anonymous
  • anonymous
can any one help me solving H9p1?
MIT 6.002 Circuits and Electronics, Spring 2007
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katieb
  • katieb
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anonymous
  • anonymous
1 = (sqrt (1 / (C * L))) / (2 * pi)
anonymous
  • anonymous
H9P1: 1) Natural frequency=sqrt(1/(L*C))/(2*pi)= 0.25 2)Initial energy stored=½*L*iL^2 + ½*C*vC^2= ½*L*iL^2=70 3)No change in energy so answer is 70 Joules 4)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. Therefore, 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. 5)vC(t) = -L*2*pi*f*sin(2*pi*f*t). vC(5) = -219.9 6)no change in iL = 0 7)C * dv/dt = i Integrating both sides gives: C*int(dv) = int(i dt) Change in v = 1/C * A vC = vC_Previous + 1/C*A = 0 8)½ * L*iL(5) ^2 + ½ * C * vC^2 = 0
anonymous
  • anonymous
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)))

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anonymous
  • anonymous
what is f in the equation -L*2*pi*f*sin(2*pi*f*t). vC(5) = -219.9
anonymous
  • anonymous
f is frequence, the awnser of a).
anonymous
  • anonymous
and sin i source in?
anonymous
  • anonymous
No sin is SINE, just change L, f and t in the calculator course.
Qaisersatti
  • Qaisersatti
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)
Qaisersatti
  • Qaisersatti
Kindly help me with (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?
Qaisersatti
  • Qaisersatti
pt v is done :) At the time just before the impulse happens what is the voltage vC(5.0−), in Volts, across the capacitor? the answer is -47.1
Qaisersatti
  • Qaisersatti
now lest with ii & iii....working on it
Qaisersatti
  • Qaisersatti
finally done it toooo... (ii) & (iii) answers are 15,15 in my case.....
Qaisersatti
  • Qaisersatti
so if u have =30.0H and C=13.51mF. A=2/π=0.64 Coulombs at time t=5.0s. then your answers are H9P1 a) 0.250122093132 b) 15 c) 15 d) 0 e) -47.1 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
@ Qaisersatti: can u tell me the answer of 5th que in H9P1 if t=9s
Qaisersatti
  • Qaisersatti
@akash97 sure please write the complete question and also the frequency value in P1 (i)
Qaisersatti
  • Qaisersatti
@Akash97 formula of part v is vC(t) = -L*2*pi*f*sin(2*pi*f*t) put your value of L f and t here.... ur t=9
anonymous
  • anonymous
@Qaisersatti : L=190.0H and C=2.13mF. f=0.25 At t=0 the state is: vC(0)=0.0 and iL(0)=1.0. At the time just before the impulse happens what is the voltage vC(9.0−), in Volts, across the capacitor?
Qaisersatti
  • Qaisersatti
ok wait
anonymous
  • anonymous
because your question, change the Capacitor and Inductor (L) in any case, use this formula energy stored=½*L*iL^2 (b and c) vC(t) = -L*2*pi*f*sin(2*pi*f*t) for (e)
Qaisersatti
  • Qaisersatti
@Akash97 -298.3
anonymous
  • anonymous
@Qaisersatti i am still not getting green tick for 15,15 and -47.1 answers...but other answers are same as you have written above.
Qaisersatti
  • Qaisersatti
@chiragsala please write your complete question and the value of frequency in the Pt 1 (i) only then i can solve your question..:)
anonymous
  • anonymous
didnt got the f,g correct in H9p2 any body plz help
anonymous
  • anonymous
@Qaisersatti i was talking about the part b,b,e of h9p1.........as you got the answer i asked you that it was not correct for me.
anonymous
  • anonymous
@1431mahi for f the answer is : (1/R)*sqrt(L/C) for g answer is: 2*sqrt(L/C) if you got correct then pls give best response.
anonymous
  • anonymous
@1431mahi the answers are the inverse of that.. so R*sqrt(C/L) for f and .5sqrt(L/C) for g.. that's because this is a parallel RLC circuit
anonymous
  • anonymous
TY @nirvanaguy
anonymous
  • anonymous
h9p1 --- e) -15.7079632679 voltage across capacitor

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