Qaisersatti
  • Qaisersatti
Anyone HW9?
MIT 6.002 Circuits and Electronics, Spring 2007
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
ya got some answers of h9p1 .25 ???? ???? 0 ??? 0 0 0
anonymous
  • anonymous
H9P1 0.25 7.5 7.5 0 HELP 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)))
anonymous
  • anonymous
What formulas did you use for the 7.5 values? Thanks :)

Looking for something else?

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

More answers

anonymous
  • anonymous
0.5*L*I^2, because in this time the capacitor is full or consider a open circuit.
anonymous
  • anonymous
Thank you! Thank worked perfectly! Have a medal :)
anonymous
  • anonymous
LAB9 C = 126.5e-12 R = 25000 in the Fig2, change only D for 58% (H=2m, and C=1u)
anonymous
  • anonymous
@jose.oliveira2 can you show a picture as i dont understand the lab
anonymous
  • anonymous
becuase 58 % doent work for me when changin D so whats the formula please
anonymous
  • anonymous
lAB 9: FIRST ENTER ANSWERS C=126.5e-12,R=25000 AND THEN IN THE DIAGRAM WHERE YOU SEE PROPERTIES (square(0,3,30k,50),change value of 50 to 58.Then do a Transient analysis and click "check" button
anonymous
  • anonymous
And the H9P1 Q5 is -23.5619449019 the count is -L*2*pi*f*sin(2*pi*.f*t) 2*pi*f = w0 -15*2*pi*0.25*sin(2*pi*.25*5) = -23.5619449019 And the LAB is
1 Attachment
anonymous
  • anonymous
thank you and can you say the formula for solving 2 and 3 as my variables are diffrent
anonymous
  • anonymous
H9P1 a) 0.250122093132 b) 7.5; the formula is 0.5*L*I^2, because capacitor is full c) 7.5; the formula is 0.5*L*I^2, because in this time, consider the capacitor on open circuit 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)))
anonymous
  • anonymous
Hi jose, H9P2 f and g did not work with me? could you please help

Looking for something else?

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