@MrAI the switch is open for a long time and it closes on t=0.

@t_n_k Thanks a lot, i understand now how things are, what i dont understand though is why the equivalent without the source is wrong, is that a wrong practice ? Do you think i am unaware of a crucial theory path ? If yes can you give me a link to study about it ?

As you can probably see there are many methods one can apply to solving the problem you posted.

My impression is that you may have been taught a general method which is based on a standard form solution for 2nd order systems. Your teacher may be bypassing some of the details of a methodical solution from first principles. There's not much point in attempting to teach you something new - rather it would be better if you reviewed what you have been taught to ensure you have fully grasped the concepts.

Also keep in mind that an equivalent circuit must exactly reflect the true circuit conditions with all driving soures accounted for. Removing a driving source may allow one to define the natural circuit response parameters but one can't substitute driven responses (such as the inductor current response in your example) back into the undriven equivalent circuit and expect to obtain a sensible answer.

Also more by way of a general comment, an equivalent circuit can in some circumstances "hide" one or more circuit variables by virtue of the reductions applied.

 

Hi,

Yeah, and i was also thinking of bringing in the dual circuit to look at which might be interesting too.

Also, i meant to mention that what made me originally think that he wanted to do an analysis AFTER the switch was opened was that he was talking about the elements being in series, and that would be the view we would take if the switch was opened, because then all the elements are in series and we'd want to know how to do the series analysis.

But with the switch being CLOSED at t=0 then the two branches are in PARALLEL which of course is different, so we can not look at it as a series circuit at least not the way it stands in its original form. This made me think that maybe it would be worthwhile to look at a typical RLC where all the elements are actually in parallel (R, L, and C with no other resistors and all those elements connect to each other in parallel). Not exactly the same, but might help a little to work with a simpler parallel circuit first.

Oh yeah, there is another view we can take here. We can note that the current source causes a capacitor current of 3 amps and inductor current of 0 amps at t=0, so we can look at the circuit as a series circuit if we remove the current source and assume that the circuit is sourceless except for the initial current in the capacitor, which would be 3 amps. This might not work though.

 

I am sorry for the late reply but i just got back from my holidays and i was busy. Thanks a lot for your help i appreciate it a lot

 

Hi,

Yeah, and i was also thinking of bringing in the dual circuit to look at which might be interesting too.

Also, i meant to mention that what made me originally think that he wanted to do an analysis AFTER the switch was opened was that he was talking about the elements being in series, and that would be the view we would take if the switch was opened, because then all the elements are in series and we'd want to know how to do the series analysis.

But with the switch being CLOSED at t=0 then the two branches are in PARALLEL which of course is different, so we can not look at it as a series circuit at least not the way it stands in its original form. This made me think that maybe it would be worthwhile to look at a typical RLC where all the elements are actually in parallel (R, L, and C with no other resistors and all those elements connect to each other in parallel). Not exactly the same, but might help a little to work with a simpler parallel circuit first.

Oh yeah, there is another view we can take here. We can note that the current source causes a capacitor current of 3 amps and inductor current of 0 amps at t=0, so we can look at the circuit as a series circuit if we remove the current source and assume that the circuit is sourceless except for the initial current in the capacitor, which would be 3 amps. This might not work though.

As you probably are aware from your own analysis of this problem, the representative characteristic polynomial for the second order system has the form

\(LC \lambda^2+(R_1+R_2)C \lambda + 1 = 0\)

Where R1 & R2 are the respective resistances in the capacitor and inductor branches. Clearly this indicates the system response may be viewed as that typical of a series arrangement of the circuit elements.

The general (over-damped) response would take the form

\(f(t)=C+A1 e^{-\lambda_1 t}+A2 e^{-\lambda_2 t}\)

where λ1 & λ2 are the polynomial roots.

The location of the current source obviously has a significant impact on the system response. One can propose a range of individual values of R1 & R2 which still add to the same value of 14Ω and therefore resolve to the same characteristic polynomial roots [-5,-2] , but which produce different forced responses.

The resulting response for the given system with the values of R1=10Ω & R2=4Ω is interesting insofar as there is no exponential component in the response linked to the characteristic polynomial root of -2. The only root present in the response is seen as the -5t exponent. One could swap the R1 and R2 values to make R1=4 and R2=10 and the opposite outcome would result, whereby the exponent(s) in -5t would vanish and only the -2t exponent(s) would appear.

Other combinations of R1 & R2 adding to 14Ω will produce responses which do include terms involving both exponents of -2t and -5t. The source modifies the "natural" response.

I doubt one can propose a simultaneous initial condition with a non-zero capacitor current and zero inductor current in a source-less series RLC circuit. The two conditions are irreconcilable.

 

As you probably are aware from your own analysis of this problem, the representative characteristic polynomial for the second order system has the form

\(LC \lambda^2+(R_1+R_2)C \lambda + 1 = 0\)

Where R1 & R2 are the respective resistances in the capacitor and inductor branches. Clearly this indicates the system response may be viewed as that typical of a series arrangement of the circuit elements.

The general (over-damped) response would take the form

\(f(t)=C+A1 e^{-\lambda_1 t}+A2 e^{-\lambda_2 t}\)

where λ1 & λ2 are the polynomial roots.

The location of the current source obviously has a significant impact on the system response. One can propose a range of individual values of R1 & R2 which still add to the same value of 14Ω and therefore resolve to the same characteristic polynomial roots [-5,-2] , but which produce different forced responses.

The resulting response for the given system with the values of R1=10Ω & R2=4Ω is interesting insofar as there is no exponential component in the response linked to the characteristic polynomial root of -2. The only root present in the response is seen as the -5t exponent. One could swap the R1 and R2 values to make R1=4 and R2=10 and the opposite outcome would result, whereby the exponent(s) in -5t would vanish and only the -2t exponent(s) would appear.

Other combinations of R1 & R2 adding to 14Ω will produce responses which do include terms involving both exponents of -2t and -5t. The source modifies the "natural" response.

I doubt one can propose a simultaneous initial condition with a non-zero capacitor current and zero inductor current in a source-less series RLC circuit. The two conditions are irreconcilable.

Hi,

I had my doubts too.

Today i cant see your equations however as something about this site has changed.

 

True...

 

Yes - looks like the new software can't integrate Latex script. May be temporary.

 

Hello,

Jrap is already notified about the Latex problems.

Bertus

 

Thanks Bertus. Looks like it may be fixed.

Hello,

Jrap is already notified about the Latex problems.

Bertus

 

Hello there once more
Taking under consideration all the interesting things you told me about i tried once more to solve the circuit using the same method with which i failed and i think i got good results, except the last part. here printscreens of my efforts and my problem stated on the last one. I believe this occurs because i have poor understanding of the way current flows and voltage + and - because i had this problem again while trying to solve a problem with an ac circuit (should i post this here as well or open a new thread ? if i have to start a new thread what name do you think that would describe well my problem ?)

http://prntscr.com/4hof6f
http://prntscr.com/4hohlw
http://prntscr.com/4hohtr
http://prntscr.com/4hoiek

 

Hi JoyAm,

The current relationship at the driven node is

\(I_{\small{C}}+I_{\small{L}}=3 \ A\)

or

\(I_{\small{L}}=3-I_{\small{C}} \)

Which accounts for the error you note in the final check for the inductor current.

 

Yes, but i cant understand why it happens this way what i have in my mind is that current goes this way http://prntscr.com/4hx3u3 but it must be wrong.
I had similar problem with this http://prntscr.com/4hx46q
http://prntscr.com/4hx4au
Can you explain me the theory behind it ?

 

Hi,

Why would you draw the cap current coming out of the cap like that in your first link? The only source is the 3 amp current source, so that must supply both L and C. If the circuit is under damped, current can flow out of the cap at a later time, but this circuit is not like that it is over damped, and we would still be better off initially drawing the current arrow as going into the cap.

 

That makes sense, thank you MrAI, could you please check out the second problem i attached ?

 

Hi,

What was the question there?

 

The right answer is the on i have put a "tick" next to
what i frist tried was the one i have marked with "X"
But i dont understand why my first attempt was wrong .

 

Hi,

I think you should show your questions a little clearer so they are not so hard to read. Maybe you should try to figure this one out for yourself because it looks like just an error in the math or the procedure. Go over it again and see if you can find out why this happened.
There is also an upload mechanism on this site so you dont have to use third party sites to post images, and that makes it easier for others to read over.

 

The right answer is the on i have put a "tick" next to
what i frist tried was the one i have marked with "X"
But i dont understand why my first attempt was wrong .

You should post the entire question and answers. Some tests are written for you to chose the best answer of the group. Upload any schematic to your post.

After you post, tell us your choice and why you made that choice.

That will help the members to better diagnosis where your going astray.

 

With reference to post #32 it's fairly clear as to where JoyAm has made the error.
It seems the Thevenin equivalent voltage across terminals b & d is required.
That is ...

Vth=Vbd.

If one follows a circuit path between points b & d the choice might be

b->a->c->d

or

Vth=Vba+Vac+Vcd

Having regard to the assigned polarities on the schematic & the circuit topology this reduces to

Vth=-Vo+0.2*Vo*(4-2j)+0 = -Vo+0.2*Vo*(4-2j)=(-1+0.8-0.4j)*Vo=(-0.2-0.4j)*Vo

Note that Vcd is a trivial difference (=0) since c & d are the same node.

There seems to be a lack of understanding on JoyAm's part as to how one sums potential differences as one moves around a circuit path - having regard to the assigned polarities of those potential differences.

 

Hi,

I think you should show your questions a little clearer so they are not so hard to read. Maybe you should try to figure this one out for yourself because it looks like just an error in the math or the procedure. Go over it again and see if you can find out why this happened.
There is also an upload mechanism on this site so you dont have to use third party sites to post images, and that makes it easier for others to read over.

You should post the entire question and answers. Some tests are written for you to chose the best answer of the group. Upload any schematic to your post.

After you post, tell us your choice and why you made that choice.

That will help the members to better diagnosis where your going astray.

Thank you for your advice i will try making more clear questions next time.

With reference to post #32 it's fairly clear as to where JoyAm has made the error.
It seems the Thevenin equivalent voltage across terminals b & d is required.
That is ...

Vth=Vbd.

If one follows a circuit path between points b & d the choice might be

b->a->c->d

or

Vth=Vba+Vac+Vcd

Having regard to the assigned polarities on the schematic & the circuit topology this reduces to

Vth=-Vo+0.2*Vo*(4-2j)+0 = -Vo+0.2*Vo*(4-2j)=(-1+0.8-0.4j)*Vo=(-0.2-0.4j)*Vo

Note that Vcd is a trivial difference (=0) since c & d are the same node.

There seems to be a lack of understanding on JoyAm's part as to how one sums potential differences as one moves around a circuit path - having regard to the assigned polarities of those potential differences.

With reference to post #32 it's fairly clear as to where JoyAm has made the error.
It seems the Thevenin equivalent voltage across terminals b & d is required.
That is ...

Vth=Vbd.

If one follows a circuit path between points b & d the choice might be

b->a->c->d

or

Vth=Vba+Vac+Vcd

Having regard to the assigned polarities on the schematic & the circuit topology this reduces to

Vth=-Vo+0.2*Vo*(4-2j)+0 = -Vo+0.2*Vo*(4-2j)=(-1+0.8-0.4j)*Vo=(-0.2-0.4j)*Vo

Note that Vcd is a trivial difference (=0) since c & d are the same node.

There seems to be a lack of understanding on JoyAm's part as to how one sums potential differences as one moves around a circuit path - having regard to the assigned polarities of those potential differences.

Thanks a lot t_n_k, you "saved the day" once more , now i think i have better understanding on what to sum and what to subtract
Thanks again