Question

can any one help me solving H9p1?

Answer 1

1 = (sqrt (1 / (C * L))) / (2 * pi)

Answer 2

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

Answer 3

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)))

Answer 4

what is f in the equation -L*2*pi*f*sin(2*pi*f*t).
vC(5) = -219.9

Answer 5

f is frequence, the awnser of a).

Answer 6

and sin i source in?

Answer 7

No sin is SINE, just change L, f and t in the calculator course.

Answer 8

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)

Answer 9

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?

Answer 10

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

Answer 11

now lest with ii & iii....working on it

Answer 12

finally done it toooo...

(ii) & (iii) answers are 15,15 in my case.....

Answer 13

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)))

Answer 14

@ Qaisersatti: can u tell me the answer of 5th que in H9P1 if t=9s

Answer 15

@akash97 sure please write the complete question and also the frequency value in P1 (i)

Answer 16

@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

Answer 17

@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?

Answer 18

ok wait

Answer 19

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)

Answer 20

@Akash97 -298.3

Answer 21

@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.

Answer 22

@chiragsala please write your complete question and the value of frequency in the Pt 1 (i) only then i can solve your question..:)

Answer 23

didnt got the f,g correct in H9p2 any body plz help

Answer 24

@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.

Answer 25

@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.

Answer 26

@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

Answer 27

TY @nirvanaguy

Answer 28

h9p1 ---
e) -15.7079632679 voltage across capacitor