0
\$\begingroup\$

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?

\$\endgroup\$
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.