Inductors Charging

Urmi Roy

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 realised 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 existance?

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

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

I understand that it's the apparant 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

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 accross 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

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.

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?


Does the bolded part have any special implication?

The Electrician

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.

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.

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

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.

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....apparantly,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 accross 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 accross 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

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

Download the file "966.pdf"

You say: "Vrms=Vr+Vl (Vr and Vl are the potential drops accross 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.

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"

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

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.


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

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

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.