Question

Still strugling with h9p1 part b,c,e ? can anyone help ...here are my values for the problem....L=90.0H and C=4.5mF.
A=2/π=0.64 Coulombs at time t=5.0s.

Answer 1

give me also iL(0)

Answer 2

Part1
The natural frequency is sqrt(1/(L*C)) / (2*pi), 0.25.
Part2
When you look at the energy stored in the system, you look at energy stored in both the
inductor and the capacitor. At the initial time, there is no energy stored in the capacitor.
Thus, the initial energy stored is ½*L * iL^2 + ½*C*vC^2 = ½ * L * iL^2 = 70.
Part3
There is no change in energy since there is no resistor to consume energy. Therefore, the
energy stored is still 70 Joules.
Part4
At the time just before the impulse happens, the current through the inductor is found in the
following way:
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.
Part5
From the previous part, we know that vC(t) = -L*2*pi*f*sin(2*pi*f*t).
vC(5) = -219.9
Part6
At the time just after the impulse, the capacitor is shorted, and the inductor is open.
Therefore, all the charges go to the capacitor at the time of change. Therefore, there is no
change in iL = 0.
Part7
At the time just after the impulse, the voltage looks like the following:
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
Part8
The total energy stored after the impulse is, using the equation from before of ½ * L*iL(5) ^2
+ ½ * C * vC^2 = 0.

Answer 3

it cant be more detailed....