Charging of Inductors: Explained Without the Duality

[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?

 

[Urmi Roy]

I tried to take a picture on my camera...

 

[The Electrician]

OK, but this doesn't show how they calculated Irms.

 

[Urmi Roy]

As you must have seen,the I in the above derivation is what they call rms current.

There is no separate derivation for current,but in regard to power calculation,they say:

"The principal current I can be resolved into two components
(i) A component Ia in phase with voltage. This is called the active or real or wattfull component.
Ia=Icos(phi)
(ii)A component Ir at right angles to V
This is called the reactive or quadrature or wattless component.
Ir=Isin(phi)
I=root over (Ia squared + Ir squared)

Actual Power(P)=VIa = VIcos(phi) =VI(R/Z)

Quadrature power=VIr = VIsin(phi) =VI(X/Z),where X= reactance of inductor."

I'll send a picture as soon as possible.

 

[Urmi Roy]

Here's a picture of the page.

 

[The Electrician]

This picture is too blurry to read. It would help to take the picture in bright light, perhaps outdoors.

Make sure it's readable before you post it.

 

[Urmi Roy]

I took it on my web cam,so I don't know if I could do any better...anyway,I assure you I copied the stuff in my book accurately in my last post,so you could see that...anyway,after reading all these derivations from my book,did you notice anything wrong about them,or would you like to stress anything from them that would aid in my concept?

I'll try to send another picture soon anyway.

 

[Urmi Roy]

Pictures...

 

[The Electrician]

In the image you've attached to post #71, down near the bottom it says:

"At any instant, applied voltage
V = VR + VL (Refer fig 5.5)
Applied voltage V = IR + jIXL = I(R + jXL)"

But I notice that some of the variables have an overline above them, and in fact the I in I(R + jXL) should be overlined, but I can't do that without using tex. Did your text explain earlier that a voltage or current variable with an overline represented a magnitude (RMS for voltage and current; magnitude for impedance)? If so, then the last equation above should be:

\text{Applied voltage }\overline{V} = IR + jIX_L = \overline{I}(R + jX_L)

and strictly it probably should be:

\text{Applied voltage }\overline{V} = \overline{I} R + j\overline{I} X_L = \overline{I}(R + jX_L)

The next line says:

V/I = R + jX_L = \overline{Z}

but it should be:

\overline{V}/\overline{I} = R + jX_L = \overline{Z}

This image is very blurry in the top part, but I think I see a line that says:

\overline{I} = \text{Effective value of circuit current}

"Effective value" means the same as RMS value.

They have made some typographical errors and left out overlines in places where it leads to confusion because when they use V and I without overlines they apparently mean "instantaneous value".

So if we put in the overlines where they belong, then these two lines make sense:

\text{At any instant, applied voltage }V = V_R + V_L\text{ (Refer fig 5.5)}

\text{Applied voltage }\overline{V} = \overline{I} R + j\overline{I} X_L = \overline{I}(R + jX_L)

In the first equation, V and I refer to instantaneous values, and the second equation they refer to RMS values.

Now do you see why I wanted pictures? Even though you said "I assure you I copied the stuff in my book accurately in my last post", and I'm sure you did your best, I wouldn't have known about the typographical errors involving the overlines without a picture.

 

[Urmi Roy]

So if we put in the overlines where they belong, then these two lines make sense:

\text{At any instant, applied voltage }V = V_R + V_L\text{ (Refer fig 5.5)}

\text{Applied voltage }\overline{V} = \overline{I} R + j\overline{I} X_L = \overline{I}(R + jX_L)

In the first equation, V and I refer to instantaneous values, and the second equation they refer to RMS values.

The first equation must be a phasor addition,not a simple addition,I suppose...(sorry if this is too obvious)...

Also,if the overlined quantities are actually the rms quantities,we shouldn't say they're the magnitudes of the current and voltage...since for one,the magnitude of voltage or current isn't the same as the rms value...and the magnitude of instantaneous current or voltage varies with time (the total voltage drop in ac circuit may even be greater than the applied voltage, I once heard!)...

Now do you see why I wanted pictures?...

I really think you're awesome,not only for predicting that something like this might be wrong in my book,but also for the help you provided in getting my understanding of inductors right. I am sincerely grateful to you...especially now that I must be one of the few first years to be aware of the errors in the book,and,more importantly, to know the correct thing.

 

[The Electrician]

The first equation must be a phasor addition,not a simple addition,I suppose...(sorry if this is too obvious)...

If when you say 'first equation', you mean this one:

(There's some kind of tex error I can't fix)

\text{At any instant, applied voltage } V = V_R + V_L\text { (Refer fig 5.5)}

then, no, that's not a phasor addition. It is a simple addition of the instantaneous volages.

Have a look later on where I discuss the third image.

This equation is an addition of phasor quantities:

\text{Applied voltage }\overline{V} = \overline{I} R + j\overline{I} X_L = \overline{I}(R + jX_L)

Also,if the overlined quantities are actually the rms quantities,we shouldn't say they're the magnitudes of the current and voltage...since for one,the magnitude of voltage or current isn't the same as the rms value...and the magnitude of instantaneous current or voltage varies with time

Be careful here. In common usage the RMS value and the effective value are the same thing. Magnitude may have different meanings depending on how the person who is using the term chooses to define it. Often, it is used to mean RMS value. It's best to add an adjective to the word "magnitude" to be sure what is meant. If instantaneous value is meant, then say "instantaneous magnitude".

Have a look at the first image I've attached. This is from a circuit theory text, and the author makes it clear that he intends the word magnitude to mean RMS value.

In this book instantaneous values are represented by lower case letters:

v(t) and i(t) are the instantaneous voltage and current.

Bold face capital letters are used to represent RMS values, rather than overlined letters. Phasors are usually the RMS values:

V is the RMS voltage; I is the RMS current.

Non-boldface capital letters are used to mean something other than RMS, with a subscript denoting the meaning. For example, in the image Vm means maximum voltage, or "peak" voltage in this expression:

V = Vm*sin(wt + theta)

So, magnitude of voltage could be the RMS value, if that's what the person who is writing chooses it to mean, but the writer should define it if that's what he intends. But, it could also mean "instantaneous" value. Once you fully understand all this, you will be able to tell from the context what is meant if the writer has neglected to define his terms.

Be sure to download that file, 966.pdf, I mentioned in post #64 and #66. I think it will help you.

(the total voltage drop in ac circuit may even be greater than the applied voltage, I once heard!)...[/B]

If you say it a little differently then it is true. If you have a series connection of R and L (or R and C, or R, L and C), it is possible for the sum (not the RMS sum, or phasor sum, but the simple addition) of the RMS voltages across the individual components to be larger than the applied voltage.

I've attached three images. The first is from a textbook. The second and third are of a physical setup. The second image shows a 1 uF capacitor and 3k ohm resistor connected in series, and with an applied voltage of 120V RMS @ 60 Hz, from the wall outlet. I've connected 3 probes from an oscilloscope to the circuit so that the first channel (orange) shows the total applied voltage, the second channel (blue) shows the voltage across the capacitor, and the third channel (purple) shows the voltage across the resistor.

I've used a capacitor in the circuit rather than an inductor, because an inductor of the same impedance would be 7 henries, and I don't have one that large.

The third image shows the oscilloscope display. On the right edge you can see the RMS values of the three voltages. The total applied voltage is 120V RMS. The voltage across the capacitor is 77.7V RMS, and the voltage across the resistor is 92.0V RMS.

You'll notice that the line voltage (orange) isn't a perfect sine wave; it's somewhat flattened on top. This leads to a resistor voltage that is also distorted, but the capacitor voltage is more nearly perfect because the capacitor attenuates the higher harmonics.

Notice that by simple addition, the total of the voltage across the capacitor and the voltage across the resistor is 77.7 + 92.0 = 169.7, substantially more than 120. But if you use phasor addition, you would calculate SQRT(77.7^2 + 92.0^2) = 120.42, very close to the total applied voltage.

The scope traces are shown with a scale factor of 50V (instantaneous) per major division, so at any instant of time you can see that the voltage shown by the orange trace is the simple sum of the blue and purple voltages. This is how it must be in a series circuit.

The first equation:

\text{At any instant, applied voltage }v(t)_{applied} = v_C(t) + v_R(t)\text{ (Refer fig 5.5)}

where I've changed some of the variable typefaces, expresses this fact.

 

[Urmi Roy]

I think I'm all muddled up...as you said,the phasor addition of the voltages across the capacitor(or any reactive member) is always equal to the magnitude of the applied voltage...but at the same time,in a series connection the applied voltage is equal to simple sum of the voltages across the two!How is this possible?

In the picture of the textbook,the very first paragraph says that the V and I with the complex notation are not the voltage and current ...that's an interesting way to put it...

 

[The Electrician]

I think I'm all muddled up...as you said,the phasor addition of the voltages across the capacitor(or any reactive member) is always equal to the magnitude of the applied voltage...but at the same time,in a series connection the applied voltage is equal to simple sum of the voltages across the two!How is this possible?

You've left out important words in this. Think about something I said in the previous post: "It's best to add an adjective..."

What you've said should have a few more words to be unambiguously true:

"I think I'm all muddled up...as you said,the phasor addition of the RMS voltages across the capacitor(or any reactive member) and the resistor is always equal to the RMS magnitude of the applied voltage...but at the same time,in a series connection the applied instantaneous voltage is equal to simple sum of the instantaneous voltages across the two!How is this possible?"

It's possible because that's how the two kinds of voltages add in a series circuit.

It's important to make the distinction between RMS voltages and instantaneous voltages. They add in different ways in a series circuit.

In the previous post, where I said "Notice that by simple addition, the total of the voltage across the capacitor and the voltage across the resistor is 77.7 + 92.0 = 169.7, substantially more than 120", the 77.7 and 92.0 are RMS voltages, as is the applied 120V, so they don't add up properly with simple addition, but they do with phasor addition.

On the other, in the scope capture, the voltages which you can read off the traces are the instantaneous voltages, and they do add properly with simple addition.

In the picture of the textbook,the very first paragraph says that the V and I with the complex notation are not the voltage and current ...that's an interesting way to put it...

The first paragraph doesn't say "the V and I with the complex notation are not the voltage and current"; it says "These complex quantities are not voltage and current; voltage and current are given in equations 24-2 and 3." Equations 24-12, 24-2 and 3 are not visible, so we don't know for sure what he's talking about.

 

[Urmi Roy]

Oh...so simple addition is applicable when we add the instantaneous values of the quantities...whereas all this time we're essentiall been talking of rms quantities...for which only phasor addition will work...I think I get it.