## Qaisersatti 3 years ago Anyone HW9?

1. hrene

ya got some answers of h9p1 .25 ???? ???? 0 ??? 0 0 0

2. Jose.oliveira2

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)))

3. Jone133

What formulas did you use for the 7.5 values? Thanks :)

4. Jose.oliveira2

0.5*L*I^2, because in this time the capacitor is full or consider a open circuit.

5. Jone133

Thank you! Thank worked perfectly! Have a medal :)

6. Jose.oliveira2

LAB9 C = 126.5e-12 R = 25000 in the Fig2, change only D for 58% (H=2m, and C=1u)

7. hejå

@jose.oliveira2 can you show a picture as i dont understand the lab

8. hejå

becuase 58 % doent work for me when changin D so whats the formula please

9. machu

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

10. Jose.oliveira2

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

11. hejå

thank you and can you say the formula for solving 2 and 3 as my variables are diffrent

12. Jose.oliveira2

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)))

13. amer83