Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Qaisersatti

  • 3 years ago

homework 11 please help....

  • This Question is Closed
  1. Jose.oliveira2
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    H11P1 a) 1/(RC*C) + RL/L b) (thats the formula = 1/(2*pi*f)^(2)*L) c) (thats the formula = 1/(RC*C) + RL/L) d) (thats the formula = 1/(2*pi*f)^(2)*L) e) c) (thats the formula = 1/(RC*C) + RL/L) H11P2 a) (The Formula = vm*0.1) b) Cs / Rp c) Cs / Rp d) 0.1*VM(t) H11P3 a)C b)R c)L d ? e) L f) 1000 g) 1 h ?

  2. Abhaypratap
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    H11P3 d) 4

  3. Jose.oliveira2
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    tks

  4. Jose.oliveira2
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    h = 16

  5. brunofornari
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Jose.oliveira2 the formula doesn't work, or I'm doing it wrong. My values are: L≈0.65mH, RL≈4.0Ω, RC≈490.0kΩ and the tunning capacitors are: 46.0416895277pF and 17.0912179362pF could you help me plz?

  6. Jone133
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have the same values as brunofornari, and I can't get the results to check ... Any help? Also f=860kHz, So I guess for H11P1b): 1/(2*pi*f)^(2)*L = 1/(2*pi*860k)^(2)*0.65m= 2.22616175526e-17 = 0.0000261p am I doing it right?

  7. brunofornari
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Jone133 No.. your question is: 1/(((2*pi*860k)^(2))*0.65m) = 5.26902190593e-11 = 52.6902190593pF

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy