# Fancy math on a derivation

#### KevinEamon

Joined Apr 9, 2017
281
Haha indeed...

Ok well I imagine the values would be in an arrangement similar to:-
R = r +j0
C = 0r - jxc
L = 0r + jxl

#### WBahn

Joined Mar 31, 2012
25,913
Haha indeed...

Ok well I imagine the values would be in an arrangement similar to:-
R = r +j0
C = 0r - jxc
L = 0r + jxl
What is Xc?

What is Xl?

If I give you a 33 uF capacitor and tell you that it is being operated at 60 Hz, what is the reactance?

I gave all of these two you in an earlier post!!!

Z_R = R
Z_C = 1/(jωC)
Z_L = jωL

If you aren't to the point of know these, then you aren't ready to be even attempting the question in this problem. Take a step back and learn this much before proceeding. Otherwise you just staying busy digging the hole you are in deeper and deeper.

#### KevinEamon

Joined Apr 9, 2017
281

#### WBahn

Joined Mar 31, 2012
25,913
Okay, so we should be at the point of agreeing that the impedance of the three components are

Z_R = R
Z_C = 1/(jωC)
Z_L = jωL

So let's set that aside for the moment.

Now, consider the following circuit:

For the moment, assume we don't know what each of the three impedances are. We'll get to that late.

In terms of Z1, Z2, and Z3, what is the total impedance between A and B?

#### KevinEamon

Joined Apr 9, 2017
281

#### WBahn

Joined Mar 31, 2012
25,913
Okay. Now, using any of those forms that you like, substitute in the appropriate values of Z1, Z2, and Z3 expressed in terms of the components in the actual circuit for this problem.

Then massage it until you have it expressed in rectangular form, namely

Zt = A + jB

where A and B are expressed in terms of the angular frequency and the component values, so {ω, R, L, C}.

#### KevinEamon

Joined Apr 9, 2017
281
I'v become stuck in some mathemtical web of madness. Though I think I might be getting close...the story so far....

#### KevinEamon

Joined Apr 9, 2017
281
Sorry I think I've become stuck after this above I can of course flip everything again to get Zt rather than it's inverse. But that's about it. I think I'm missing some mathmatical tools I require.

#### WBahn

Joined Mar 31, 2012
25,913
I'v become stuck in some mathemtical web of madness. Though I think I might be getting close...the story so far....
Your mathematical web of madness is purely, 100% one of your own making and the fact that you couldn't get out of it is because you still refuse to track your damn units!

You got off to an okay start, although you need to clean up one misconception.

An impedance has a resistive part and a reactive part. Both of these are real numbers. So, in general, we have

Z = R + jX

The impedance of an inductor is

Z_L = jωL = R + jX

Matching up terms, that means that the resistive part of the impedance is zero and that the reactive part is

X_L = ωL

There's no 'j' there.

The reactance of a capacitor is then

X_C = -1/(ωC)

because when you write the impedance of a capacitor in rectangular form, that is what the reactance has to be to match up.

Now, remember how I am constant harping on you about tracking your units and asking if they are working out?

Let's see what that tells us here.

We know that impedance, resistance, and reactance all have units of Ω. So let's mark up your work indicating the units on each term.

Everything is working out fine until we get to the equation that is boxed in red.

The first term in the numerator (as far as units go) is a resistance multiplied by an inverse resistance, so the result is unitless. That's a good thing because that term is getting added to a pure number which is inherently unitless, so we can add them. The denominator is a resistance, so the quantity in the parentheses has units of 1/Ω. That gets multiplied by Zt, which has units of Ω, so the entire left hand side is unitless. But the right hand size is an inductive impedance and so it has units of Ω.

Your units don't work out. Therefore EVERY bit of time and effort spent working on this after this point is pure wasted time and effort.

As soon as you get to the equation in the red box where the units don't work, look at the equation before it and confirm that the units in it did work out (and in this case they do). That means you made an algebra error going from one to the other. There is simply no point going any further until you identify it and correct it.

How many times am I going to have to show you where you went off the rails and spent who knows how much time chasing down a rabbit hole that is guaranteed to lead to a wrong answer simply because you steadfastly refuse to use units properly.

I have said it time and time again -- we all make math errors on an all-too-frequent basis; it goes along with being human. But most of those errors will screw up the units. If we will simply bother to check them, we can usually find and fix those errors immediately after making them. But if we don't, then at best we waste time and effort where there is now justifiable reason to. That results in wrong answers which results in lower grades and, later, getting fired or not catching an error that ends up getting someone seriously injured or killed and, to top that off, getting prosecuted for gross negligence, either criminally or civilly, and relying on the jury to buy the excuse that you didn't have time to track units properly, so sorry.

xox

#### KevinEamon

Joined Apr 9, 2017
281
Ahaha I didn't actually know I could do track my units with these things, or believe me! I would have done it...

Ok stupid question - Why is the capacitor an inverse resistance?
I mean I get the fact that Xc = (-1/(ωC)) and thanks for pointing out the error with the j part. I need to study that more to understand. I'll probably read over your notes again a few times before the penny drops...

And I understand that if you moved it to the denominator, it would become a (ωC^-1)
But then the resistor was on the denominator too and that isn't an inverse... hmmm

Is it just because capacitors are always in negative values? Like -j20. Is that why its units are inverse Omhs?

I'm gonna go back here and look at this again - tracking my units this time...

#### WBahn

Joined Mar 31, 2012
25,913
Ahaha I didn't actually know I could do track my units with these things, or believe me! I would have done it...

Ok stupid question - Why is the capacitor an inverse resistance?
The capacitor's reactance doesn't have units of inverse resistance.

Let's figure out the units of

Xc = (-1/(ωC))

The units of ω are inverse seconds (1/s).

The units of capacitance can be determined from the defining relation

Q = CV

So C has units of charge per volt, or C/V.

Don't let the two different 'C's confuse you. The first C is the variable name commonly used for capacitance, the second C is the SI abbreviation for the unit of charge, namely the coulomb. Since we like single letter variable names AND single letter unit abbreviations and since we only have so many letters available, there's going to be duplication and we just have to learn to deal with it.

If you find yourself getting confused, then use something else. When all is said and done, the unit of charge is not necessarily the coulomb (just as the unit of distance is not necessarily the meter -- it could be feet). So feel free to be generic and say that the unit of capacitance is (charge/voltage). Similarly, the unit of current can simply be (charge/time) instead of amperes or A.

So ωC has units of C/(Vs) which is the same as (C/s)/V. But a C/s is an ampere, so we have that ωC has units of A/V, which is inverse resistance (since resistance is V/A).

However, our capacitive reactance has ωC in the denominator, so the units on the capacitive reactance is V/A, or units of resistance.

Now, go look at the equation where I indicated inverse resistance. You will see that there is a factor of ωC, and NOT 1/(ωC).

You want to get to the point where you can identify the common units quickly.

For instance, the units of RC (resistance times capacitance) is time. You should be able to show this from the definitions of R and C.

V = IR, so R has units of volts/amperes (V/A).

Q = CV, so C has units of coulombs/volt (C/V), or ampere-seconds/volt (As/V).

RC therefore has units of (V/A)(As/V) = s (time).

You don't have to memorize this stuff -- understand it.

You SHOULD have the fundamental defining equations memorized, however.

Any time I need the units of inductance (other than just "henries") I have to figure it out, but I don't have to ask someone or go to the internet to do so. I know that the defining relation for inductance is

V = L(di/dt)

The means that the units are

(voltage) = L (current/time)

So L has units of (voltage-time)/current, or Vs/A.

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#### KevinEamon

Joined Apr 9, 2017
281

#### WBahn

Joined Mar 31, 2012
25,913
What happened to all the 'j's? You can't just drop them and pretend they don't exist.

The right hand side of the equation is dimensionally consistent.

My system isn't letting me insert the ohms symbol right now, so let's use m (i.e., meters).

1/m is the same as m^-1

It is NOT the same as 1/(m^-1) because that is the same as simply m.

If all else fails, write it out explicitly.

$$\frac{1}{\Omega^{-1}} \; = \; \frac{1}{\( \frac{1}{\Omega}$$}\)

Now multiply the numerator and the denominator by omega.

Don't lose site of what you are trying to do. You want to express the total impedance as a resistance plus a reactance. That means you want Zt on one side of the equation by itself and nowhere on the other side.

Well, look at your first equation. That's what you have. So get the 'j's back in the left side where they belong and go from there. Since the Zt is already isolated -- leave it where it is and only manipulate the other side.

The goal of the manipulations of the other side are to get it into rectangular form, namely

Zt = Rt + jXt

So first get everything over a common denominator and write both numerator and denominator in rectangular form. So that will give you something like

Zt = (A + jB) / (C + jD)

What do you think the next step after than should be?

#### KevinEamon

Joined Apr 9, 2017
281
I still have no idea how to get that equation into the form you want Wbahn. I'm now several layers of abstraction deep and my initial question far from sight.

Proceeding from the point you mentioned, given those variables I would do something like this

#### WBahn

Joined Mar 31, 2012
25,913
I still have no idea how to get that equation into the form you want Wbahn. I'm now several layers of abstraction deep and my initial question far from sight.

Proceeding from the point you mentioned, given those variables I would do something like this

View attachment 152205
Two points:

Assuming your work above is correct, we just have to cross the bridge from your expression for Zt earlier to a form in which we can identify A, B, C, and D.

Once again, check the units.

What do we know about the units of A, B, C, and D?

We don't know exactly what they are, but we can say the following:

A and B have to have the same units in order for the terms in the numerator to be added together. The same holds true for the units of C and D. Furthermore, the ratio of the units of numerator to the denominator have to work out to units of resistance. With that in mind, let's look at your result above and see if it makes sense.

Let's call the units of A and B the fictitious N (for numerator) and for C and D we'll use D (for denominator). We could use anything, such as volts for the numerator and amps for the denominator, but N and D will work.

What are the units on the three terms of your numerator and what is the units of the denominator and what is the ratio?

numerator: ND + ND + N²D²
denominator: D²

Do you see a problem?

Does seeing that problem pointed out by the units help you track down the error more easily?

You REALLY need to get absolutely religious about units. I mean fanatical. You need to do it far more than most. The reason is simple -- your algebra skills are so poor that you are constantly making mistakes that would likely fail you in 8th grade algebra. Let units come to your rescue. Check them in EVERY line of your work. Without fail. The result will be that your algebra weakness will slap you in the face with immediate feedback allowing you to spot the specific mistakes you are making right after you make them. By fixing them right then and there, while the thought processes that led you to make them are still fresh, you will be able to identify and correct the deficiencies with the result being that your algebra skills will improve dramatically in a relatively short period of time.

#### KevinEamon

Joined Apr 9, 2017
281
I think you're right Wbahn. I've actually started to do it unconsciously now, thinking about my units as I go along. Part of our electrical module was on the per unit system, which made a whole lot more sense to me because you're always chatting about units. It's weird that the concept of a unitless unit... made tons more sense because I've thought so much about units... lul

Anyways I figured out this bad boy. See this equation at point 1, it gets fed back into the equation at point 3. At resonance the imaginary, of point 3, is zero, which is how he set point 4 equal, to each other.
Two freaking days I've been beating my head against a wall with that...sigh

My algebra skills have come a long.... long way. They used to be much worse believe me. I suppose I should have added those variables up there, in that last pic, rather than setting them as multiples.... Anyways, most people come into my degree, having done math for A level(High school). I didn't, I done a diploma in science. Math was on the course but not focused enough for my needs. I've been running behind, almost from the day I started the degree.
But i'll get there, I always do...

Thanks for the help as always my friend, I'd be lost without you

#### WBahn

Joined Mar 31, 2012
25,913
Anyways I figured out this bad boy. See this equation at point 1, it gets fed back into the equation at point 3. At resonance the imaginary, of point 3, is zero, which is how he set point 4 equal, to each other.
Two freaking days I've been beating my head against a wall with that...sigh
I think you might have missed Post #5 where I said:

The second equation is simply the impedance (which is purely resistive) at the resonant frequency.

So they aren't going FROM equation 1 TO equation 2, but more likely there is an earlier equation giving the impedance at ANY frequency (so it would have ω in it) and then they ask what the resonant frequency is (probably defined as the frequency at which the reactance goes to zero) and come up with equation 1. They then plug that back into the prior equation to come up with equation 2, which is only good at that one frequency.

But why didn't they simplify equation 2 to just L/(RC)?

#### KevinEamon

Joined Apr 9, 2017
281
I think you might have missed Post #5 where I said: The second equation is simply the impedance (which is purely resistive) at the resonant frequency.
I did read that Wbahn. I wasn't sure what it meant exactly. As I say, I miss things. People tend to talk to me at a certain level, assuming I know and can understand everything, at that level. When I can be a bit thick tbh... usually about really simple things...
but more likely there is an earlier equation giving the impedance at ANY frequency
The full picture is on post #8 of this thread and I don't think it is, though I could be wrong
But why didn't they simplify equation 2 to just L/(RC)?
He did it's right at the end there.

#### WBahn

Joined Mar 31, 2012
25,913
I did read that Wbahn. I wasn't sure what it meant exactly. As I say, I miss things. People tend to talk to me at a certain level, assuming I know and can understand everything, at that level. When I can be a bit thick tbh... usually about really simple things...

The full picture is on post #8 of this thread and I don't think it is, though I could be wrong

He did it's right at the end there.
Go back and review his work (Post #8) again, because it is EXACTLY what I described.

The center column (the left hand column of work) starts with the basic description of the circuit topology and then ends at the bottom with the equation for the impedance that is good for any frequency.

The top of the right column is the direct result of setting the imaginary part to zero. That results in the next to last line which gives the frequency at which this happens. The final line is simply plugging this frequency back into the real part of the general impedance equation. Only the real part is used because it is already known that the imaginary part will be zero at that frequency).

EDIT: Fixed mistake pointed out by The Electrician.

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#### The Electrician

Joined Oct 9, 2007
2,778
Go back and review his work (Post #8) again, because it is EXACTLY what I described.

The center column (the left hand column of work) starts with the basic description of the circuit topology and then ends at the bottom with the equation for the impedance that is good for any frequency.

The top of the right column is the direct result of setting the imaginary part to zero. That results in the next to last line which gives the frequency at which this happens. The final line is simply plugging this frequency back into the real part of the general impedance equation. Only the real part is used because it is already known that it will be zero at that frequency).
Don't you mean to say: "Only the real part is used because it is already known that the imaginary part will be zero at that frequency)"?