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Parallel RL circuits

  1. Mar 5, 2015 #1


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    This query comes from not being able to answer some one else's query :oops:
    it show's I have been away from textbooks and classrooms way too long ( see my signature )
    I have forgotten my basics

    These RL circuits

    RL cct.GIF
    Now I realise A and B are just the same with the RL rotated
    and following that C would be the same as B with the load inserted below the RL
    For C ... What difference is there if the load is say
    1) antenna
    2) some other circuit on a separate module ( possibly/probably capacitively coupled)

    or if circuit B and the load was parallel across the source and the RL network ?

    Bouncing around a number of www sites infers that the parallel RL is used as a form impedance matching
    between the source and load ?
    I cannot find specific comments for another use ... is there one ?
    I was considering a filter ....

    with ref to ....
    With an ac signal applied to it, the parallel RL circuit shown below offers significant impedance to the flow of current. This impedance will
    change with frequency, since that helps determine XL, but for any given frequency, it will not change over time.

    Then gets into a mass of complex impedance calc's as most sites do

    one place seen was in an FM phase modulator circuit ....


    I haven't played with phase modulation circuits ... so not too sure of the process there

    a little help in understanding please :smile:

  2. jcsd
  3. Mar 5, 2015 #2

    jim hardy

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    Seems to me that "mass of calc's" was unnecessary.

    Why they let it get more complex than this, their own starting point,

    is a mystery to me.

    You'll remember this from 1960s....

  4. Mar 5, 2015 #3


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    @jim hardy haha .... why do people do that to us poor non-mathematical people
    and 1+1=2 also depends on who you are ... a statistician, mathematician, an engineer or a philosopher :wink:

    But we digress
    Any help with understanding what this RL circuit is doing ?

  5. Mar 5, 2015 #4

    jim hardy

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    Circuit C ?

    I'd treat it as a voltage divider.
    Vout = Vin X (Rload)/(Rload + (R1 parallel L1))

    and plot Vout/Vin for frequencies where XL<< either resistor to XL>> either resistor.

    At DC, Vout = Vin
    and when XL>> either resistor, Vout = Vin X Rload/(R1+Rload)
    If one sets Rload = K*R1
    there's some frequency where XL = R1, ω = R1/L1, so one could derive a family of curves of attenuation vs frequency.for various K's

    Sound plausible?

    old jim
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