# Complex Impedance and Polar Form

Discussion in 'Math' started by pianolife, Jan 5, 2013.

1. ### pianolife Thread Starter New Member

Nov 15, 2012
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0
Dear members of all about circuits, I've got a question about Complex impedance and polar form.
I've read it in your ebook, and I'm practicing with some exercises. But here I am asked to write a kind of "general" formula, without values, and I just wanted to ask if this is correct.
What is the impedance Z between terminals A and B of the networks shown below? Express your answers in polar form.

All the best and happy new year!

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2. ### MAOR19 New Member

Jan 5, 2013
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i think it's supposed to be " |z|= sqrt (R^2 + Xc^2)

3. ### pianolife Thread Starter New Member

Nov 15, 2012
28
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Oh yes, MAOR19, no j under the square root! Thank you so much!

What about the rest? Is that correct?
Thank you!

4. ### MAOR19 New Member

Jan 5, 2013
3
0
it's also supposed to be Z1*Z2/(Z1+Z2)

5. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
Here is were checking the units comes into play.

You final answer is Z1/Z2. What are the units on this? Nothing - it is dimensionless. But you are looking for an effective impedance, so your answer has to have units of impedance (namely ohms), yet it doesn't. So you KNOW this answer is incorrect.

Always track your units throughout the work.
Always verify that the units work out.

6. ### pianolife Thread Starter New Member

Nov 15, 2012
28
0
Sorry I was totally wrong in the last message.
I just tried to go on with this problem, but I can not simplify them further...
I am really stuck

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7. ### WBahn Moderator

Mar 31, 2012
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I haven't walk through your work in detail, but you appear to be mixing two different definitions of capacitive reactance.

Some people, particularly those that don't ever deal with complex impedance, like to define both inductive and capacitve reactance as positive values and then memorize a bunch of formulas that have minus signs here and there. The better way to deal with this is to simply define reactance as the imaginary part of the complex impedance.

The impedance of R, L, and C are
$
Z_R = R

Z_L = j\omega L

Z_C = \frac{1}{j\omega C} = j \left( \frac{-1}{\omega C} \right)
$

The reactance of each is then simply

$
X_R = 0

X_L = \omega L

X_C = - \frac{1}{\omega C}
$

This lets you work with the generic notion of reactance and relieves you of having to remember whether to use a minus sign in your formulas depending on whether a particular reactance happens to turn out inductive or capacitive -- the sign is carried as part of the reactance value itself.

8. ### pianolife Thread Starter New Member

Nov 15, 2012
28
0
You are right, WBahn, I don't get which way would be easier (or more efficient) to work out this problem...
Talking with the lecturer, he would use jΩ's values, but the further I go, the messier it gets... I mean, I get to a point when I could multiply the denominator of the fraction for its complex conjugate, in order to get a real number, at least in the denominator.
But all this work seems really confusing, even if it is eventually right.

9. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
I firmly believe that it is better to let a reactance be a reactance and be done with it. If it happens to be negative, then it is capacitive and vice versa.

So for this network we have

$
Z_T = Z_1 || Z_2
\text{where}
Z_1 = R_1 + jX_C
Z_2 = R_2
\text{thus}
Z_T = \frac{Z_1 Z_2}{Z_1 + Z_2}
Z_T = \frac{(R_1 + jX_C)R_2}{R_1 + jX_C + R_2}
Z_T = R_2\frac{R_1 + jX_C}{(R_1 + R_2) + jX_C}
\text{putting the numerator and denominator in polar form}
Z_T = R_2\frac{\sqrt{R_1^2 + X_C^2} \angle \tan^{-1} \left( \frac{X_C}{R_1} \right) }{\sqrt{(R_1 + R_2)^2 + X_C^2} \angle \tan^{-1} \left( \frac{X_C}{R_1+R_2} \right)}
Z_T = R_2\sqrt{\frac{R_1^2 + X_C^2}{(R_1 + R_2)^2 + X_C^2} } \angle \left[ \tan^{-1} \left( \frac{X_C}{R_1} \right) - \tan^{-1} \left( \frac{X_C}{R_1+R_2} \right) \right]
\text{now you can substitute in}
X_C = - \frac{1}{\omega C}
\text{to get}
Z_T = R_2\sqrt{\frac{R_1^2 + {\left( \frac{1}{\omega C} \right)}^2}{(R_1 + R_2)^2 + { \left( \frac{1}{\omega C} \right) }^2} } \angle \left[ \tan^{-1} \left( \frac{-1}{\omega R_1 C} \right) - \tan^{-1} \left( \frac{-1}{\omega (R_1+R_2)C} \right) \right]
\text{It's important to note that the angles are in the fourth quadrant, not the second.}
\text{You can play a number of games to rearrange this in order to highlight one aspect or another. For instance:}
Z_T = R_2\sqrt{\frac{1 + (\omega R_1 C)^2 }{1 + (\omega (R_1 + R_2) C)^2} } \angle \left[ \tan^{-1} \left( \frac{-1}{\omega R_1 C} \right) - \tan^{-1} \left( \frac{-1}{\omega (R_1+R_2)C} \right) \right]

$

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10. ### pianolife Thread Starter New Member

Nov 15, 2012
28
0
Wow! Thank you so much WBahn, I wasn't expecting such a huge help! Thank you so much!

Since it looks like you are really "on the ball" with this subject, I am just guessing if is worthy substituting (R1+R2) with Rtot.

Just, possibly, to make it more "legible"...

Thank you very much, I owe you at least a pint

11. ### WBahn Moderator

Mar 31, 2012
17,454
4,701
You can, but then you have to be sure to clearly define what Rtot is. Don't assume that "it's obvious" -- the correct name for obvious things that aren't define is "undefined"!

12. ### pianolife Thread Starter New Member

Nov 15, 2012
28
0
My ignorance is more then defined