confused about reactance

Thread Starter

mogadeet

Joined May 1, 2007
19
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
 

recca02

Joined Apr 2, 2007
1,214
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.
 

bloguetronica

Joined Apr 27, 2007
1,372
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.
 

Ron H

Joined Apr 14, 2005
7,014
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
Why do you have the minus sign in the equation? Take it out and your equation is correct.
 

Papabravo

Joined Feb 24, 2006
13,503
... Since I is a logaritmic function, I' is a positive function that tends to zero. ...
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.
 

bloguetronica

Joined Apr 27, 2007
1,372
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.
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.
 

Papabravo

Joined Feb 24, 2006
13,503
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.
 

bloguetronica

Joined Apr 27, 2007
1,372
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.
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.

Well, I didn't assumed that was correct.
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.
 

bloguetronica

Joined Apr 27, 2007
1,372
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.
 

Ron H

Joined Apr 14, 2005
7,014
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.
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).
 

bloguetronica

Joined Apr 27, 2007
1,372
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).
I was refering to a situation where a constant potential was suddenly applied. Am I missing something here?
 

Ron H

Joined Apr 14, 2005
7,014
I was refering to a situation where a constant potential was suddenly applied. Am I missing something here?
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)
 

bloguetronica

Joined Apr 27, 2007
1,372
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)
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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