# RL Circuits

Discussion in 'Homework Help' started by EJR5, Jan 20, 2015.

1. ### EJR5 Thread Starter New Member

Jan 20, 2015
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0
can anyone help on RL series to parallel impedance matching I have the values for XS and RS and need to get RP and XP. I need to show these can be expressed the same using algebra.

Apr 5, 2008
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Jan 20, 2015
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4. ### WBahn Moderator

Mar 31, 2012
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4,788
Could you be a bit more explicit in what you are looking for. Not sure just what you mean by "matching" in this context.

How about stating a problem and showing what you have done so far on it.

5. ### EJR5 Thread Starter New Member

Jan 20, 2015
4
0
I have an RL circuit that is currently in series so I know z=RS+XS I need to make this into parallel to get RP and Xp. I have the values for R
I have a series RL circuit with a known impedance I need to convert this into a parallel circuit with the same equivalent impedance. I know Z=RS+XS for a series RL circuit and I have the values of RS & XS. I need to find RP & XS I have seen the following equations in textbooks and on the web but need to know the algebraic method that was used to get to these equations:
Rp = (RS²+XS²)/RS

Xp = J(RS²+XS²)/XS

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,377
494
The equivalent resistance of two resistors in parallel (say R1 is in parallel with R2) is:
$
Requivalent=\frac{R1*R2}{R1+R2}
$

So. Rs+Xs=z. z is known to you. Rp is given to you.
All you have to do is:
$
z=\frac{Rp*Xp}{Rp+Xp}
$

You know z, you know Rp. Solve for Xp.

7. ### EJR5 Thread Starter New Member

Jan 20, 2015
4
0
I know RS and XS need to find RP and XP so I have two unknowns

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,377
494
Then pick any Rp, say 1 kOhm and solve for Xp.

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I use admittance to simplify the process:

For the parallel case:

$G_p=\frac{1}{R_p}-j\frac{1}{X_p}$

For the series case:

$G_s=\frac{1}{R_s+jX_s}=\frac{R_s-jX_s}{R_s^2+X_s^2}=\frac{R_s}{R_s^2+X_s^2}-j\frac{X_s}{R_s^2+X_s^2}$

Clearly, if Gs and Gp are equivalent, then equating real and imaginary components leads to:

Real Part:
$\frac{1}{R_p}=\frac{R_s}{R_s^2+X_s^2}$

And

Imaginary part:
$\frac{1}{X_p}=\frac{X_s}{R_s^2+X_s^2}$