LC circuit analysis problems with 2 voltage sources

[qpham26]

Homework Statement

. In the following transient circuit, assume at t<0, the circuit is at steady state.
Find v0(t) for t>0
http://sphotos-b.xx.fbcdn.net/hphotos-ash3/c0.0.354.354/p403x403/546839_509413575744829_42997155_n.jpg

Homework Equations

The Attempt at a Solution

The first thing that I was suggested to do was to always find i_L(t)
so since we have 2 source of voltage. I will be using superposition method.

First let use the 12V source only
R_eq for the circuit will be: (2||6) + (3+1||3)
Use R_eq to find the current source due to the 12V and by using current divider rule we can obtain i_L(0^-)

then move on to the 3V source
at this point, I am having trouble, so when we short out the 12V, will the current just go through the wire where the 12V used to be and skip the right side of the circuit?
but if it is so then i_L(0^-) for this source will be 0?, which doesn't seem right to me.

ok, assume that we actually figured out the value for i_L(0^-) then to find i_L(∞) we only have to deal with the right side of the circuit.
so i_L(∞) will be equal to the current going through the 1Ω ?

and R_th for t>0 will be 3+(1||3) this can be use to find time constant.

so suppose we got our equation for i_L(t)
to find V_o(t) we can try to find i_o(t) from current divider rule
i_L(t)= i_o(t) X (3/4)

and once i_o(t) is obtained, we multiply that by 3Ω to get V_o(t)

thanks for your time.

 

[gneill]

Homework Statement

. In the following transient circuit, assume at t<0, the circuit is at steady state.
Find v0(t) for t>0
http://sphotos-b.xx.fbcdn.net/hphotos-ash3/c0.0.354.354/p403x403/546839_509413575744829_42997155_n.jpg

Homework Equations

The Attempt at a Solution

The first thing that I was suggested to do was to always find i_L(t)
so since we have 2 source of voltage. I will be using superposition method.

First let use the 12V source only
R_eq for the circuit will be: (2||6) + (3+1||3)
Use R_eq to find the current source due to the 12V and by using current divider rule we can obtain i_L(0^-)

then move on to the 3V source {24V source?}
at this point, I am having trouble, so when we short out the 12V, will the current just go through the wire where the 12V used to be and skip the right side of the circuit?

Yes.

but if it is so then i_L(0^-) for this source will be 0?, which doesn't seem right to me.

It is correct. The 12V source, situated where it is, effectively isolates the "right hand side" of the circuit from the "left hand side". Nothing that happens to the left of that supply can change the potential across that supply; it's an ideal source and will source or sink ANY current required to maintain that potential.

 

[rude man]

Yes.
It is correct. The 12V source, situated where it is, effectively isolates the "right hand side" of the circuit from the "left hand side". Nothing that happens to the left of that supply can change the potential across that supply; it's an ideal source and will source or sink ANY current required to maintain that potential.

Right. This is a 'trick' question!

 

[aralbrec]

The Attempt at a Solution

The first thing that I was suggested to do was to always find i_L(t)
so since we have 2 source of voltage. I will be using superposition method.

First let use the 12V source only
R_eq for the circuit will be: (2||6) + (3+1||3)
Use R_eq to find the current source due to the 12V and by using current divider rule we can obtain i_L(0^-)

The left side of the circuit is isolated from the right side because of the position of the 12V source. The current through the inductor is governed by the right side circuit only, which contains an equivalent resistance (3 + (1||3)).

then to find i_L(∞) we only have to deal with the right side of the circuit.
so i_L(∞) will be equal to the current going through the 1Ω ?

Yes. You may want to consider if anything has actually changed by opening the switch.

 

[qpham26]

The left side of the circuit is isolated from the right side because of the position of the 12V source. The current through the inductor is governed by the right side circuit only, which contains an equivalent resistance (3 + (1||3)).



Yes. You may want to consider if anything has actually changed by opening the switch.

but for t<0, before the switch is open, don't we need to include the 2Ω and 6Ω?

as for calculating the time constant, then we can just remove everything on the left side.

 

[aralbrec]

but for t<0, before the switch is open, don't we need to include the 2Ω and 6Ω?

If you cut the circuit in half through the 12V source and separated the left side from the right side, then connected two separate 12V sources to each of the two separated halves, nothing would change.