Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Homework 9 Pt 1 (ii,iii & v) Pt 2 (iv,v,vii) any help please

MIT 6.002 Circuits and Electronics, Spring 2007
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Question 1 2. 70 3. 70 5. -219.9 Question 2 didn't Complete please tell me answer of question 2 i,ii,iii,vi
if correct please click best response
H9P1 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)))

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Reviwe 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)))
At the time just before the impulse happens what is the current iL(5.0−), in Amperes, through the inductor? Equation???
iL(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.
My 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...
Thats the same values and questions in my, i don´t know what happens.
try this energy stored=½*L*iL^2 (b and c) vC(t) = -L*2*pi*f*sin(2*pi*f*t) for (e)
@Jose.oliveira2 thanks brother its already done...thanks

Not the answer you are looking for?

Search for more explanations.

Ask your own question