Electrical – Proof of formula to convert from parallel to series impedance

impedance

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):

\$Z_s=R_s+jX_s\$

and the complex representation of admittance:

\$Y_p=G_p+jB_p\$

which gives:
\$\frac{1}{Z_p}=\frac{1}{R_p}+j\frac{1}{X_p}\$, then, with some algebric manipulation:
\$Z_p=\frac{R_pX_p}{X_p+jR_p}\$

After that, I do \$Z_s=Z_p\$, meaning:

\$R_s+jX_s=\frac{R_pX_p}{X_p+jR_p}\$

To bring the complex term to the numerator, I multiply by the conjugate of the denominator:

\$R_s+jX_s=\frac{R_pX_p}{X_p+jR_p}*\frac{X_p-jR_p}{X_p-jR_p}=\frac{R_pX_p^2-jR_p^2X_p}{R_p^2+X_p^2}=\frac{R_pX_p^2}{R_p^2+X_p^2}+j\frac{-R_p^2X_p}{R_p^2+X_p^2}\$

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?

Best Answer

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.

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