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olimpiacaj
 2 years ago
Best ResponseYou've already chosen the best response.01 = (sqrt (1 / (C * L))) / (2 * pi)

machu
 2 years ago
Best ResponseYou've already chosen the best response.4H9P1: 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

Jose.oliveira2
 2 years ago
Best ResponseYou've already chosen the best response.1H9P1 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)))

hejå
 2 years ago
Best ResponseYou've already chosen the best response.0what is f in the equation L*2*pi*f*sin(2*pi*f*t). vC(5) = 219.9

Jose.oliveira2
 2 years ago
Best ResponseYou've already chosen the best response.1f is frequence, the awnser of a).

Jose.oliveira2
 2 years ago
Best ResponseYou've already chosen the best response.1No sin is SINE, just change L, f and t in the calculator course.

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.2My 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
 2 years ago
Best ResponseYou've already chosen the best response.2Kindly 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
 2 years ago
Best ResponseYou've already chosen the best response.2pt 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
 2 years ago
Best ResponseYou've already chosen the best response.2now lest with ii & iii....working on it

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.2finally done it toooo... (ii) & (iii) answers are 15,15 in my case.....

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.2so 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)))

akash97
 2 years ago
Best ResponseYou've already chosen the best response.0@ Qaisersatti: can u tell me the answer of 5th que in H9P1 if t=9s

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.2@akash97 sure please write the complete question and also the frequency value in P1 (i)

Qaisersatti
 2 years ago
Best ResponseYou've already chosen the best response.2@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

akash97
 2 years ago
Best ResponseYou've already chosen the best response.0@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?

Jose.oliveira2
 2 years ago
Best ResponseYou've already chosen the best response.1because 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)

chiragsala
 2 years ago
Best ResponseYou've already chosen the best response.1@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
 2 years ago
Best ResponseYou've already chosen the best response.2@chiragsala please write your complete question and the value of frequency in the Pt 1 (i) only then i can solve your question..:)

1431mahi
 2 years ago
Best ResponseYou've already chosen the best response.0didnt got the f,g correct in H9p2 any body plz help

chiragsala
 2 years ago
Best ResponseYou've already chosen the best response.1@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.

chiragsala
 2 years ago
Best ResponseYou've already chosen the best response.1@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.

nirvanaguy
 2 years ago
Best ResponseYou've already chosen the best response.5@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

happysingh
 2 years ago
Best ResponseYou've already chosen the best response.0h9p1  e) 15.7079632679 voltage across capacitor
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