Question

In the circuit (see attachment),steady-state conditions have been reched at t=0.At t=0 switch is thrown to position 2.Solve for V(t) for "t" larger than "0" ?

Answer 1

Don't forget within a branch with an inductance, there's a continuity of current. That why i(0⁺)=i(0⁻)=i(0) !

Answer 2

I do have possible answers (see attachment) but I don get any of this,comparing my work with the answers.

Answer 3

There is a problem in your solutions, first v(t) is not sinusoidal. Have you sure it's the correct exercise ?

Answer 4

It is correct exercise.I did eliminate the first one.What about another four possible answers.

Answer 5

I made a mistake the solutions aren't correct. And the exponential is not 2 but -2. I will remake my demonstration and I post it here. The answer is whether (2) or (5).

Answer 6

It's the (2).

Answer 7

Thank you !!!!!!

Answer 8

I just find too bad we can edit a message on OpenStudy. I will make the remark to developpers. To be not confused I've preferred to delete my previous messages, and I give you the right one (with demonstration) here :

Before t=0
\[v(t)=E_{generator}-R_1.i(t)\]
Steady-state supposes v(t)=0 so :
\[I=E_{generator}/R_1=20/4=5=i(t=0^+)\]

At t=0⁺
\[v(t)=40.e^{-2.t}-R_2.i(t)=L.di(t)/dt\]
We have a linear differential equation :
\[di(t)/dt +(R_2/L).i(t)=(40/L).e^{-2.t}=20.e^{-2.t}\]
Solution = Homogeneous Solution + Particular Solution
\[HomogenousSolution=i(t)=A.e^{-(R_2/L).t}=A.e^{-5.t}\]
We need to find a particular solution which works :
\[\lambda'(t).e^{-5.t}=20.e^{-2.t}\]
\[\lambda'(t)=20.e^{3.t}\]
so
\[ParticularSolution=\lambda(t).e^{-5.t}=(20/3).e^{-2.t}\]
Then A=? At t=0⁺ I=5 A
\[i(t=0^+)=5=A.+ (20/3)\]
\[A=-5/3\]
\[i(t)=(-5/3).e^{-5.t}+(20/3).e^{-2.t}\]
\[v(t)=L.di(t)/dt=2*di(t)/dt=2*[(25/3).e^{-5.t} - (40/3).e^{-2.t}]\]

The result is (2) !!! I've tried to be clearest, like that you could understand.

Answer 9

Thank you so much,very profesional.