confused about reactance

Discussion in 'General Electronics Chat' started by mogadeet, May 21, 2007.

  1. mogadeet

    Thread Starter Member

    May 1, 2007
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    hi all

    i'm having trouble making sense of impedance. bear with me a second: i'm going to start slow because i want to be sure of asking this question properly.

    let's say an ideal inductor, with inductance L, has leads A and B, and that B is grounded, so that B's voltage level by definition is always zero. let V(t) be the voltage level of A at time t, and let I(t) be the current going through the inductor at time t, measured in such a way that a current entering through A and exiting through B is taken as positive.

    in this situation is it not the case that

    V(t) = (-1) * L * I'(t)

    where I' is the derivative of I ...? if i'm already mistaken about this part there's no point asking the rest of my question, so i'll leave it there for now and hopefully someone can offer some enlightenment.

    peace
    stm
     
  2. recca02

    Senior Member

    Apr 2, 2007
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    the voltage v(t) in the above equation is the back emf developed in the coil due to rate of change of current with time .it is not the voltage applied across inductance.
    is it what u wanted to ask?

    edit: if a varying current is produced due to voltage v then due to changing current in coil the magnetic flux linked with it tends to change. the emf induced according to lenz's law tends to oppose this change hence it opposes the cause producing it which
    in this case happens to be the voltage applied to it. hence a negative voltage is induced in coil given by the above formula.
    for a non changing current inductance is zero hence there shud not be any reactance in case of dc. ideally dc shud pass without any opposition from inductor.
     
  3. cumesoftware

    Senior Member

    Apr 27, 2007
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    In a inductance, the current tends to be bigger along with time (this explains why a inductance block AC and lets DC through). Since I is a logaritmic function, I' is a positive function that tends to zero. So it makes sense that V(t) increases along with time.
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    I thought I(t) was an exponential function in a series RL circuit. That would make I'(t) an exponential function as well. Or did I miss something since I was an undergraduate?
     
  5. cumesoftware

    Senior Member

    Apr 27, 2007
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    Well, I didn't assumed that was correct.
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    You didn't assumed what was correct?
    You made a rediculous statement and you need to defend it or retract it.
     
  7. recca02

    Senior Member

    Apr 2, 2007
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    since log and exp are inverse functions its possible to mistake one for other.
     
  8. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Why do you have the minus sign in the equation? Take it out and your equation is correct.
     
  9. cumesoftware

    Senior Member

    Apr 27, 2007
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    Didn't understand. Why defending it?
     
  10. Papabravo

    Expert

    Feb 24, 2006
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    In a series RL circuit, I is an Exponential function. I' is a also an Exponential function. It does not tend to zero, but to the value of the forcing function.

    That logarithm and exponential are inverses is irrelevant, and with even a rudimentary understanding of mathematics and physics it is pretty hard to mistake one for the other and then pretend like you can't figure out what I'm talking about.
     
  11. cumesoftware

    Senior Member

    Apr 27, 2007
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    Ok. I was wrong (if you need me to admit it in public).
    I didn't mistake an exponential function with a logaritmic function (I know each one and how they are represented). Nevertheless, they are pretty hard to mistake one for another.
    Here is what happened: I THOUGHT I(t) WAS A LOGARITMIC FUNCTION.

    And if you think I was taught some rudiments about mathematics, think again. Most guys wished to have the rudiments I have so they can finish their degree. Also, and since I'm writing about myself, because I don't write my english very well, it doesn't mean I am dumb. English is not my native language, and I don't have to know about englis to finish my degree. I don't even have the obligation to know english.

    I don't have to prove nothing to you. Maybe, it is you that have to prove something, since you are so eager to raise your voice and show your knowledge by showing others how dumb they are.

    EOS (end of story)

    But then again, we are here to help and not to discuss these things that have nothing to do with electronics.
     
  12. Papabravo

    Expert

    Feb 24, 2006
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    This is not about you or me or first languages or second languages. If we do not draw attention to these things then others will go away from the discussion with incorrect ideas and information. If you look back, I did not start the process with emphasis; that came later when it seemed that I was unable to make myself clear. If you want to participate in the global marketplace of ideas you need to expect occasioanal critical comments on your ideas and statements. I never commented on any of your personal traits nor do I intend to. My remarks were directed soley at your written statement. Nothing personal in that, and no reason for you to be defensive.
     
  13. cumesoftware

    Senior Member

    Apr 27, 2007
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    Well, I don't know why or where you infered that I managed to confuse (somehow) a logaritmic function with an exponential one, I didn't even said that. The only thing I said after my first statement and before this discussion was that I might be wrong.

    Also, I don't have to defend my statements, even if they are correct. Most of the time I won't bother to respond. I'm not in trial here. My only concern is to give correct answers if I can, or not answer at all if I am uncertain or don't know.
     
  14. cumesoftware

    Senior Member

    Apr 27, 2007
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    It makes sense that current increases exponentially throught an inductor. I just reminded that the magnetic field builds up depending on the current that is passing through the inductor, and mutually, the magnetic field will cause more current to flow. Since we are in this "positive feedback" situation, it makes sense that I is exponential.
     
  15. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Actually, with a voltage step input, the current in an ideal inductor increases linearly. In the real world, where the inductor has series resistance, the current will increase exponentially, and the exponent is negative (no positive feedback).
     
  16. cumesoftware

    Senior Member

    Apr 27, 2007
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    I was refering to a situation where a constant potential was suddenly applied. Am I missing something here?
     
  17. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Apparently you are. I said the situation was a "voltage step input". This is the same as "a situation where a constant potential was suddenly applied."
    For the LR series circuit, see this site.
    For a pure inductor, the current waveform is a ramp.
    i(t)=1/L*integral(V*dt)
     
  18. cumesoftware

    Senior Member

    Apr 27, 2007
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    i(t) = - (1/L) x int (V) dt = - (1/L) * V * t (if V0 is 0 and V1 = V) (taken from V (t) = -L x dI / dt). My mistake, it is a ramp.
     
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