Hi, I'm seeking help for analysis of the following circuits.

1) I have intended to use Thevenin's Theorem to convert the left hand side circuit to simpler circuit and removing the dependent Source
Gotten Vth= 1V, Rth=40 ohms

2) I can't figure out the initial conditions of the below circuit.
Is the initial condition when t=0- = 0A, same goes with di(0-)/dt = 0A.

I figure out that i would require the law of computing the Voltage expression via multiplying the Inductor with the rate of change of current w.r.t time
(V = L * di/dt)

Seeking some hints on how to tackle this question

 

The initial conditions is that you have one loop: 80V with R1 with R2 with 80i. Since it is a single loop, you can find current i and then find dependent voltage source 80i.

 

The initial conditions is that you have one loop: 80V with R1 with R2 with 80i. Since it is a single loop, you can find current i and then find dependent voltage source 80i.

Yes, i got that, I used Mesh Analysis to form i1(80V loop) to be 0.00625A while i2 (80i) loop) to be 0.025A.
Afterwards i used Thevenin's Theorem to find Vth=1 (Open Circuit Voltage at 80ohm and 80i worked out back to be 1V) and Rth = 40 ohms.

Is there anywhere did my calculation went wrong?

Initial Conditions of the Inductor should be 0A since the switch is not closed as such no current and voltage is able to be passed through?

How about the steady state response?

 

Hi,

To solve for the initial condition on the left side circuit there is only one loop. However, it's a static result anyway and the ic's for the right hand side are both zero so you may not have to solve for that anyway and just handle it as a two source circuit
I mention this because the left hand side is dependent on the current 'i' and when you connect the load, that 'i' will change and will go to 80/160 amps for the first instant, so any equivalent circuit should you decide to do it that way will have to include 'i' in the result or else it wont be able to respond to the change in load, where the load is 40 ohms in parallel with 30H in parallel with 1/60 F.

The steady state solution is just with time 't' going to infinity. In this case can you see what role the inductor plays in finding this value? Remember how inductors respond to DC current/voltage.

Do you have to follow any specific method or can you solve this any way you want to?

[LATER]
Actually worded a bit better, the left side is static and linear so there should be an equivalent circuit possible but this could be a current source. I'll wait to see what you want to do here and how you normally solve these.
1v does not seem correct though because if you got a current i1 with a short in the original circuit then you must get that same current i1 with the new equivalent circuit with a short in the same place. Calculate the open circuit voltage and short circuit current and go from there.

 

Hi,

To solve for the initial condition on the left side circuit there is only one loop. However, it's a static result anyway and the ic's for the right hand side are both zero so you may not have to solve for that anyway and just handle it as a two source circuit
I mention this because the left hand side is dependent on the current 'i' and when you connect the load, that 'i' will change and will go to 80/160 amps for the first instant, so any equivalent circuit should you decide to do it that way will have to include 'i' in the result or else it wont be able to respond to the change in load, where the load is 40 ohms in parallel with 30H in parallel with 1/60 F.

The steady state solution is just with time 't' going to infinity. In this case can you see what role the inductor plays in finding this value? Remember how inductors respond to DC current/voltage.

Do you have to follow any specific method or can you solve this any way you want to?

[LATER]
Actually worded a bit better, the left side is static and linear so there should be an equivalent circuit possible but this could be a current source. I'll wait to see what you want to do here and how you normally solve these.
1v does not seem correct though because if you got a current i1 with a short in the original circuit then you must get that same current i1 with the new equivalent circuit with a short in the same place. Calculate the open circuit voltage and short circuit current and go from there.

Hi, Thanks for the advice. I've managed to solve it.

I initially did not included the 40ohm resistor on the RLC circuit.
I was advised by my friend to convert the whole circuit (excluding the L and C) and convert it to a Thevenin's Equivalent Circuit so that i will only have 1 resistor and 1 source and followed by a Source Transformation to make it a Parallel RLC Circuit. I managed to solve it this way. Thanks alot for the help!