learning basic workings of inductor

[PG1995]

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 **broken link removed** 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

 

[Ratchit]

PG,

appears across the inductor which 'pushes' back the pressure exerted by the battery and therefore current is zero.

Batteries exert voltage, not "pressure".

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?

"Behave" with respect to what?

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?!

Both the current and voltage in a RL circuit are sinusoidal, which means they have max/min, and zero values, but not at the same time. The coil takes energy from the circuit to build up its magnetic field, and releases the energy back to the circuit when the field collapes. This build up and release takes time, so the current does not track the voltage like it does in a purely resistive circuit. Voltage in a coil is -L*di/dt. When the current is a sin, the voltage is -cos. Since -cos is 90° behind the sin, the current lags the voltage. -cos(x)=sin(x-90°)

Ratch

 

[MrAl]

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 **broken link removed** 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

Hi,


Another way of looking at this is as follows...

A sine wave is like this:
sin(w*t)

but a sine wave with a phase shift is like this:
sin(w*t+TH)

and TH stands for 'theta' which is the phase shift in radians.

Now it is helpful to compare this sin(w*t+TH) with just sin(w*t).
TH can be either positive or negative. If TH is positive it means that the instantaneous voltage levels in sin(w*t+TH) occur BEFORE the levels in sin(w*t). That means with TH positive sin(w*t+TH) LEADS sin(w*t), or alternately we can say that sin(w*t) LAGS sin(w*t+TH). With TH negative it's the opposite.
On a scope picture we would see sin(w*t+TH) go through zero and start rising more positive, and we would see sin(w*t) negative and starting to rise up toward zero.

Now we have the defining equation for the inductor:
V=L*di/dt

and since the current is sin(w*t) we have:
V=L*d(sin(w*t))/dt

Taking the derivative we get:
V=L*w*cos(w*t)

and cos is phase shifted from sin, and also cos(w*t) can be written as:
cos(w*t)=sin(w*t+pi/2)

And so comparing sin(w*t) to sin(w*t+pi/2) as we did above, we see that cos is a phase shifted sine wave that is LEADING sin(w*t).

So to sum up, since cosine is a phase shifted sine wave that LEADS the non shifted sine wave, and the cosine wave is the voltage and the sine wave is the current, we can say that either voltage leads the current or current lags the voltage. If the resistance is zero, then the phase shift is 90 degrees.

 

[PG1995]

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.

This is the diagram.

 

[MrAl]

Hi PG,

There is something wrong with that diagram. The lower two waveforms are correct for the inductor, but the upper waveform seems to have nothing to do with it unless the inductor is in some kind of circuit, in which case you'll have to show us the circuit.

The lower two waveforms show what would be the voltage across the inductor and the current through the inductor if the source voltage was applied directly across the inductor and the 'scope' was synced to the current waveform at the zero crossing.

 

[PG1995]

Sorry for the belated thanks, Ratch, MrAl. I was them for today!

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.

Please help me with the above queries so that I can conceptual understand the inductor.

Regards
PG

 

[MrAl]

Hi,


When the voltage is first applied, the inductor looks like an open circuit so the current is zero, but the derivative could be non zero at 0+.
When you say "short circuit" you have to specify what it is a short circuit to, DC or AC. It's not a short to AC but it is a short to DC. But for the circuit shown you would think of it as shorted out so that you can calculate the final current level.

The simplest expression of the inductor is:
v=L*di/dt

so you dont have to worry about the flux and all that. The energy comes from photons which the field is said to be made up of. The moving electrons can be said to produce photons, but in fact it is really the photons moving the electrons. The total energy stored is 1/2 L*i^2.

 

[PG1995]

Thanks a lot, MrAl.

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

 

[MrAl]

Hi,

That voltage curve is not drawn correctly.
Also, energy is not consumed by the inductor it is stored.

If you integrate the voltage over time, multiply by the current, then divide by 2, you'll get the energy at any time t.
You dont have to do that though. You can just take the current and square it, divide that by 2, then multiply by the inductance.