Resonance and Q Factor in True Parallel RLC Circuits

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  • Опубліковано 2 лис 2024

КОМЕНТАРІ • 21

  • @segos2010
    @segos2010 2 роки тому +6

    I am a teacher at a technical college in South Africa. You answered a question that I always had regarding the use of series and parallel resonant circuit formula. Thank you for your video.

  • @notsocrazyjohn5348
    @notsocrazyjohn5348 4 роки тому +1

    Thank you so much. I feel so lucky to have found this site. Cheers John

  • @irishchocolate3872
    @irishchocolate3872 2 роки тому +1

    I wish Hewlett Packard continued to make RPN scientific calculators. They have a distinct advantage in electrical engineering over algebraic machines.

  • @hawaiiskjortaa
    @hawaiiskjortaa 3 роки тому +12

    Who else was nervous over the sudden integral between 5:16-6:00

  • @zdzisawk5198
    @zdzisawk5198 2 роки тому +1

    I love your lectures. Thank you so much

  • @mtopangaprimo2556
    @mtopangaprimo2556 Рік тому +1

    May I ask a question?.Is the advanced formula for all non neglible resistors on Paraller.or the formula changes with circuit setup?.am Stuck on this.
    Thanks for the help as well as for the videos you upload.

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Рік тому

      The "advanced" formula is specifically for this particular circuit arrangement.

  • @aljosagraovac8308
    @aljosagraovac8308 2 роки тому +1

    How to get the formula for fo? i tried with impendance but i can't get that formula. for Resonance part (imaginary part must be 0 for resonance) i get something like iwL/(R^2 + (wL)^2) + wC = 0
    because of that i get a "-" which makes me problems... Any help?

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  2 роки тому +2

      Kaiser has a derivation in his book, pg 5-27: books.google.co.uk/books?id=nZzOAsroBIEC&lpg=PP1&pg=SA5-PA27#v=onepage&q&f=false which may be helpful

    • @aljosagraovac1918
      @aljosagraovac1918 2 роки тому

      @@EngineersAcademyLTD thank you, i will look into it.

  • @gambitillmind
    @gambitillmind 3 роки тому +1

    Amazing brother

  • @FlurinKoenz
    @FlurinKoenz Рік тому +1

    How do you derive Rd = L/(RC)? Thanks a lot!

    • @EngineersAcademyLTD
      @EngineersAcademyLTD  Рік тому +2

      The derivation is not very nice, but simply put it is expressing RD as the parallel arrangement of R+jXL and -jXC, substituting the expression for f0 for frequency where it arises in the reactive terms, and then simplifying.

    • @FlurinKoenz
      @FlurinKoenz Рік тому +1

      @@EngineersAcademyLTDI just figured it out: R_d = Re{Z}. The derivation was not too bad. Thanks a lot for you video, your help and your fast response!

  • @mahdigaben2963
    @mahdigaben2963 5 років тому +2

    Thank you sir!

  • @saudonmars
    @saudonmars 5 років тому +4

    Thank you

  • @yiansansoon3093
    @yiansansoon3093 4 роки тому +2

    may I ask why in this advance formula need to put (R^2) /(L^2) ?

    • @konstantinparchevsky2031
      @konstantinparchevsky2031 4 роки тому +3

      Impedance of this circuit is calculated in the same way as resistivity in the DC circuit (L and R are in series and LR are in parallel with C):
      Z = (XL+R) || XC,
      where XL=iwL, XC=1/(iwC) are the impedance of the coil and the capacitor respectively, w=2 pi f is the cyclic frequency, f is the frequency, i^2=-1 is the imaginary unit, R1||R2=R1 R2/(R1+R2) is the resistivity/impedance of resistors connected in parallel.
      By definition, resonance occurs when current I and voltage U are in phase, which means that at resonance Z is real. So, to find the resonance frequency we need to solve equation
      Im[Z]=0
      with respect to w, which gives the required formula.

  • @AkashKumar-jc9kb
    @AkashKumar-jc9kb 3 роки тому +1

    Thanks 👍

  • @dnowbrg6080
    @dnowbrg6080 5 років тому

    ty