1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Vout/Vin of an RL filter

  1. May 28, 2014 #1
    I need to find Vout/Vin for an RL filter. The filter in question is set up as low pass filter with a resistor of R=9 ohms, a inductance of 76mH and angular frequency of 1.6ωc.

    To my knowledge Vin should be the vector sum of the voltages of each component so the equation would be Vin=√((IR)2+(IωL)2).

    When trying to solve this problem I have set Vout as both the voltage of the inductor and the voltage over the resistor but neither of these has brought the correct answer
     
  2. jcsd
  3. May 28, 2014 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi stude. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Your formula for the magnitude of Vin is right. Vout is the voltage across R, i.e., between one end of the resistor and ground.

    How could you recognize your answer as probably being correct, even if you weren't told the "correct" answer? (Remember, the variable here is ω.)
     
    Last edited by a moderator: May 6, 2017
  4. May 28, 2014 #3
    I'm not sure what you're asking with the last line, it's online homework site so it tells me I'm wrong, I just don't know what to do to be right.
     
  5. May 28, 2014 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Maybe you aren't correctly expressing Vout/Vin? What are you writing? There won't be an "I" involved.

    Or maybe they want the answer in dB? What exactly is the question the computer asks?

    Quite possibly they are expecting you to specify the phase shift as well as magnitude change?
     
  6. May 28, 2014 #5
    It asks "what is the ratio between the output and input voltage amplitudes" and then "Vout/Vin=" followed by a blank to answer in
     
  7. May 29, 2014 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Show your work in detail. We can find your mistake if we see what you have done. Note that the frequency is 1.6 times the cut-off frequency. What does it mean?


    ehild
     
    Last edited: May 29, 2014
  8. May 29, 2014 #7
    Vout/Vin=Vr/√(Vr2+Vl2)

    Vout/Vin=(IR)/√((IR)2+(IωL)2)

    Vout/Vin=(I*9)/√((I*9)2+(I*1.6*ωc*0.076)2)

    ωc=1/RC, I made the assumption that I could use 1/RL because my book doesn't actually address this kind of problem.

    Vout/Vin=(I*9)/√((I*9)2+(I*1.6*1/(9*0.076)*0.076)2)

    Vout/Vin=0.999
     
  9. May 29, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member


    That is wrong. At the critical frequency, the reactance is equal of the resistance. If it is an RC circuit, XC=1/(ωC), and from XC=R ωc=1/(RC) follows.

    If it is an RL circuit, XL=ωL, from XL=R, ωc=R/L follows.

    ehild
     
  10. May 29, 2014 #9
    That was the trick, thanks a lot
     
  11. May 29, 2014 #10

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You are welcome.


    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted