# Series Parallel Circuit Impedance Calculation

#### arb-92

Joined Mar 30, 2013
2
Hi,
College years being far (45+ years...) away, I was attempting to brush up on circuit impedance calculation, checking my understanding by redoing the examples provided on the on-line textbook.
Well, I got stuck at one point: every intermediate steps check but the final one! What am I doing wrong? Probably very dumb mistake...
(see attached)
Many thanks!
ARB

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#### MrAl

Joined Jun 17, 2014
8,612
Hi,

So this isnt really homework then, you're just doing it at home?

I do not think that the way you are doing it is the right way to go about these more complex circuits. I dont know if the book is wrong or what, but although your calculations work out correctly, you dont get the same result as the book because the procedure being used will not lead to the right result.

My question would be, is the book showing that way of doing it? Because if so, then they must be showing HOW to do it one way, then actually DOING it another way completely. In other words, the 818 value is the right result for this circuit, but if you do it the way that is being explained here you'll get the 721 result which is not correct for this circuit.

The best way to handle these AC problems is to work with complex numbers. They are not that hard to deal with and can solve every one of these kinds of circuits rather than just some.

1. The first step is to convert all the reactive elements into their complex forms.
2. The second step is to combine those forms either in series or parallel.
3. The final step is to simplify the result.

If this is not really homework then i can give you more information. If you have never seen this way of doing it then you'll need more help but you'll see how general this method is once you start using it.

#### WBahn

Joined Mar 31, 2012
26,398
You neglected two things. You (and the textbook, it appears) are treating all reactances as positive quantities. But that means that you must separately keep track of which reactances are inductive and which are capacitive, which you don't appear to be doing. Then you need to keep track of resistance and reactance separately since they add in different ways.

So your next to last line gives you the magnitude of the impedance, but it loses the distinction between the resistance and the reactance and it also captures nothing regarding the inductive/capacitive nature of the reactance. You need all of that information in order to combine it with the reactance of C1 to get the final result.

The simplest and most bullet proof way of dealing with impedance is via complex impedances. But if your math background does not include complex numbers, then this is difficult. The best solution then is to learn how to work with complex numbers. Then everything works out very beautifully. I can show you how that would work in this case and then you can study up on complex arithmetic until you can duplicate that work.

#### arb-92

Joined Mar 30, 2013
2
Oh, I see : so, there is no short-cut to full complex number representation and manipulation in the present case.
I think I can handle the maths (never had to use complex numbers in my work-life actually..).
Thanks for putting me back on track !
ARB

#### WBahn

Joined Mar 31, 2012
26,398
You can certainly slug through things without ever using complex numbers, but I think it is much more difficult, you have to keep track of stuff in an ad hoc fashion, and it is much more prone to error. Using complex impedance you simply combine things exactly as you would good ole resistors in DC circuits and you let the math take care of everything else.

In this case you simply have Z_C1 + [ (Z_L1 + Z_C2) || (R) ]

where

Z_L = jωL
Z_C1 = 1/(jωC1)
Z_C2 = 1/(jωC2)

ω = 2πf

#### MrAl

Joined Jun 17, 2014
8,612
Oh, I see : so, there is no short-cut to full complex number representation and manipulation in the present case.
I think I can handle the maths (never had to use complex numbers in my work-life actually..).
Thanks for putting me back on track !
ARB
Hi,

It would also be interesting to see how your 'book' worded it. DId they word it correctly or maybe not, and that would throw you off too. But yes, complex numbers work better for AC analysis than anything else i've seen so far unless it is the simpler problems with one L and one C and maybe one R, or something like that. Once you move into circuits with multiple L and C and R you definitely want to turn to complex numbers, and this helps by a huge amount once you get into multi phase circuits too.

If you want to go over some examples we can do that here too. You'll see how easy it is once you do a few of these, mainly because you follow a general procedure that applies to all circuits.

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#### RBR1317

Joined Nov 13, 2010
677
...every intermediate steps check but the final one!
So why did your method work until the very last step? The formula for combining a pure resistance with a pure reactance is the square root of the sum of the squares, but only works because there is a 90 degree phase angle between resistance and reactance. However, Z" is neither a pure resistance nor reactance, and treating it as though it were just gives the wrong answer when combining it with XC1. That's the reason complex arithmetic is necessary.