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  #1  
Old 10-20-2011, 02:20 PM
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PG1995 PG1995 is offline
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Default learning workings of inductor

Hi

Please have a look on the attachments. The question "Q" is about the diagram on this page. I don't understand why the voltage on the inductor goes to -2.5V when the square wave is at 0V. The square wave was pushing the current in clockwise direction before it jumped to 0V, so the inductor would try maintain the current in the same direction which means the same polarity as the square wave generator. Please help me with it. Thanks.

Page 1: http://forum.allaboutcircuits.com/at...1&d=1319119739
Page 2: http://forum.allaboutcircuits.com/at...1&d=1319119739
Page 3: http://forum.allaboutcircuits.com/at...1&d=1319119739
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  #2  
Old 10-20-2011, 02:32 PM
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Yes, the inductor will try to maintain the current in the same direction, so think about how it would do that. When the inductor is absorbing energy the positive polarity is at the end where the current is entering the inductor (the same as a resistor for example). When the inductor is supplying energy (trying to maintain the current) then the positive polarity has to be at the end where the current is leaving the inductor (like a battery). This means the relative polarity across the inductor becomes negative at the end of the inductor where the current is entering, as you observed.

If the inductor voltage stayed positive where the current is entering the inductor when the square-wave goes to zero, then the inductor voltage would be opposing the current flow, not helping it continue.
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  #3  
Old 10-20-2011, 02:34 PM
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Simply remove square wave source. And know you will see that the coil voltage is pushing the current in clockwise direction.
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Old 10-20-2011, 03:48 PM
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Thank you, Carl, Jony.

Please have a look on the attachment. You can find my two questions there. Please help me with them. Thanks a lot.

Regards
PG
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Old 10-20-2011, 07:29 PM
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The waveforms you are asking about are shown next to your question. Don't understand that question.

Inductor current lags the voltage because that's what inductors do. They resist the flow of current, so cause it to lag. Alternately capacitor current leads the voltage.
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Old 10-21-2011, 02:07 AM
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Quote:
Originally Posted by PG1995 View Post
Thank you, Carl, Jony.

Please have a look on the attachment. You can find my two questions there. Please help me with them. Thanks a lot.

Regards
PG
Hi

I think my question wasn't clear enough. Let me try again.

Please have a look here. It shows the voltage appearing across the inductor for the corresponding voltage of the square wave generator source. So, I was inquiring that if it is possible to have a diagram which shows the AC source voltage for the corresponding voltage appearing across an inductor. Is this clear now? Thanks.

Regards
PG

PS: I though I should ask the question still another way. You can have a look here.
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File Type: jpg relationbetweeninduction.jpg (125.1 KB, 11 views)

Last edited by PG1995; 10-21-2011 at 12:55 PM.
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Old 10-21-2011, 12:56 PM
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Hi

Electric current through a DC RL circuit ("R" stands for resistor and "L" stands for inductor) reaches its maximum value after 5τ (where "τ" is RL time constant and equals L/R). For example, in this RL circuit, when switch is closed, momentarily equal and opposite voltage appears across the inductor which 'pushes' back the pressure exerted by the battery and therefore current is zero. But then the voltage across the inductor starts dying away and current starts increasing. If you an ammeter connected in series in the circuit and a voltmeter connected parallel with the inductor, their readings will reflect this. After 5τ, voltage appearing across the inductor is 0 and current through it is maximum.

But I'm having difficulty conceptualizing the phenomenon when you replace DC with an AC source to create AC RL circuit such as this one.

Here is what I think. The voltage of an AC source is continuously changing values. For example, when voltage starts rising from 0 towards positive peak value, how would inductor behave? In my opinion as the AC voltage builds up (going from 0 towards +ve peak value), so does the voltage appearing across the inductor but this voltage should equalize (which means it cancels the pressure pressure exerted by the AC source) the voltage of the source. But this would mean that there won't be any current in the circuit as voltage goes from 0 towards the positive peak value because the voltage appearing across the inductor is opposite and equal. But once the voltage of AC source has reached the positive peak, it will start declining toward 0. Then, what?! Please help me. Thanks.

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PG
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  #8  
Old 10-21-2011, 01:29 PM
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This may help you:
http://www.falstad.com/circuit/e-induct.html

You will need to have Java installed in order to see the circuit work.
If you do not have Java installed, you can download/install it for free here:
http://www.java.com/en/download/index.jsp

There is a switch at the top center of the circuit. Click on the switch to change the path of current, and see what happens to the voltage/current through the inductor.

The yellow dots are current flow. The faster the yellow dots move, the greater the current. Wire color has meaning; gray is 0v, green is positive, red is negative.

You can change the power supply to something other than AC.
Right-click on the battery on the left. Select Edit, to the right of Waveform, select A/C, then OK.
Make sure that the switch is in the left position. You will then see the waveforms on the bottom display change, and the current flow through the circuit change.

This simulator will help you answer most of your questions, if you experiment with it enough. There are quite a few example circuits, and you can build circuits of your own, and/or modify the example circuits.

If you have modified a circuit and don't know why it does or does not work, then right-click on a black background area, click File -> Export Link, copy the text from the box by pressing Ctrl+C, and post the link on here.
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  #9  
Old 10-21-2011, 01:55 PM
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As you should know the voltage across the coil is proportional to the rate of change in current
V = L * dI/dt
This also mean that to produce voltage across an inductance the applied current must change.


And some further DC analysis
Quote:
Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it?
In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear!
A mysterious electric field somewhere inside the inductor! Where did that come from?
It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’
So let us now try to figure out exactly how the induced voltage behaves when the switch is closed. Looking at the inductor charging phase, the inductor current is initially zero. Thereafter, by closing the switch, we are attempting to cause a sudden change in the current. The induced voltage now steps in to try to keep the current down to its initial value(zero).
So we apply ‘Kirchhoff’s voltage law’ to the closed loop in question. Therefore, at the moment the switch closes, the induced voltage must be exactly equal to the applied voltage, since the voltage drop across the series resistance R is initially zero (by Ohm’s law).
As time progresses, we can think intuitively in terms of the applied voltage “winning.” This causes the current to rise up progressively. But that also causes the voltage drop across R to increase, and so the induced voltage must fall by the same amount (to remain faithful to Kirchhoff’s voltage law).
That tells us exactly what the induced voltage (voltage across inductor) is during the entire switch-closed phase.
Why does the applied voltage “win”? For a moment, let’s suppose it didn’t. That would mean the applied voltage and the induced voltage have managed to completely counter-balance each other — and the current would then remain at zero. However, that cannot be, because zero rate of change in current implies no induced voltage either! In other words, the very existence of induced voltage depends on the fact that current changes, and it must change.
We also observe rather thankfully, that all the laws of nature bear each other out. There is no contradiction whichever way we look at the situation. For example, even though the current in the inductor is subsequently higher, its rate of change is less, and therefore, so is the induced voltage (on the basis of Faraday’s/Lenz’s law). And this “allows” for the additional drop appearing across the resistor, as per Kirchhoff’s voltage law!
And before we proceed to AC analysis kept in mind that when you connect the coil to the DC voltage source, at first step the voltage across the coil reach its maximum value equal to Vsource, but the current is equal zero amps. And when current in the coil reach his maximum value the voltage across the coil reach zero volts.
So we can say that current in the coil lags the voltage.



This imagine will help you understand behavior of the coil using this equation:

V = L * dI/dt

when we replace DC source with AC source.


Last edited by Jony130; 10-21-2011 at 04:32 PM.
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  #10  
Old 10-22-2011, 08:15 AM
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Thank you. But sorry, I'm still stuck.

Could you please tell me if the diagram below is correct? Perhaps, I can derive some understanding from it if it's correct. Thanks.

Here is the diagram.
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