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Parallel RL circuits

  1. Mar 5, 2015 #1

    davenn

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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 ....
    http://amrita.vlab.co.in/?sub=1&brch=75&sim=322&cnt=1
    Theory:
    With an ac signal applied to it, the parallel RL circuit shown below offers significant impedance to the flow of current. This impedance will
    image001%20%281%29.gif
    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 ....

    trnsfig3.gif

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


    a little help in understanding please :smile:

    cheers
    Dave
     
  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,
    image003%283%29.png

    is a mystery to me.

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

    https://totient.wordpress.com/2008/03/19/a-funny-derivation/
     
  4. Mar 5, 2015 #3

    davenn

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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 ?


    Dave
     
  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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