# Capacitor and inductor in series - common impedance calculation

Discussion in 'General Electronics Chat' started by fektom, May 20, 2013.

1. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Hello everybody!

I try to get a result of a complex number arithmetic, but I can't.
I don't want to use scientific calculator, I want to do step by step as described on http://www.allaboutcircuits.com/vol_2/chpt_2/6.html

The example I want to solve is on http://www.allaboutcircuits.com/vol_2/chpt_5/4.html

The problem is with ZL and ZC2 which are in series.
ZL = 245.04<90°
ZC2 = 1768.4<-90°

If I want to add up, I have to convert them to rectangular form.
ZL = 0+j245.04
ZC2 = 0-j1768.4
ZL-C2= 0-j1523.36

And here comes the question: how to convert this back to polar form?

$\sqrt{0 square +(-1523.36 square)}$ = 1523.36
From this the polar form would be:
1523.36 < arctan 1523.36/0
However, dividing a number with 0 makes my calculator to give error message. It's ok, because in elementary we learned, that dividing with 0 is foolish.
In the mean time, somehow I have to get the polar form, but how, if division with 0 is impossible?

If you have an idea, and a correct result for common impedance of ZL and ZC2 please let me know!

2. ### tshuck Well-Known Member

Oct 18, 2012
3,527
675
If the real part is 0, you know the polar form has an angle of 90°. The sign depends on the value of the imaginary portion of the complex number, so, a net capacitance is at -90°, where a net inductance is at a positive 90°, no calculation necessary.

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3. ### crutschow Expert

Mar 14, 2008
16,903
4,608
Dividing by zero is not foolish. Any number divided by zero is infinity and the arctangent of infinity is 90°.

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4. ### fektom Thread Starter New Member

Aug 20, 2012
16
0
Ok!
So the final impedance is going to be more capacitive than inductive.

Last edited: May 22, 2013
5. ### WBahn Moderator

Mar 31, 2012
20,437
5,838
And hopefully in trigonometry they taught you to not let your calculator do your thinking for you.

What is the tangent of ±90°?

BTW: I applaud you for wanting to walk through the steps instead of just throwing the right calculator at the problem. You are on the right track and just need to think about the fundamentals that apply to the problem (and that most calculators don't embody).

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