Complex Impedance - Real and Imaginary question

Discussion in 'General Electronics Chat' started by pnew, Jan 22, 2013.

1. pnew Thread Starter New Member

Jan 22, 2013
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0
Hello, I have used the forums regularly, but only as an observer who has benefited from the knowledge. However, i'm trying to find an answer to a question that is no doubt simple.

When deriving the analytical solution for a parallel RC complex impedance there is a method of calculating the real and imaginary components of the complex impedance:

Real:
(R x XC^2) / (R^2 + X^2)

Imaginary:
(R^2 x XC) / (R^2 + X^2)

As far as I am aware, these values are measured in Ohms, so what exactly do they represent? If we put some figures in it may help me to explain.

Say R = 5Ω, C = 100uF and f=100Hz.

Therefore, XC = 15.9Ω

Now, according to the above formulas the real and imaginary parts are;
Real = 4.55Ω
Imag = 1.42Ω

How can the real part be less than the resistance (real part of the impedance?) And if it is, what exactly does it mean?

Or are they not measured in ohms, but rather a ratio?

I hope this post makes sense. Thank you.

2. Papabravo Expert

Feb 24, 2006
10,340
1,850
Do you mean:

R + j0 in parallel with 0 + jXc

If so then:
The units of the real part are (Ohms^3)/(Ohms^2) or Ohms
The units of the imaginary part are also (Ohms^3)/(Ohms^2) or Ohms

When you compute the magnitude of a complex quantity you square both parts, add them together, take the square root and the units are still Ohms.

If you go back to DC circuits you will remember that the parallel combination of two resistors is always less than either of the two resistors. AC circuits behave the same way. If you look at the magnitude, it works out to 4.75 Ohms, which is consistent with the result from DC circuits.

Last edited: Jan 22, 2013
3. pnew Thread Starter New Member

Jan 22, 2013
3
0

Yes, I meant R + j0 in parallel with 0 + jXc

Yes, I can calculate that the magnitude is 4.75 Ohms, and that the resistance in parallel is always less, but i'm still confused as to what the real term actually is.

If I have a resistance of 5Ω (R + j0), then why is the real part of the complex impedance 4.55Ω and not 5Ω?

I just don't understand how the real part can be less than 5Ω? I understand the methods of calculating the magnitude, phase angle etc. but this is vexing me! All textbooks, references I have found state that the real part is the resistance, but it is not...it's only a part of the resistance.

If I wanted to plot the spectroscopic real impedance, I thought it should be frequency independent, but when I calculate the real and imaginary parts they both change with frequency. Does the real part represent a ratio between the real and imaginary components relative to how much current flows through that element?

I fear I may be over thinking things here, and getting myself more confused than necessary.

4. pnew Thread Starter New Member

Jan 22, 2013
3
0
Hang on, I may have answered this myself.

I think I was/ am overlooking the fact that the real impedance of the parallel-RC circuit which i presented is in fact the magnitude, but represented in the cartesian form rather than polar?

Is this a case of the phrase 'real' being used in two contexts?