
    13,440
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    Recheck your last step.
    Then think about what Imax is.
     
  13. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
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    My mistake I=V/R+e^_(R/L) *t ampear
    At t=0
    I(0)=V /R amp
    At t=infinity
    I=V/R
     
    Last edited: Sep 5, 2016
  14. wayneh

    Expert

    Sep 9, 2010
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    Right, so you can rearrange the result to:
    I(t) = I(0) • [1 -exp(-t•R/L)]

    You can further define the time constant tau as L/R, giving:
    I(t) = I(0) • [1 -exp(-t/tau)]
     
  15. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
    11
    I am confused,
    We have equations
    I(t) =V/R +e ^-(R /L) *t
    How it become I(t) =I(0).[1-e ^-(R /L)*t]?
     
  16. wayneh

    Expert

    Sep 9, 2010
    13,440
    4,275
    My bad. You had an earlier error I missed. I was going to draw this out for you but I found a nice exposition of it online.

    See here.
     
  17. vead

    Thread Starter Well-Known Member

    Nov 24, 2011
    701
    11
    I looked that is good. Also I have equation for i(t) given in text book. and I can find out voltage across resistor and inductor. Look at attached page
     
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