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Problem 7A of The Electronics of Radio asks the reader to find the relationships between resistance and reactance values in any arbitrary series/parallel circuit and their counterparts in a parallel/series equivalent circuit with the same impedance; i.e., given a parallel circuit with resistance Rp and reactance Xp, what is the relationship between those values and the the resistance Rs and reactance Xs of a series circuit with the same impedance, and vice-versa.

As a starting point for solving the problem, the author reminds the reader that Qs= Xs/Rs and Qp= Rp /Xp, and that if the impedances of these circuits are to be the same, then these Q values must be same. Therefore, I thought it was logical and correct to conclude that, for example, Xs=(RpRs)/Xp.

This felt easier than it should be, so I looked it up on the internet and found relationships like Xs=(Xp Rp2)/(Xp2+Rp2) (Source) . I can’t figure out how to get from the aforementioned relationships to that equation, and the websites I can find don’t go through all the intermediate steps either. Could anybody please help me understand this problem?

enter image description here

(Fig 8.5c is only relevant for the later parts of the problem)

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    \$\begingroup\$ I am aware of and am fairly certain that I understand all of the concepts that you have mentioned; I read through the entire chapter on phasors and copied out all the important equations to make sure I got the concepts in my head. However, what I seem to be having trouble with is connecting it all together, and would like to be able to understand the flaw in the reasoning for my initial solution. \$\endgroup\$
    – Architect
    Commented Dec 23, 2021 at 5:45
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    \$\begingroup\$ Do it via complex notation: \$R_s+jX_s=\frac{jX_pR_p}{R_p+jX_p}\$, then rationalise and equate real parts, and imaginary parts. \$\endgroup\$
    – Chu
    Commented Dec 23, 2021 at 9:11

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Chu explains it well in a comment; perfectly and with wonderful brevity!

You know that in the parallel case it must be that \$Z_p=R_p \mid\mid jX_p = \frac{R_pjX_p}{R_p+jX_p}\$ and that in the series case it must be that \$Z_s=R_s+jX_s\$. If these are supposed to be equal, then:

$$\begin{align*} R_s+jX_s&=\frac{R_pjX_p}{R_p+jX_p} \\\\ &=\frac{R_pjX_p}{R_p+jX_p}\cdot \frac{R_p-jX_p}{R_p-jX_p} \\\\\ &=\frac{R_p X_p^{\,2}+j R_p^{\,2}X_p}{R_p^{\,2}+X_p^{\,2}} \\\\ &=\frac{R_p X_p^{\,2}}{R_p^{\,2}+X_p^{\,2}}+j\frac{R_p^{\,2}X_p}{R_p^{\,2}+X_p^{\,2}} \end{align*}$$

I think you can see the obvious from here.

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  • \$\begingroup\$ Thank you for this, I’ll study this and figure it all out. One question, how do I reconcile this with my understanding that Q=Xs/Rs=Rp/Xp, because if I follow this, then the relationship ends up being much simpler than it is as you have shown, but this is what I inferred from the way the problem was presented. What is the error with that equation? \$\endgroup\$
    – Architect
    Commented Dec 23, 2021 at 11:19
  • \$\begingroup\$ @Architect It's just a hypothetical (made up by someone who likes to make up such things for purposes that aren't stated yet by what you've read) ratio of the real to the imaginary parts. In the series case it's really easy to see that this is RS/XS. In the parallel case, as you can see from the last equation at the bottom, you have to take the ratio of the first to the second part. The denominators of each entirely cancel out and the factors in each of the numerators then devolve into the ratio of XP/RP. I'm sure you can see that fact. \$\endgroup\$
    – jonk
    Commented Dec 23, 2021 at 17:57

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