So, I tried to find a proof to the formulas to convert from parallel to series reactive circuit, which I found in the ARRL Handbook:
\$R_s=\frac{R_pX_p^2}{R_p^2+X_p^2}\$ and \$X_s=\frac{R_p^2X_p}{R_p^2+X_p^2}\$
where \$R_s\$ and \$X_s\$ are the series resistance and reactance to match the impedance of a parallel circuit with \$R_p\$ and \$X_p\$ resistance and reactance.
The resistances are positive and real values and the reactances are signed real values (positive for inductive, and negative for capacitive).

I decided to use the complex representation of impedance (I use j for i because it seems to be the convention in electronics):
and the complex representation of admittance:
which gives: \$\frac{1}{Z_p}=\frac{1}{R_p}+j\frac{1}{X_p}\$, then, with some algebric manipulation:
After that, I do \$Z_s=Z_p\$, meaning:
To bring the complex term to the numerator, I multiply by the conjugate of the denominator:
Now, to set real and imaginary parts equal, we get:
\$R_s=\frac{R_pX_p^2}{R_p^2+X_p^2}\$ and \$X_s=-\frac{R_p^2X_p}{R_p^2+X_p^2}\$

The first formula is fine, but what is wrong with the second formula? Why is there that extra "-" in front?


It must be like \$\dfrac{1}{Z_p} = \dfrac{1}{Rp} + \dfrac{1}{j.X_p}\$. In your case, the imaginary \$j\$ is in numerator.

  • \$\begingroup\$ How is that? I'm pretty sure that by substituting B_p by 1/X_p, (the definition of susceptance) the j stays in the numerator. \$\endgroup\$ – nc404 Apr 9 '16 at 14:29
  • \$\begingroup\$ In that case your Bp must be different. OK, let two resistors R1 and R2 be in parallel. Net resistance is given by 1/Rp = 1/R1 + 1/R2. You know reactance is given by j.X and resistance by R. Like in case of just resistors, when a resistor with reactive element are in parallel, it must be 1/Rp = 1/R + 1/(j.X) \$\endgroup\$ – user3219492 Apr 9 '16 at 17:45

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