# Impedance

#### wazev

Joined May 26, 2020
8
If you have a common circuit with 1 resistor, 1 capacitor, and 1 inductor. Inousoidal voltage and current is applied with angular frequency, ω.
Then you want to use the jω-method and express the total complex Impedance, Z, of the above circuit in terms of R,L,C and ω. Answer in the form Z = a + bj . Now, I know that you can usually use the "Z = Sqr(R^2 + (XL - XC)^2) formula to solve for impedance, but I'm not sure what they mean with the form "Z = a + bj"; if anyone could give me guidance in how to solve these types of problems or give a bit more insight in what they mean with "Z = a + bj" this would be really helpful. Currently studying engineering and got this question in the book, now the downside is that this book don't have solutions for all problems within it, which is why im turning towards you guys. No values for R, C, or L was given.

#### Papabravo

Joined Feb 24, 2006
13,721
Any complex number can be represented as the sum of one real number, "a", and one imaginary number, "jb", where "b" is a real number and "j" is the imaginary unit. In your example we would write the following:

$Z\;=\;R+j\omega L+\frac{1}{j\omega C}$

Can you do the algebra required to put that expression into the standard form?

#### wazev

Joined May 26, 2020
8
I'm not exactly sure how to do that. I could do it with actual values there, but I find it difficult when it's all like that. Something I definetly need to work on.

If I recall two "j" equals"-1", so maybe Z = R +wL * -1 + 1/wC?

#### Papabravo

Joined Feb 24, 2006
13,721
I'm not exactly sure how to do that. I could do it with actual values there, but I find it difficult when it's all like that. Something I definetly need to work on.

If I recall two "j" equals"-1", so maybe Z = R +wL * -1 + 1/wC?
Nonsense, you don't need no stinkin' values. "j", the imaginary unit is a shorthand notation for the SQUARE ROOT of -1. Since there is no real number that you can square to get -1, we use the letters "i"(Mathematicians) or "j"(Electrical Engineers) as a stand-in. To get the above expression into the standard form you can multiply the term for the capacitor by "j/j" which equals 1 which doesn't change anything. What happens when you do that?

#### WBahn

Joined Mar 31, 2012
25,757
If you have a common circuit with 1 resistor, 1 capacitor, and 1 inductor.
Which "common circuit" containing those three elements are you referring to? All three in series? All three in parallel? One of the three configurations in which one element is in series with the parallel combination of the other two? Or one of the the three configurations in which one element is in parallel with the series combination of the other two?

I'm not exactly sure how to do that. I could do it with actual values there, but I find it difficult when it's all like that. Something I definetly need to work on.
That's definitely something you need to work on. It's called "algebra" and it is a core fundamental skill for any engineering discipline -- you need to not only be able to do it, you need to get downright good at it.

If I recall two "j" equals"-1", so maybe Z = R +wL * -1 + 1/wC?
I don't know what " two 'j' equals '-1' " means or how that matches what you tried to do with it.

"j" is defined as the value that, when squared, yields -1. in other words

j² = -1

or

j = sqrt(-1)

If you are studying engineering, then hopefully you have some calculus under your belt. Is that the case? How far have you gotten? Have you been introduced to LaPlace transforms yet?

You pretty much need to forget equations like "Z = Sqr(R^2 + (XL - XC)^2)". They are for people that don't have the mathematical background to work with complex impedance and they are extremely limiting. That one is particularly troublesome because if treats both inductive and capacitive reactance as positive quantities and embed the phase relationship as a minus sign in the equation, which is extremely limiting.

Before you go further, I strongly recommend taking a step back and getting comfortable working with complex numbers.

#### MrAl

Joined Jun 17, 2014
7,591
If you have a common circuit with 1 resistor, 1 capacitor, and 1 inductor. Inousoidal voltage and current is applied with angular frequency, ω.
Then you want to use the jω-method and express the total complex Impedance, Z, of the above circuit in terms of R,L,C and ω. Answer in the form Z = a + bj . Now, I know that you can usually use the "Z = Sqr(R^2 + (XL - XC)^2) formula to solve for impedance, but I'm not sure what they mean with the form "Z = a + bj"; if anyone could give me guidance in how to solve these types of problems or give a bit more insight in what they mean with "Z = a + bj" this would be really helpful. Currently studying engineering and got this question in the book, now the downside is that this book don't have solutions for all problems within it, which is why im turning towards you guys. No values for R, C, or L was given.
Hi,

I think i know what you are after here i'll through out some info for you see what you think.

First, you should realize that this kind of notation a+b*j is used in AC circuit analysis and is used to represent AC quantities that are sinusoidal. When b is non zero there is a phase shift and when b is zero there is no phase shift.

Ok, so if we have a resistor R1 the value in in Ohms, so the impedance written out is:
Z=R1+0*j
so the imaginary part is zero and the real part is R1 so in the form a+b*j we would have:
a=R1
b=0
Sound simple? It is.

Now let's do a resistor R1 and cap C1.
The capacitor impedance by itself is:
Zc=1/(j*w*C)
and again the resistor by itself is:
Zr=R1
Now for Zc we want to show the imaginary part as "j" times something so we multiply top and bottom by 'j' so we get a 'j' in the top:
(1*j)/((j*w*C)*j)
and you should know that j*j=-1 so we end up with:
j/(-w*C)
and then multiply top and bottom by -1 and get:
-j/(w*C)
so we end up with:
Zc=-j/(w*C)

Now that we have both the Zr and Zc, what if we have a resistor in series with a capacitor. Since impedances in series add, we get:
Zr+Zc=R1+(-j/(w*C1))
or simply:
Z=R1-j*(1/(w*C1))
and now this is in the form a+j*b where:
a=R1
b=-1/(w*C1)

If the resistor was in parallel with the capacitor we would have to do this:
Z=Zr*Zc/(Zr+Zc)
and then simplify the whole thing to get it into the form a+j*b.
We had to do it that way because impedances in parallel combine in the same way that two resistors combine:
Z=1/(1/Z1+1/Z2+1/Z3+...+1/Zn)

Also of interest is the magnitude and phase shift:
|Z|=sqrt(a^2+b^2)
Ph=atan2(b,a)
and note that atan2() is the two argument inverse tangent function not the single argument version. You can sometimes simplify it down to the single argument version but that comes later once you have actual values.

Now a voltage that is represented as a+b*j is also interesting because the imaginary part means there will be a phase shift in the voltage wave.

If any of this is not clear let me know.

#### BobaMosfet

Joined Jul 1, 2009
1,108
Nonsense, you don't need no stinkin' values. "j", the imaginary unit is a shorthand notation for the SQUARE ROOT of -1. Since there is no real number that you can square to get -1, we use the letters "i"(Mathematicians) or "j"(Electrical Engineers) as a stand-in. To get the above expression into the standard form you can multiply the term for the capacitor by "j/j" which equals 1 which doesn't change anything. What happens when you do that?
Not nonsense, I sympathize with the poster, but this thread has enlightened me... particulary WBahn's explanation, and Mr. AI.... _thank you_

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#### Papabravo

Joined Feb 24, 2006
13,721
Not nonsense, I sympathize with the poster, but this thread has enlightened me... particulary WBahn's explanation, and Mr. AI.... _thank you_
Focusing on values is just an excuse for ignoring the algebra. I sympathize as well which is why I tried to help in the fist place. The TS had shown some effort and I was trying to lead him to the solution of his own problem without just spoon feeding an answer. I did make an assumption (almost always a bad thing to do) that the TS was talking about a series circuit. Wbahn was correct in pointing out that the TS never specified the circuit they were talking about.

This is a forum, you ask a question, you get an array of answers. We can decide that only certain people can answer, but this place becomes a great deal less interesting if that is to be the case.

#### MrAl

Joined Jun 17, 2014
7,591
Focusing on values is just an excuse for ignoring the algebra. I sympathize as well which is why I tried to help in the fist place. The TS had shown some effort and I was trying to lead him to the solution of his own problem without just spoon feeding an answer. I did make an assumption (almost always a bad thing to do) that the TS was talking about a series circuit. Wbahn was correct in pointing out that the TS never specified the circuit they were talking about.

This is a forum, you ask a question, you get an array of answers. We can decide that only certain people can answer, but this place becomes a great deal less interesting if that is to be the case.
Hi,

Maybe we can deduce that from the formula that the TS did actually give?
I dont know for sure, but i did not do that either i just gave a couple common examples in the hopes that it would help.

#### Papabravo

Joined Feb 24, 2006
13,721
Hi,

Maybe we can deduce that from the formula that the TS did actually give?
I dont know for sure, but i did not do that either i just gave a couple common examples in the hopes that it would help.
The big thing that distinguishes this thread from others in the HH sub forum is that the TS did show some work.

#### wazev

Joined May 26, 2020
8
Thank you all for the great answers, it's certainly set me in the right path. And to answer one asked question, the C was in series with the Resistor, but in parallel with the L.

Just if I could ask one more question. What would the limiting absolute values of Z be if ω → 0 and ω → ∞ respectively ?

#### wazev

Joined May 26, 2020
8
Hi,

I think i know what you are after here i'll through out some info for you see what you think.

First, you should realize that this kind of notation a+b*j is used in AC circuit analysis and is used to represent AC quantities that are sinusoidal. When b is non zero there is a phase shift and when b is zero there is no phase shift.

Ok, so if we have a resistor R1 the value in in Ohms, so the impedance written out is:
Z=R1+0*j
so the imaginary part is zero and the real part is R1 so in the form a+b*j we would have:
a=R1
b=0
Sound simple? It is.

Now let's do a resistor R1 and cap C1.
The capacitor impedance by itself is:
Zc=1/(j*w*C)
and again the resistor by itself is:
Zr=R1
Now for Zc we want to show the imaginary part as "j" times something so we multiply top and bottom by 'j' so we get a 'j' in the top:
(1*j)/((j*w*C)*j)
and you should know that j*j=-1 so we end up with:
j/(-w*C)
and then multiply top and bottom by -1 and get:
-j/(w*C)
so we end up with:
Zc=-j/(w*C)

Now that we have both the Zr and Zc, what if we have a resistor in series with a capacitor. Since impedances in series add, we get:
Zr+Zc=R1+(-j/(w*C1))
or simply:
Z=R1-j*(1/(w*C1))
and now this is in the form a+j*b where:
a=R1
b=-1/(w*C1)

If the resistor was in parallel with the capacitor we would have to do this:
Z=Zr*Zc/(Zr+Zc)
and then simplify the whole thing to get it into the form a+j*b.
We had to do it that way because impedances in parallel combine in the same way that two resistors combine:
Z=1/(1/Z1+1/Z2+1/Z3+...+1/Zn)

Also of interest is the magnitude and phase shift:
|Z|=sqrt(a^2+b^2)
Ph=atan2(b,a)
and note that atan2() is the two argument inverse tangent function not the single argument version. You can sometimes simplify it down to the single argument version but that comes later once you have actual values.

Now a voltage that is represented as a+b*j is also interesting because the imaginary part means there will be a phase shift in the voltage wave.

If any of this is not clear let me know.

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#### wazev

Joined May 26, 2020
8
Hi,

I think i know what you are after here i'll through out some info for you see what you think.

First, you should realize that this kind of notation a+b*j is used in AC circuit analysis and is used to represent AC quantities that are sinusoidal. When b is non zero there is a phase shift and when b is zero there is no phase shift.

Ok, so if we have a resistor R1 the value in in Ohms, so the impedance written out is:
Z=R1+0*j
so the imaginary part is zero and the real part is R1 so in the form a+b*j we would have:
a=R1
b=0
Sound simple? It is.

Now let's do a resistor R1 and cap C1.
The capacitor impedance by itself is:
Zc=1/(j*w*C)
and again the resistor by itself is:
Zr=R1
Now for Zc we want to show the imaginary part as "j" times something so we multiply top and bottom by 'j' so we get a 'j' in the top:
(1*j)/((j*w*C)*j)
and you should know that j*j=-1 so we end up with:
j/(-w*C)
and then multiply top and bottom by -1 and get:
-j/(w*C)
so we end up with:
Zc=-j/(w*C)

Now that we have both the Zr and Zc, what if we have a resistor in series with a capacitor. Since impedances in series add, we get:
Zr+Zc=R1+(-j/(w*C1))
or simply:
Z=R1-j*(1/(w*C1))
and now this is in the form a+j*b where:
a=R1
b=-1/(w*C1)

If the resistor was in parallel with the capacitor we would have to do this:
Z=Zr*Zc/(Zr+Zc)
and then simplify the whole thing to get it into the form a+j*b.
We had to do it that way because impedances in parallel combine in the same way that two resistors combine:
Z=1/(1/Z1+1/Z2+1/Z3+...+1/Zn)

Also of interest is the magnitude and phase shift:
|Z|=sqrt(a^2+b^2)
Ph=atan2(b,a)
and note that atan2() is the two argument inverse tangent function not the single argument version. You can sometimes simplify it down to the single argument version but that comes later once you have actual values.

Now a voltage that is represented as a+b*j is also interesting because the imaginary part means there will be a phase shift in the voltage wave.

If any of this is not clear let me know.
So, we don't actually do anything with the inductor, correct?

#### MrAl

Joined Jun 17, 2014
7,591
So, we don't actually do anything with the inductor, correct?
Hi, sorry for some reason i did not see this reply until now.

When you have multiple components you try to reduce things in series and things in parallel.
For the R in series with C, you would first calculate the a+j*b for that, then put that in parallel with the a+j*b for the inductor (which would not have any real part) and the combination will lead to a total a+j*b which will be the total complex impedance.
For impedances in series:
Z=Z1+Z2
for two impedances in parallel:
Z=Z1*Z2/(Z1+Z2)
and for N impedances in parallel:
Z=1/(1/Z1+1/Z2+1/Z3+...+1/ZN)
where ZN is the last impedance to be put in parallel with the others.

#### WBahn

Joined Mar 31, 2012
25,757
Thank you all for the great answers, it's certainly set me in the right path. And to answer one asked question, the C was in series with the Resistor, but in parallel with the L.

Just if I could ask one more question. What would the limiting absolute values of Z be if ω → 0 and ω → ∞ respectively ?
You already have a thread in Homework Help on that question. You need to show YOUR best attempt to answer it. Don't just try to sneak it into another thread to trick someone into doing your homework for you.