Circuit Analysis Problem #5 - Guidance Needed

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brad sue
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Homework Statement


Hi,
I have problem doing this problem:
It is the problem #5 on the attachment.( I was not able to remove problem#4)

Please can you give me some guidance?

Thank you
B


Homework Equations





The Attempt at a Solution



I know that for t>0( steady state) the inductor is like a short and the capacitor like an open circuit.

Do I need to replace the inductor by a short and cancel the part with the capacitor and then do a KVL and solve for V(t)?

I was wondering if I need to use the initial value (just before the switch closes).

Thank you
B.
 

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brad sue said:

Homework Statement


Hi,
I have problem doing this problem:
It is the problem #5 on the attachment.( I was not able to remove problem#4)

Please can you give me some guidance?

Thank you
B


Homework Equations





The Attempt at a Solution



I know that for t>0( steady state) the inductor is like a short and the capacitor like an open circuit.

Do I need to replace the inductor by a short and cancel the part with the capacitor and then do a KVL and solve for V(t)?

I was wondering if I need to use the initial value (just before the switch closes).

Thank you
B.

Steady state is not for t>0, is for t equal infinity. For 0 < t < infinity you have the transient. That is what the problem asks for.
For t > 0 you must use the properties of the inductor and the capacitor: [tex]v_L = L\frac{di_L}{dt}[/tex] and [tex]i_C = C\frac{dv_C}{dt}[/tex]. Use Kirchoffs laws to form a second order differential equation, involving either [tex]i_L[/tex] or [tex]v_C[/tex].
Since you have a nonhomogeneous equation, the general solution is the sum of the solution of the homogeneos equation and one particular solution.
The particular solution can be obtained using the steady state, when the inductor is a short circuit and the capacitor an open circuit.
The general solution involves two integration constants. In order to obtain their values, you use the initial conditions.