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!

Electrical Circuits Reactive Networks Question

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data
    e3C9ZR3.png

    2. Relevant equations
    At low voltage frequency (inductors) L become short circuit and (capacitors) C becomes open circuit
    At high voltage frequency (inductors) L become open circuit and (capacitors) C becomes short circuit

    3. The attempt at a solution

    Part A
    At low frequency, all the capacitors are open, so no current cant run beyond Rs. Hence current at the load is zero so voltage is also zero

    Part B
    At high frequency, all the inductors are open, so the only path for the current is through Rs and around all the loops at the top where all the capacitors use to be (skipping all resistors)

    So now it is the voltage supply Rs and the load in series, so a voltage divider can be used to workout the voltage across the load

    V1 = V(Supply) x R1 / (R1 + R2)

    v(Load) = 10 x 35 / (35 + 30) = 5.384615

    Part C
    I am assuming the angular frequency has no effect on the voltage as T = 1/f and omega = 2PIf so the frequency cancels out
    This is most likely wrong

    Part D
    Using the same value from Part C, I would use P = V^2/R where R is 35

    I think I have the first two parts correct and the last two parts wrong
    any help?
     
  2. jcsd
  3. Jan 30, 2016 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    At low frequencies, current won't see a path through the capacitors, so why wouldn't current flow through the resistors to reach the load?
     
  4. Jan 30, 2016 #3
    Because of the short circuits were all the inductors use to be, thus the resistors and the load are isolated
     
  5. Jan 30, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    At low frequencies the inductors will quickly curtail any signal that tries to take the resistor route; They'll look like shorts to ground along the path.

    Parts (c) and (d) look like you're supposed to know something significant about the "unit cell" behavior, since there are three identical filter sections in cascade. I can identify them as so-called bridged T filters. Perhaps there's a property of them that allows you to use transfer function approach?
     
  6. Jan 30, 2016 #5
    So am I right to assume since the signals are "shorted to ground", that after current passes through Rs and the first resistor, it is "curtailed" by any inductors and no other resistors or the load receives any current, which is why I put zero.

    In regards to C and D, don't think I've learnt about any of that yet but thanks
     
  7. Jan 30, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    Yes.
     
  8. Jan 30, 2016 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    That is correct, and an improvement on how you explained it earlier.

    Any idea how you might be expected to attack this problem? Have you looked at something similar to it in class, perhaps calculated the bridged-T resonant frequencies?
     
  9. Jan 30, 2016 #8
    Alright thanks for the help and clarification :)

    Nope, no idea at all on these, going to review my notes and lectures and see if theres anything I've missed
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted