A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

Jose.oliveira2
 2 years ago
Best ResponseYou've already chosen the best response.1H11P1 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 ?

brunofornari
 2 years ago
Best ResponseYou've already chosen the best response.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?

Jone133
 2 years ago
Best ResponseYou've already chosen the best response.0I 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= 2.22616175526e17 = 0.0000261p am I doing it right?

brunofornari
 2 years ago
Best ResponseYou've already chosen the best response.0@Jone133 No.. your question is: 1/(((2*pi*860k)^(2))*0.65m) = 5.26902190593e11 = 52.6902190593pF
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.