Charging of Inductors: Explained Without the Duality

[The Electrician]

(In other words,I can't picture what happens in the first few instants after the voltage ha been applied till the establishment of current.)

Do you understand how it is possible for the current to be zero, but at the exact same instant of time the rate of change of the current can be non-zero?

If the current begins a linear ramp up of current, starting out at zero amps at time tzero, the "back EMF" can be non-zero even if the current is zero.

The value of the current and the value of its rate of change don't have to be the same. This is a critical thing for you to understand. It is at the root of your puzzlement.

 

[Urmi Roy]

...just figuring things out...by the way,the value of Ldi/dt doesn't have to be equal to the applied voltage at that instant,does it?(if it was,there would never be any current during the charging of inductor in dc circuit...but there is..the only effect if inductor here is to inhibit the current,not to cancel it

This analysis,after what The Electrician said in the last post and readind the page I referred to,seems similar to the charging of inductor in a dc circuit...except here,the voltage is 'switched on and off' repeatedly.

 

[The Electrician]

...just figuring things out...by the way,the value of Ldi/dt doesn't have to be equal to the applied voltage at that instant,does it?(if it was,there would never be any current during the charging of inductor in dc circuit...but there is..the only effect if inductor here is to inhibit the current,not to cancel it

In fact, L*di/dt is always equal to the voltage impressed across the inductor. If there is no resistance in the circuit, then L*di/dt is a constant if a constant DC voltage is applied to the inductor. In that case, the current in the inductor increases in a linear ramp forever.

One of the quotes you gave from the AAC book, says:

""If a source of electric power is suddenly applied to an unmagnetized inductor, the inductor will initially resist the flow of electrons by dropping the full voltage of the source. As current begins to increase, a stronger and stronger magnetic field will be created, absorbing energy from the source. Eventually the current reaches a maximum level, and stops increasing."

The lead-up to this doesn't explicitly say that the example involves a circuit with no resistance at all, but just prior they did give an example of an inductor shorted on itself with superconducting wire. They are aware that current would circulate forever.

If an ideal DC voltage source (one with zero internal resistance and that can supply infinite current) is applied to an inductor with no internal resistance, and if there is no other resistance in the circuit, it's not true that "Eventually the current reaches a maximum level, and stops increasing." This only happens if there is resistance in the circuit somewhere.

If there is a resistor in series with the inductor, then it's still true that the voltage applied to the inductor is equal to L*di/dt at all times. But in this case, as the current in the inductor increases, the resistor drops more and more of the voltage, so that the voltage left to be applied to the inductor is less, and the current approaches a limit, a "maximum level".

At tzero, the resistor drops no voltage, because the current is still zero, and the total source voltage is applied to the inductor, and L*di/dt is exactly equal to the applied voltage; the slope of the current is initially the same as it is when there is no resistor. But the slope immediately begins decreasing, and the current waveform in this case is the usual exponential approaching a limit.

Rather than trying to make sense of the point of view that the "back EMF" is opposing the applied voltage and therefore determining the inductor current, it's better to take the current as the thing which determines the "back EMF".

The current assumes that value that just makes the "back EMF" equal to the applied voltage. The current is proportional to the integral of the applied voltage, and assumes the exact value needed to produce a "back EMF" equal to the applied voltage. This is because saying that the current is proportional to the integral of the applied voltage is the same thing as saying that the "back EMF" of the inductor is proportional to the di/dt of the inductor current.

 

[Urmi Roy]

In fact, L*di/dt is always equal to the voltage impressed across the inductor. If there is no resistance in the circuit, then L*di/dt is a constant if a constant DC voltage is applied to the inductor. In that case, the current in the inductor increases in a linear ramp forever.

The fact that the "current can increase for ever in the absence of a resistance" must be due to the fact that the zero resistance of the circuit can ideally allow infinite current...thus,even if we have a fixed potental applied,to the circuit,the current may be infinite..like in superconductors...right?

One of the quotes you gave from the AAC book, says:

If there is a resistor in series with the inductor, then it's still true that the voltage applied to the inductor is equal to L*di/dt at all times. But in this case, as the current in the inductor increases, the resistor drops more and more of the voltage, so that the voltage left to be applied to the inductor is less, and the current approaches a limit, a "maximum level".

That means since the voltage across the ideal inductor decreases,this inductor,with zero resistance cannot allow for infinite current anymore...and the final current is determined but the resistance as per Ohm's law.

At tzero, the resistor drops no voltage, because the current is still zero, and the total source voltage is applied to the inductor, and L*di/dt is exactly equal to the applied voltage; the slope of the current is initially the same as it is when there is no resistor. But the slope immediately begins decreasing, and the current waveform in this case is the usual exponential approaching a limit.

This decrease in slope is due to the limiting effect of the reseistance..right?

Rather than trying to make sense of the point of view that the "back EMF" is opposing the applied voltage and therefore determining the inductor current, it's better to take the current as the thing which determines the "back EMF".

We could put it like this,perhaps...the applied voltage causes the current to increase from zero independantly (meaning the role of applied voltage is not affected by anything else)...and when this current reaches the inductor,it produces a magnetic field...at the same time,the current has a certain instantaneous rate of change (di/dt),which determines the back emf of the inductor...cause and effect sort of thing.

 

[Urmi Roy]

Tell me one thing,In the equation for an ideal inductor woth no resistance,

V is proportional to the rate of change of current.

In a circuit with only a resistance,

the applied voltage is proportional to the current produced...

the fact that the voltage is proportional to the rate of change of current...is it due to the fact that no resistance can allow for the current to reach infinite value,so appliying even a constant voltage means that the current is changing all the time.

This could explain the fact that the current at any time is proportional to the integral of V...it would mean that the current at the moment is the result of the total work done by the voltage source over a period of time.

In other words,in an inductor,work is done to increase the current and not to overcome the resistance (as in a resistance)...and the current established at any time is the result of the total work done over some time.

 

[The Electrician]

...the applied voltage causes the current to increase from zero independantly (meaning the role of applied voltage is not affected by anything else)...and when this current reaches the inductor,it produces a magnetic field...at the same time,the current has a certain instantaneous rate of change (di/dt),which determines the back emf of the inductor...cause and effect sort of thing.

Delays in the current flow out of the source are of the order of nanoseconds, and such delays play no part in ordinary network analysis. They are not considered at all in this context.

The applied voltage will not cause any current unless the inductor is connected to it. There's no such thing as a delay until the "...current reaches the inductor...". Current doesn't begin to flow out of the wire connected to the voltage source independently of the inductor, and after a delay, "...reaches the inductor...". Current won't flow out of the source unless and until the inductor is connected to it.

 

[The Electrician]

Tell me one thing,In the equation for an ideal inductor woth no resistance,

V is proportional to the rate of change of current.

In a circuit with only a resistance,

the applied voltage is proportional to the current produced...

the fact that the voltage is proportional to the rate of change of current...is it due to the fact that no resistance can allow for the current to reach infinite value,so appliying even a constant voltage means that the current is changing all the time.

The fact that the current is unbounded with time doesn't necessarily imply that the voltage is proportional to di/dt; it could be proportional to SQRT(di/dt), or some other functional relationship.

The terminal characteristic of an inductor is v(t) = L*di/dt and that implies that if a constant voltage is applied, the current will increase without limit. The fact that the current will increase without limit doesn't imply any particular functional relationship.

This could explain the fact that the current at any time is proportional to the integral of V

As I said, the fact that the current is unbounded with the application of a constant voltage doesn't necessarily imply that v(t) = L*di/dt; it could happen with other functional relationships.

...it would mean that the current at the moment is the result of the total work done by the voltage source over a period of time.

In other words,in an inductor,work is done to increase the current and not to overcome the resistance (as in a resistance)...and the current established at any time is the result of the total work done over some time.

But, as it happens, the current is proportional to the integral of the applied voltage, and this does mean that knowing the current at any time, we can calculate the total work done; it's work = L*i(t)^2/2, energy stored in the inductor.

Everything you need to know about inductors is contained in the terminal characteristic:

v(t) = L*di/dt

Trying to find reasons why this is so is to enter the realm of electronic philosophy, which I generally find not to be helpful.

Better to just accept that v(t) = L*di/dt and learn how circuit behavior follows inevitably from that.

 

[Urmi Roy]

The fact that V proportional to di/dt,in any way, does seem to say that 'the appiled voltage causes a change in current' (putting it simply,without going in too much into the reasons,as you advised)..even if the functional relation had been different,there would still have been a direct relation between the voltage and current,implying that their existence depend directly on each other ...similarly,in a pure resistor,the fact that voltage proportional to current does seem to say that the voltage causes current...if I put it this way,somehow it becomes easy for me to visualise.

If this this is not entirely wrong,I think I might have proceeded to understand inductors,a little bit.

Also,after reading the article in AAC,I formed a better understanding of how ain inductor behaves in a plain dc circuit ...and using what you said,about the current in a purely inductive circuit increasing forever for a fixed voltage to fine-tune the concepts provided by AAC,...I think I could imagine a purely inductive ac circuit as a dc circuit being turned on and off successively (in a fancy sinusoidal way)...do you think this okay?

 

[b.shahvir]

b Shahvir,I read through your post on the other thread...please clarify some points about it...



‘t (time)’ > 0 towards it’s peak value, the rate of change of magnetic flux becomes more pronounced ad the current wave now starts going out of phase w.r.t the voltage wave.
...why does the rate of change of magnetic field become more pronounced as soon as it becomes t>0...like at t+dt?

If you concentrate on the negative half cycle of the voltage vs. current waveform, you will find that the phase shifting is due to the ‘discharge’ of electrical energy by the inductor back to the power source as the magnetic field collapses (or rises, whatever may be the case), which precisely occurs when the input voltage wave has already reversed itself. ...does that mean the discharge of the inductor occurs only when the applied voltage reverses itself?

This charge/discharge action of an inductor occurs due to the finite time taken my the magnetic field to rise or collapse……as force fields cannot change it’s states abruptly at ‘t = 0’ as per the laws of Physics ...PLease explain this in a little more detail...I'm finding it difficult to visualise,even after reading part B of your post.

Dear Urmi,

As I had mentioned earlier, it is quite difficult to visualize working of an inductor as the parameters involved are magnetic field & current flow (unlike capacitors which involve accumulation of charge and voltage!)

It would be easy if you mention (in brief) exactly what factors about inductor behavior makes you uneasy the most...or else the thread would go on till infinity.

In passing i wud again mention, that if a pure sinusoidal voltage is applied to an 'ideal' inductor, there will be absolutely no phase difference or delay between applied volts and current as at t = 0, the current is just starting to flow and would be negligibly small to cause any significant creation of magnetic field. Also at t = 0, both applied volts and current would be immediately in phase as enough magnetic field is not developed to cause 'magnetic inertia.' As time (and hence voltage wave progresses further, the magnitude of voltage wave increases causing more current to flow, which in turn now causes significant increase in magnetic field) This increase in magnetic field creates back emf opposing further flow of source current causing cyclic current charge discharge process(both in positive and negative halves just like in an ideal capacitor)

Now, this cyclic rhythm slowly progresses with time, eventually forcing a complete 90 deg phase shift (delay, in layman terms) between the applied voltage and current waves.

As for part B, I suggest you go thru it once more slowly and you might understand...although I must state that my post might not be scientifically accurate or relevant as this is based on personal way of understanding a theoretical concept. Actually I've attempted to suggest that change of state cannot happen at t = 0, but requires a finite elapse of time i.e at t > 0, etc.

Pls. feel to discuss further if reqd.

Regards,
Shahvir

 

[Urmi Roy]

As I said, I seemed to have reached near a conclusion,especially by comparing what happens in a dc circuit with only inductor while switching on and off (I used the word seems,as I am not sure of anything yet!)...please refer to post 43.

 

[Urmi Roy]

I think I've found an analogy to explain why the inductor's back emf does not cancel the effect of the applied voltage at any instant...suppose we have an elastic band,and we attach one end to a wall...we pull at it and at any instant,the applied force is equal to the resisting force in the band.
However,the work done by the applied force to increase the length of the band is slowly getting stored in the band...which gets released as soon as we withdraw the applied force...in the same way,then the applied voltage causes an increase in current,the back emf is always opposing this,but the applied voltage,in the course of time is doing work in buiding a current...and as soon as we withdraw the applied voltage (reduce it to zero),the accumulated work,in the form of current flows back to us.

Is that okay?

 

[The Electrician]

I think I've found an analogy to explain why the inductor's back emf does not cancel the effect of the applied voltage at any instant...suppose we have an elastic band,and we attach one end to a wall...we pull at it and at any instant,the applied force is equal to the resisting force in the band.
However,the work done by the applied force to increase the length of the band is slowly getting stored in the band...which gets released as soon as we withdraw the applied force...in the same way,then the applied voltage causes an increase in current,the back emf is always opposing this,but the applied voltage,in the course of time is doing work in buiding a current...and as soon as we withdraw the applied voltage (reduce it to zero),the accumulated work,in the form of current flows back to us.

Is that okay?

Mechanical analogs can be helpful in understanding electrical components. There are two standard analog systems. See:

http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/Analogs/ElectricalMechanicalAnalogs.html

 

[b.shahvir]

I think I've found an analogy to explain why the inductor's back emf does not cancel the effect of the applied voltage at any instant...suppose we have an elastic band,and we attach one end to a wall...we pull at it and at any instant,the applied force is equal to the resisting force in the band.
However,the work done by the applied force to increase the length of the band is slowly getting stored in the band...which gets released as soon as we withdraw the applied force...in the same way,then the applied voltage causes an increase in current,the back emf is always opposing this,but the applied voltage,in the course of time is doing work in buiding a current...and as soon as we withdraw the applied voltage (reduce it to zero),the accumulated work,in the form of current flows back to us.

Is that okay?

I think this is a good analogy ...but then again, understanding goes as far as the power of an individual's imagination. This being especially true for electrical technology!

 

[Urmi Roy]

I think this is a good analogy ...but then again, understanding goes as far as the power of an individual's imagination. This being especially true for electrical technology!

That's good!

Noone said anything about post 43...

 

[b.shahvir]

The fact that V proportional to di/dt,in any way, does seem to say that 'the appiled voltage causes a change in current' (putting it simply,without going in too much into the reasons,as you advised)..even if the functional relation had been different,there would still have been a direct relation between the voltage and current,implying that their existence depend directly on each other ...similarly,in a pure resistor,the fact that voltage proportional to current does seem to say that the voltage causes current...if I put it this way,somehow it becomes easy for me to visualise.

If this this is not entirely wrong,I think I might have proceeded to understand inductors,a little bit.

Also,after reading the article in AAC,I formed a better understanding of how ain inductor behaves in a plain dc circuit ...and using what you said,about the current in a purely inductive circuit increasing forever for a fixed voltage to fine-tune the concepts provided by AAC,...I think I could imagine a purely inductive ac circuit as a dc circuit being turned on and off successively (in a fancy sinusoidal way)...do you think this okay?

Yes this is OK. However, sinusoids are cyclic waves causing continuous voltage to be impressed across the inductor whenever flow of electrons (current) is persistantly opposed by back emf developed in the inductor during every cyclic variation of voltage, current and magnetic field.

 

[The Electrician]

Also,after reading the article in AAC,I formed a better understanding of how ain inductor behaves in a plain dc circuit ...and using what you said,about the current in a purely inductive circuit increasing forever for a fixed voltage to fine-tune the concepts provided by AAC,...I think I could imagine a purely inductive ac circuit as a dc circuit being turned on and off successively (in a fancy sinusoidal way)...do you think this okay?

I'm not sure what you're asking. You say "I think I could imagine a purely inductive ac circuit as a dc circuit being turned on and off successively (in a fancy sinusoidal way)", and then you ask "do you think this okay?". Do I think it's okay in what sense?

You can imagine any voltage waveform you wish, applied to an inductor, and you can solve the differential equation for the current waveform. I suppose that would be okay, wouldn't it?

 

[Urmi Roy]

I'm not sure what you're asking. You say "I think I could imagine a purely inductive ac circuit as a dc circuit being turned on and off successively (in a fancy sinusoidal way)", and then you ask "do you think this okay?". Do I think it's okay in what sense?

Actually,the understanding I have formed of inductor brhaviour in ac is only after I first got the behaviour of inductors in dc circuit clear in my head..in which AAC helped.
Then I realized that this could be explained by a dc circuit being turned on and off...as an analogy.

I just wanted to confirm if this view was consistent with what actually happens in ac circuit...if there were any considerations I had left out.

 

[The Electrician]

When you say "a dc circuit being turned on and off successively", that sounds like what could be described as a "unipolar square wave". That means that for a repetitive interval the voltage is some constant value, Vdc, and then for a interval it's zero, then Vdc again, ad infinitum.

Such a wave form has a DC component, which regular sinusoidal AC does not.

The current produced in an inductor by the waveform you propose would consist of a series of ramps (after the decay of a possible transient component) when the voltage is non-zero, interspersed with a series of constant current segments. But, the current would still increase without limit in the long term.

It's not the same as ordinary AC, which has no DC component, applied to an inductor.

If you applied a "bipolar square wave", where the applied voltage was Vdc for a repetitive interval, and -Vdc for equal invervals, then the current in the inductor would consist of a triangle wave, with possibly a decaying exponential transient component, depending on just where in the cycle you applied the bipolar square wave.

This last waveform is more consistent with what happens in a circuit with an ordinary sinusoidal AC waveform applied.

I'm not sure if all this helps your understanding of what happens in an AC circuit, but let's hope it does.

 

[Urmi Roy]

Supposing I've at last understood something of inductors,I'd like to come to 'reactance of inductors'...without which the topic is incomplete.

I've spent the last few days wondering and analysing this...

First,if we start off with the all famous equation:
Ldi/dt = V,
we have Ldi=Vdt
=> LI=integral of V over the time since the voltage was applied to present moment,

If V(t)=Vsinwt (sinusoidal),
as a result of integration method,we get current I(t)=(V/wL)sin(wt-90)
Here,I is the current that has developed over the time,since the voltage was applied.

Now, if we look at Ohm's law,V=IR,and compare with the above eauqtion,it seems that the wL plays the same role in inductors as R does in simple resistances.

 

[Urmi Roy]

The point is,I don't think the wL actually plays the same role as the resistance,and therefore is not the analogue of resistance because:
1. In Ohm's law,the voltage directly determines the current ,whereas in an inductor,the voltage determines the rate of change of current,so we can't just say the current due to the voltage is reduced by a factor of wL,like we would in case of a resistance.

2. wL is just a result of integration...does it have to have a physical existence?

3. The inductor doesn't exactly oppose the current (on the long run)...since,as The Electrician said,for a pure inductor with a dc supply attached,the current would increase forever.

Also,some queries in regard to impedance...
1.the wL is supposed to determine the amount of power loss there is in the circuit due to the 'storage of energy' in the inductor...but in an ac circuit,the energy is continuously taken and given back by the inductor...so really,there isn't any loss at all.

2.The reactive power due to the inductor is supposed to be VIsin(phi),where Vand I are the RMS values...why do we take RMS values (on the long run,power loss is zero,so even if we do get a certain value of RMS,it wouldn't make sense.)

3. I don't understand why we assign the 'sin(phi) ' part of the expression...as if the reactive power is a component of total power (it might make sense for the phasor representation,but it doesn't make sense physically,it seems).

 

[b.shahvir]

3. I don't understand why we assign the 'sin(phi) ' part of the expression...as if the reactive power is a component of total power (it might make sense for the phasor representation,but it doesn't make sense physically,it seems).

Actually, here total power might mean 'Apparent power' and not necessarily only 'wattful' or 'active' power. Since reactive power is a component of Apparent power (total power) and reactive amps are ideally 90 deg out of phase hence representation by sin(phi).

 

[Urmi Roy]

Actually, here total power might mean 'Apparent power' and not necessarily only 'wattful' or 'active' power.

I understand that it's the apparent power that is mentioned in the formula...but again,the word 'component' may be meaningful in case of vectors and phasors,which have directions...but in reality..or in the actual physical world,the current,power etc. can't have directions!

Also,could you have a look at the rest of my questions?

 

[b.shahvir]

I understand that it's the apparent power that is mentioned in the formula...but again,the word 'component' may be meaningful in case of vectors and phasors,which have directions...but in reality..or in the actual physical world,the current,power etc. can't have directions!

Also,could you have a look at the rest of my questions?

Current has magnitude as well as direction.
Could you pls. ennumerate in brief the points you are after.

 

[Urmi Roy]

Let me first start off with my basic problem:

In an inductor,there is no loss of power on the long run...whatever it takes away,it gives back.
Now,still,we define something called the 'reactive power',which is VIsin(phi)...but I just can't make sense of why we consider the RMS voltage and current,when there is,infact no loss of energy.

Also,it is said that the RMS voltage and current across the inductor and resistor in a RL ac circuit are equal...how?
The voltage and current in an inductor and resistor cannot be equal simultaneously,can they?

 

[The Electrician]

Let me first start off with my basic problem:

In an inductor,there is no loss of power on the long run...whatever it takes away,it gives back.
Now,still,we define something called the 'reactive power',which is VIsin(phi)...but I just can't make sense of why we consider the RMS voltage and current,when there is,infact no loss of energy.

First of all, let's talk only about single frequency sinusoidal waves.

Since we human beings usually want to do something with voltages and currents, there will ultimately be some resistance or load somewhere that will dissipate power. If we use RMS quantities then our computations are simpler when the usage of power comes into play.

Otherwise, you could use any kind of measure: peak-to-peak, peak, average.

The reactance of an inductor, wL, is just the ratio of the voltage to the current at a frequency, and does not depend on the kind of measure.

Also,it is said that the RMS voltage and current across the inductor and resistor in a RL ac circuit are equal...how?
The voltage and current in an inductor and resistor cannot be equal simultaneously,can they?

 

[Urmi Roy]

The fact that the rms of the voltage and current through an inductor are considered must imply that we are talking about one particular half cycle or something...since after the second half,when the entire energy is delivered back,the measure does not have any significance...considering it over a half cycle will allow us,perhaps to calculate the thickness of the wire needed to carry that current,during that time.

Also,'The Electrician',could you please throw some light on the analogy between resistance and reactance that I referred to in post 55?

The reactance of an inductor, wL, is just the ratio of the voltage to the current at a frequency, and does not depend on the kind of measure.

Does the bolded part have any special implication?

 

[The Electrician]

The fact that the rms of the voltage and current through an inductor are considered must imply that we are talking about one particular half cycle or something...since after the second half,when the entire energy is delivered back,the measure does not have any significance

This is not so. The RMS value of a voltage of current is a measure used when we're talking about a steady state situation. It doesn't apply to transient currents such as were discussed earlier in this thread. But if you say that the voltage at the wall socket is 120 VAC (in the U.S.), that means the RMS value is 120 volts, whether for just one full cycle, or many cycles. When you calculate the RMS value of a waveform, you must make the calculation over at least one full cycle. See:

http://www.ee.unb.ca/tervo/ee2791/vrms.htm

It has nothing to do with the direction of energy flow. Suppose you look at the voltage waveform applied to some component, a resistor, a capacitor or an inductor, but you don't look at the current. Assume you don't even know what kind of component you're using. The direction of energy flow depends on both the voltage and current, but you don't know the current (by my hypothesis), so you don't know anything about energy flows.

You can still calculate (or measure) the RMS value of the voltage even though you don't know what kind of component the voltage is applied to, or what kind of energy flows may be taking place.

Also,'The Electrician',could you please throw some light on the analogy between resistance and reactance that I referred to in post 55?

Reactance plays the same role as resistance in the following sense: If you apply a single frequency sine wave of voltage to an inductor, a very specific current will flow. The current will not be infinite as it would if the applied voltage were DC. The inductor will limit the current to a finite value. The ratio of the voltage to the current will be E/I = wL, the reactance of the inductor at the operating frequency.

When you apply a voltage to a resistor, DC or AC, the resistor limits the current, and the ratio of the voltage to the current is E/I = R. R is completely analogous to resistance; it expresses the opposition to the flow of current.

It doesn't matter if there is temporary energy storage or not. All that matters to make the analogy between resistance and reactance in AC circuits is that, when you apply a voltage to a component, the magnitude of the current is determined by the component in some manner, whether by dissipation of energy, or by back and forth exchange of energy.

Does the bolded part have any special implication?

The implication is just what I said in the previous sentence. "Kind of measure" means the method of measurement of a current or voltage. The three I mentioned were peak-to-peak, peak and average. For the measurement of AC quantities, strictly speaking the average of a sine wave is zero. When, for example, a meter is referred to as average responding (rather than RMS responding), it means to rectify the AC and then take the average.

If you apply a 120 volt peak-to-peak sine wave to an inductor, and measure the current as a peak-to-peak quantity, the ratio of the peak-to-peak voltage to the peak-to-peak current will be wL, as it will be if you measure as a peak voltage, or an average rectified voltage.

The ratio of voltage to current (for a single frequency sine wave) will always be wL, if you use the same measurement units for both voltage and current.

I explained the reason for preferring RMS in post #60.

 

[Urmi Roy]

From your explanation about the reactance and resistance,what particularly struck me was :"The current will not be infinite as it would if the applied voltage were DC. The inductor will limit the current to a finite value."

So though reactance is not exactly similar to resistance,for our mathematical convenience,we define (omega)L as reactance and say it's similar to resistance...right?

Also,as I said earlier,there is a sin(phi) when we define the reactive power in terms of VIZ,where Z =impedance,Iand V are RMS values...firstly,I presume this sin(phi) is a fixed value,and not changing (like what we usually do in analysis of RL/Rc/RLC circuits,where the arrow rotates)...

secondly,the fact that the quantity impedance being defined as the (square root of the (reactance squared + resistance squared))...is it just a ,mathematical convenience? Afterall,the currents,reactances,resistances don't really have directions and obey pythagora's theorem,right?

 

[The Electrician]

From your explanation about the reactance and resistance,what particularly struck me was :"The current will not be infinite as it would if the applied voltage were DC. The inductor will limit the current to a finite value."

So though reactance is not exactly similar to resistance,for our mathematical convenience,we define (omega)L as reactance and say it's similar to resistance...right?

I think it may be of help to you to read this:


http://zrno.fsb.hr/katedra/download/materijali/966.pdf

and any other references you can find dealing with "the impedance concept". The point of the impedance concept is that in every situation where there are physical quantities analogous to "pressure" and "flow", there is inevitably a relationship between the two which is an "impedance".

Resistance is not "exactly" similar to reactance in every respect, but in the single respect that it determines the relationship between voltage and current in a two terminal device, the similarity is exact.

Also,as I said earlier,there is a sin(phi) when we define the reactive power in terms of VIZ,where Z =impedance,Iand V are RMS values...firstly,I presume this sin(phi) is a fixed value,and not changing (like what we usually do in analysis of RL/Rc/RLC circuits,where the arrow rotates)...

secondly,the fact that the quantity impedance being defined as the (square root of the (reactance squared + resistance squared))...is it just a ,mathematical convenience? Afterall,the currents,reactances,resistances don't really have directions and obey pythagora's theorem,right?

Have you ever seen any explanation of impedance where it was claimed that "...currents,reactances,resistances..." have directions? I think you're overanalyzing.

Complex numbers can be represented as directed line segments (vectors) on the two dimensional plane. The arithmetic of vectors is the same as the arithmetic of complex numbers in the essentials that relate to analysis of electric networks. Often having an analogous representation can assist understanding, but that doesn't mean that the two things are the same. They're just analogous in some aspects, some behaviors.

The earlier reference I gave mentions Steinmetz's discovery that the differential equations of circuit theory could be solved with simple algebra of complex numbers. This doesn't mean that currents in inductors are imaginary.

It's more than just a mathematical convenience; it's a necessity for solving a circuit. The solution of networks is an excellent example of applied mathematics. A mathematical process that behaves the same way as some physical quantities is discovered, which can then can be used to determine the behavior of the physical system. As I said, this is more than a convenience; it's a necessity to analyze and design circuits.

 

[Urmi Roy]

The page isn't opening...apparently,it's damaged...is there any similar page?

Anyway,from your post,the ethos seems to be that the issue of 'impedance' is an essential mathematical tool.

Please confirm one vital thing for me...
We say that the rms current I is equal through both the resistor and the inductor...
Then, Vrms=Vr+Vl (Vr and Vl are the potential drops across the resistor and inductor at any instant respectively.)
Vrms=Ir+jIX
then the magnitude of Vrms= root over((Ir)^2 + (IX)^2)
...1.the fact that I is considered equal for both the components must be due to the fact that over time,the net charge transferred through the inductor and resistor are the same...

2..and the fact that the voltage across the inductor and resistor are always at a phase difference of 90deg,so averaged over time,the Vrms arrow must be represented as the hypotenuse of the right angeled triangle.
Sorry if this is getting too frustrating...

 

[The Electrician]

The page isn't opening...apparently,it's damaged...is there any similar page?

Back up a little and look on this page: http://zrno.fsb.hr/katedra/download/materijali/

Download the file "966.pdf"

Anyway,from your post,the ethos seems to be that the issue of 'impedance' is an essential mathematical tool.

Please confirm one vital thing for me...
We say that the rms current I is equal through both the resistor and the inductor...
Then, Vrms=Vr+Vl (Vr and Vl are the potential drops across the resistor and inductor at any instant respectively.)
Vrms=Ir+jIX
then the magnitude of Vrms= root over((Ir)^2 + (IX)^2)

You say: "Vrms=Vr+Vl (Vr and Vl are the potential drops across the resistor and inductor at any instant respectively.)"

I assume that you mean for Vrms to an applied voltage across a series combination of a resistor and an inductor. The way you've described it, Vr and Vl are the instantaneous voltages across each component. If that is so, then Vrms is not equal to Vr+Vl; Vrms is a kind of average, not equal to a sum of instantaneous voltages.

Your further say: "Vrms=Ir+jIX"

This is not so. When we speak of an RMS voltage or current, we denote a magnitude. Anytime you say Vrms, it's a magnitude. In your very next sentence: "then the magnitude of Vrms= root over((Ir)^2 + (IX)^2)" you show the proper way to find the resultant of the two individual voltage drops; a simple sum is not correct.

...1.the fact that I is considered equal for both the components must be due to the fact that over time,the net charge transferred through the inductor and resistor are the same...

Well, of course. For two components in series, the current through each is not just "considered" equal (for some mathematical convenience); it is physically identical in each component, which is indeed the same thing as saying "over time,the net charge transferred through the inductor and resistor are the same"

2..[/B]and the fact that the voltage across the inductor and resistor are always at a phase difference of 90deg,so averaged over time,the Vrms arrow must be represented as the hypotenuse of the right angeled triangle.
Sorry if this is getting too frustrating...

This description of a "phase difference of 90 degrees" is only relevant in the steady state with a single frequency sine wave applied to the series combination. A non-sinusoidal voltage waveform still has a well-defined RMS value. To get the current in the non-sinusoidal case, you have to solve the circuit differential equation for the applied voltage, and the applied RMS voltage is not related to the individual voltages according to a single right triangle analogy.

But, yes, for a single frequency sine wave you can use the right triangle analogy, although I would say "can be represented as the hypotenuse" rather than "must be represented as the hypotenuse". Triangles need not be brought into it at all. A person who knew nothing about the geometry of triangles could still solve for the resultant of the voltage across a series connected resistor and inductor using the square root of the sum of the squares formulation.

 

[Urmi Roy]

You say: "Vrms=Vr+Vl (Vr and Vl are the potential drops across the resistor and inductor at any instant respectively.)"

The way you've described it, Vr and Vl are the instantaneous voltages across each component. If that is so, then Vrms is not equal to Vr+Vl; Vrms is a kind of average, not equal to a sum of instantaneous voltages.

This is exactly what I was getting at...but my book derives it this way!I am aware that the rms voltage cannot be represented as the simple sum...I suppose my book must be wrong somewhere.

Well, of course. For two components in series, the current through each is not just "considered" equal (for some mathematical convenience); it is physically identical in each component, which is indeed the same thing as saying "over time,the net charge transferred through the inductor and resistor are the same"

The reason I'm stressing on this is because here I is the rms current...and I have a feeling that the instantaneous current in a RL circuit cannot be equal for both the resistor and inductor can it?...I mean,while the resistor is conducting current according to applied voltage,the inductor is busy trying to oppose the change in current.

 

[The Electrician]

This is exactly what I was getting at...but my book derives it this way!I am aware that the rms voltage cannot be represented as the simple sum...I suppose my book must be wrong somewhere.

Can you post a picture of the page(s) in your book where they derive this?

The reason I'm stressing on this is because here I is the rms current...and I have a feeling that the instantaneous current in a RL circuit cannot be equal for both the resistor and inductor can it?...I mean,while the resistor is conducting current according to applied voltage,the inductor is busy trying to oppose the change in current.

You are imprecise in the way you say things. You left out an important qualifier that makes all the difference.

You said: "...the instantaneous current in a RL circuit cannot be equal for both the resistor and inductor can it?"

It absolutely is the case that the instantaneous current in a series RL circuit is identical for both the resistor and inductor.


This is the property that characterizes a series circuit; the current in all the components that are in series is the same.

 

[Urmi Roy]

Can you post a picture of the page(s) in your book where they derive this?

It's difficult to post the picture of the pages,as I don't know anywhere in my campus where they give students acess to scanners...but I can copy it out and promise that I'm copying the exact thing out,if you want.

 

[The Electrician]

It's difficult to post the picture of the pages,as I don't know anywhere in my campus where they give students acess to scanners...but I can copy it out and promise that I'm copying the exact thing out,if you want.

Can't you take a picture with a cell phone camera, yours or a friend's? Or just with a regular digital camera?