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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    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.
     
  2. bertus

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    Apr 5, 2008
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  3. EJR5

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    Jan 20, 2015
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  4. WBahn

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    Mar 31, 2012
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    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
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    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
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    The equivalent resistance of two resistors in parallel (say R1 is in parallel with R2) is:
    <br />
Requivalent=\frac{R1*R2}{R1+R2}<br />

    So. Rs+Xs=z. z is known to you. Rp is given to you.
    All you have to do is:
    <br />
z=\frac{Rp*Xp}{Rp+Xp}<br />
    You know z, you know Rp. Solve for Xp.
     
  7. EJR5

    Thread Starter New Member

    Jan 20, 2015
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    I know RS and XS need to find RP and XP so I have two unknowns
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Then pick any Rp, say 1 kOhm and solve for Xp.
     
  9. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    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}

    From which your required resulting equations are readily determined.
     
  10. WBahn

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    Mar 31, 2012
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    Part of the problem likely stems from the fact that Z does NOT equal RS+XS, assuming the XS is the reactance of the inductor.

    Is it safe to assume that you are not familiar with complex impedance, but are rather working only with reactance? If so, then what you need to do is find the expressions for the magnitude of the impedance for the series circuit and the phase angle of the impedance for the series circuit. The do the same for the parallel circuit. Once you have those expressions, set the two magnitude equations equal to each other and set the two angle equations equal to each other. That will give you your two equations in two unknowns.
     
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