A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Homework 11, Part1 question c)
For f=740k
L=0.65mH
RL=4ohm
RC=490k
It's the only question left, I tried values from 4 to 10, but it wont check, can't get a working formula either... Any help? :)
anonymous
 3 years ago
Homework 11, Part1 question c) For f=740k L=0.65mH RL=4ohm RC=490k It's the only question left, I tried values from 4 to 10, but it wont check, can't get a working formula either... Any help? :)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hello C = 71.1645106214 pF Band = 5.54357213286 kHz If you second part has the value f=1180kHz C = 27.98 pF Band = 12.58kHz If you got green marks, please click in "best response" :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi plz tell me the formula for bandwidth in h11p1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the bandwidth: dw = RL/L + 1/(Rc *C) after that, find f=(w/2*PI)/(1000) If works, please click at "Best response" :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hey it's not working my values are RL≈4.0Ω, L≈0.65mH, RC≈490.0kΩ, c=46pF, f=920.0kHz & for another one 3 bit f=1030.0kHz & c=36pF plzz. give me answer...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hello gurukandula I am not sure if this values are correct. Please, check if its ok :) For 920KHz C = 46.04 pF Bandwidth = 8.03428050596 KHz For 1030KHz C = 36.73 pF Bandwidth = 9.82248626529 KHz If works, please click at "Best Response"
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.