# learning workings of inductor

#### PG1995

Joined Apr 15, 2011
818
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.

#### crutschow

Joined Mar 14, 2008
25,461
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.

• anhnha and PG1995

#### Jony130

Joined Feb 17, 2009
5,183
Simply remove square wave source. And know you will see that the coil voltage is pushing the current in clockwise direction.

• PG1995

#### PG1995

Joined Apr 15, 2011
818
Thank you, Carl, Jony.

Regards
PG

#### crutschow

Joined Mar 14, 2008
25,461

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.

#### PG1995

Joined Apr 15, 2011
818
Thank you, Carl, Jony.

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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#### PG1995

Joined Apr 15, 2011
818
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.

Regards
PG

#### SgtWookie

Joined Jul 17, 2007
22,220

You will need to have Java installed in order to see the circuit work.

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.

• PG1995

#### Jony130

Joined Feb 17, 2009
5,183
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
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:

#### PG1995

Joined Apr 15, 2011
818
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.

#### Jony130

Joined Feb 17, 2009
5,183
How can be any voltage phase-shift in this circuit when we connect the coil directly to the AC voltage source?

#### crutschow

Joined Mar 14, 2008
25,461
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.
Where is Vs applied? You need to show the circuit. Obviously Vs and VL can't be the same voltage.

What is the problem you have with the current lagging the voltage in an inductor? It seems a simple concept. As you apply a sine voltage to the inductor, the current starts to increase. It continues to increase until the voltage returns to zero at 180° and the voltage starts to go negative (since anytime you apply a voltage across an inductor the current will either be increasing or decreasing, depending upon the polarity of the voltage vs. the current direction). At that point the inductor current has reached its maximum value (the 90° point for the inductor current). Thus the inductor current lags the voltage by 90°.

#### PG1995

Joined Apr 15, 2011
818
Thank you, Jony, Carl.

The circuit I had in mind was the attached one. I didn't include a resistor because it was making things complicated to visualize.

So, was the attached diagram in my previous correct for the circuit shown? I would ask my next query after your reply. Thanks a lot for the help.

Regards
PG

#### Jony130

Joined Feb 17, 2009
5,183

#### PG1995

Joined Apr 15, 2011
818
As I said earlier your diagram is completely wrong.
Sorry. I thought perhaps showing you the circuit I had in mind would clarify the confusion. Thanks for the confirmation.

Best wishes
PG

#### PG1995

Joined Apr 15, 2011
818
I request you to keep it simple and please don't use math equations where you could have explained the phenomenon using simple words. Thanks.

When the switch is closed, no current passes through the circuit momentarily but maximum voltage would appear across the inductor. You can say, for a tiny moment, inductor acts as an open circuit.

Q1: Would the voltage appearing across the inductor be equal to Vs (voltage of source) or less? I'm asking this because the circuit also has a resistor which can also take some share of the supply voltage.

Then the voltage starts decreasing and current starts increasing. Magnetic flux would also increase along with increasing current. When voltage across inductor has reached zero value, the current and magnetic flux have reached their maximum values. At this stage, inductor acts a short circuit.

Q2: Is what I say above correct?

The inductor stores its energy in its magnetic flux. Q3: But where did it take this energy from? A resistor dissipates energy by 'extracting' energy out of electrons (electron current) passing through it and as a result electrons lose their potential.

Q4: When the magnetic flux around the inductor was 0 Wb, it was completely opposing the flow of current through it. As magnetic flux starts increasing so does the current. Shouldn't it be so that when magnetic flux is maximum, it then opposes the current fully instead?

Q5: Does the working of inductor have something to do with inertia of electrons? I think it does. Because at the start, when switch is closed, the electrons would resist the change in their state as everything does according to Newton's first law.

Regards
PG

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#### steveb

Joined Jul 3, 2008
2,436
When the switch is closed, no current passes through the circuit momentarily but maximum voltage would appear across the inductor. You can say, for a tiny moment, inductor acts as an open circuit.

Q1: Would the voltage appearing across the inductor be equal to Vs (voltage of source) or less? I'm asking this because the circuit also has a resistor which can also take some share of the supply voltage.
It is equal because there is no current, and hence there is no voltage drop on resistive elements.

Then the voltage starts decreasing and current starts increasing. Magnetic flux would also increase along with increasing current. When voltage across inductor has reached zero value, the current and magnetic flux have reached their maximum values. At this stage, inductor acts a short circuit.

Q2: Is what I say above correct?
Correct

The inductor stores its energy in its magnetic flux. Q3: But where did it take this energy from?
It came from the power supply. The voltage times the current in the power supply represents power it is supplying to the circuit. That power times the time of operation (or more accurately, power integrated over time) is the total supplied energy. That energy needs to go somewhere. Some of it goes away as heat in the resistive elements, but some of it goes into potential energy in the magnetic field of the inductor.

This is similar to me pushing a rock up a hill. The food I eat is the energy source and my body outputs power over time (energy). Some of that energy is wasted as heat (I sweat), and some of that energy goes into the gravitational potential energy of the rock (and my body mass also) which is now at a greater height.

Q4: When the magnetic flux around the inductor was 0 Wb, it was completely opposing the flow of current through it. As magnetic flux starts increasing so does the current. Shouldn't it be so that when magnetic flux is maximum, it then opposes the current fully instead?
The amount of current flowing has nothing to do with the opposition of current change. Note that the formula is V=L dI/dt. The voltage determines how fast current can change, or the rate of change of current determines the voltage. However, the value of current itself does not show up in this formula, but only the rate of change is there.

Q5: Does the working of inductor have something to do with inertia of electrons? I think it does. Because at the start, when switch is closed, the electrons would resist the change in their state as everything does according to Newton's first law.
No, forget about inertial of electrons in this context. The relevant Law of Nature here is Faraday's Law, which is very different from Newton's law. There is an underlying commonality which relates to the conservation of energy, but all laws of nature obey conservation of energy, so that doesn't help too much.

• PG1995

#### PG1995

Joined Apr 15, 2011
818
Thanks a lot, Steve.

Please have a look on the attachment. You can find my query there along with other information. Please help me with it. Thank you.

Regards
PG

#### steveb

Joined Jul 3, 2008
2,436
Please have a look on the attachment. You can find my query there along with other information.
Here, you have to be careful with fundamental definitions, and the application of calculus to calculate physically meaningful numbers.

You start off correctly by noting that energy is power integrated over time, and power is voltage times current. This is represented mathematically as follows.

$$E=\int_0^t P(\tau) {\rm d}\tau=\int_0^t V(\tau) I(\tau){\rm d}\tau$$

This is different than what you wrote. There is actually a problem with what you wrote because your formula does not even have the correct units for energy. It actually has units of power. Even if you correct this, by multiplying by time, your formula is only valid when both voltage and current are constants. In that case average values of voltage and current are equal to the voltage and current, so there is no problem. However, in general you can't find average power by multiplying average voltage times average current because the power calculation is a nonlinear formula. There are times when you can make this calculation as an approximation, but you have to be very careful.

Still, you don't even need to take this approach at all for your question. The energy formula for the coil is $$E={{1}\over{2}}L (I(t))^2$$ and it tells you the magnetic potential energy at any point in time, directly from the instantaneous value of the current at that time. Note that no integration is needed at all. So, curve 2 is the relevant curve, but the area B is not needed.

By the way, what is the integral of the coil voltage over time? It is the total magnetic flux. This is analogous to finding the total charge in a capacitor plate by integrating current over time.

$$F=\int_0^t V(\tau) {\rm d}\tau=\int_0^t V(\tau) I(\tau){\rm d}\tau$$

$$Q=\int_0^t I(\tau) {\rm d}\tau=\int_0^t V(\tau) I(\tau){\rm d}\tau$$

These formulas are a direct result of the definitions.

V=dF/dt and I=dQ/dt

• PG1995

#### PG1995

Joined Apr 15, 2011
818
Once again, I offer my thanks, Steve.

Please have a look on the attachment. If there is any confusion, please let me know so that I can clarify it. You might also need to have a look on this diagram from my previous post. Please help me. Thanks.

With best regards
PG