A community for students.
Here's the question you clicked on:
 0 viewing
Qaisersatti
 3 years ago
Homework 9
Pt 1 (ii,iii & v)
Pt 2 (iv,v,vii) any help please
Qaisersatti
 3 years ago
Homework 9 Pt 1 (ii,iii & v) Pt 2 (iv,v,vii) any help please

This Question is Closed

Nurali
 3 years ago
Best ResponseYou've already chosen the best response.0Question 1 2. 70 3. 70 5. 219.9 Question 2 didn't Complete please tell me answer of question 2 i,ii,iii,vi

Nurali
 3 years ago
Best ResponseYou've already chosen the best response.0if correct please click best response

Jose.oliveira2
 3 years ago
Best ResponseYou've already chosen the best response.3H9P1 0.250122093132 7.5 7.5 0 23.5619449019 0 0 0 H9P2 L*C L/R 1 HELP Help (1/R)*sqrt(L/C) HELP H9P3 b/(2*m) 1/(sqrt(m*1/k)) 1/(2*pi*sqrt(m*1/k)) (2*m)/(sqrt(m*(1/k)))

Jose.oliveira2
 3 years ago
Best ResponseYou've already chosen the best response.3Reviwe H9P1 a) 0.250122093132 b) 7.5 c) 7.5 d) 0 e) 23.5619449019 f) 0 g) 0 h) 0 H9P2 a) L*C b) L/R c) 1 d) sqrt(1/(L*C))/(2*pi) e) 1/(2*R*C) f) (1/R)*sqrt(L/C) g) 2*sqrt(L/C) H9P3 a) b/(2*m) b) 1/(sqrt(m*1/k)) c) 1/(2*pi*sqrt(m*1/k)) d) (2*m)/(sqrt(m*(1/k)))

olimpiacaj
 3 years ago
Best ResponseYou've already chosen the best response.0At the time just before the impulse happens what is the current iL(5.0−), in Amperes, through the inductor? Equation???

Jose.oliveira2
 3 years ago
Best ResponseYou've already chosen the best response.3iL(t) = A*cos(2*pi*f*t) + B*sin(2*pi*f*t) vC(t) = L*diL/dt = L*2*pi*f*(A*sin(2*pi*f*t) + B * cos(2*pi*f*t)) The initial conditions mean that A = 1, and B = 0, (in my case) So, iL (t) = cos(2*pi*f*t), vC (t) = L*2*pi*f*sin(2*pi*f*t) iL(5) = cos(2*pi*0.25*5) = cos(2.5*pi) = 0. If you have another values, change only.

Qaisersatti
 3 years ago
Best ResponseYou've already chosen the best response.0My Q No 1 (ii,iii & v) is not coming correct can anyone comment. The current source puts out an impulse of area A=2/π=0.64 Coulombs at time t=5.0s. At t=0 the state is: vC(0)=0.0 and iL(0)=1.0. The equation governing the evolution of the inductor current in this circuit is d2iL(t)dt2+1LCiL(t)=ALCδ(t−5.0) (ii) At the initial time what is the total energy, in Joules, stored in the circuit? (iii) At the time just before the impulse happens t=5.0− what is the total energy, in Joules, stored in the circuit? (v) At the time just before the impulse happens what is the voltage vC(5.0−), in Volts, across the capacitor? please comment...

Jose.oliveira2
 3 years ago
Best ResponseYou've already chosen the best response.3Thats the same values and questions in my, i don´t know what happens.

Jose.oliveira2
 3 years ago
Best ResponseYou've already chosen the best response.3try this energy stored=½*L*iL^2 (b and c) vC(t) = L*2*pi*f*sin(2*pi*f*t) for (e)

Qaisersatti
 3 years ago
Best ResponseYou've already chosen the best response.0@Jose.oliveira2 thanks brother its already done...thanks
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.