Charging of Inductors: Explained Without the Duality

[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.