general second order circuit problem

Discussion in 'Homework Help' started by JoyAm, Aug 22, 2014.

  1. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    Hello there, this is my first thread on this forum so please excuse me if i violate (without knowing ) any rule. Yesterday i tried to solve a couple of general second order circuit problems but i failed in all of them and it was frustrating.
    So today i decided to post one of the problems here and see if i can get any help hoping that i do the same mistake every time . I will try to upload two images with the circuit and my futile attempt to solve it.( it is hand written and in many places i haven't put the volt or amp symbols but i hope you will be able to understand it)


    http://prntscr.com/4fa06k

    http://prntscr.com/4fa0mm

    The answer is supposed to be 3*(1 - e^(-5t)) A

    Thanks in advance for your help :)
     
  2. t_n_k

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    Mar 6, 2009
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    I guess the critical piece of the puzzle is the value of di/dt at t=0.

    Consider what the voltage across the source is at t=0 when the switch is closed.

    All the source current flows into the left hand branch at t=0. With the capacitor initially uncharged the voltage across the RC branch (and hence the source) will be 30V [10 ohms and 3A].

    With no current flowing (at t=0) in the RL branch the 30V must be initially supported entirely by the 2H inductor. In other words

    L \frac{di_{ \small{L}}}{dt}=30

    or

    \frac{di_{ \small{L}}}{dt}=\frac{30}{L}=\frac{30}{2}=15

    This will lead you to the correct solution.
     
    Last edited: Aug 22, 2014
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  3. JoyAm

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    Aug 21, 2014
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    Thanks a lot sir, i will check it out on my own and see if i have any more questions, although i doubt it, you were very helpful :)
     
  4. JoyAm

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    Aug 21, 2014
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    Can you explain me why at t=0 all the current goes to the left branch ?

    Maybe because on t=0- the iL(0-)=0 so it will remain on t=0 and because of that all the 3 amps will go on the left ?
     
    Last edited: Aug 22, 2014
  5. shteii01

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    What are you trying to find?
     
  6. Jony130

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    Feb 17, 2009
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    Yes, your answer is correct.
     
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  7. JoyAm

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    Aug 21, 2014
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    I was trying to find the iL(t) and then the Vc(t) which gives me troubles as well, i will check it out again later and update this thread with new thoughts and probably questions
     
  8. JoyAm

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    Aug 21, 2014
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    I think i understand how to get the iL(t) but what about the vc(t)?
    vc(t)=14iL(t)+L*(diL/dt) ?
     
  9. t_n_k

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    That would give you the voltage across the current source rather than the capacitor. What will the capacitor branch current ic(t) be? The capacitor voltage will be

    v_c (t)=\frac {1 }{ C} \int{i_c (t) dt
     
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  10. JoyAm

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    Aug 21, 2014
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    That makes Uc(t)=60t+12*e^(-5t), right ? the answer is supposed to be
    12*(1 - e^(-5t))
     
  11. t_n_k

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    The supposed answer is right. You need to check your method. Post your working and we can trace likely errors.
     
  12. t_n_k

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    I would also raise a small point about the original question formulation - with respect to the schematic circuit. Perhaps a point for discussion with the person who formulated the problem.
    It's probably better "practice" to place the switch across the current source rather than in series with the current source. Prior to t=0 the parallel switch would be closed thereby shorting the current source. At t=0 the switch would be opened and the ensuing circuit response would be the same.
    Can you think of a reason why this alternative configuration for the current source switching is more logical?

    Not sure why the angry red face has appeared at the top of this post - I'm certainly not annoyed about anything?????
     
  13. MrAl

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    Jun 17, 2014
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    Hi,

    What is being solved for here?

    Is it just the capacitor voltage or the voltage at the top of the network too, or something else?

    Opening up a current source is an unusual thing to do, but then so is shorting out a voltage source, yet we "do" both these things when we use source superposition.
    I guess when we do that we assume that we do something else too.

    For shorting a voltage source we short the two terminals where the source connects, but actually open circuit the source itself. The place where the source connected must continue to look like a zero impedance.

    For opening a current source, we open the connection to the source but then short out the source itself.
    The original current source connection must look like an infinite impedance so we cant just short it out, so we must open circuit its terminals and then short out the source itself...open circuiting it keeps it an infinite impedance, while shorting out the source itself sets the current itself to zero.

    We cant just open circuit it either though, because there's no way to open circuit a constant current source, given the definition of a current source.
     
    Last edited: Aug 23, 2014
  14. JoyAm

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    Aug 21, 2014
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    Hello MrAI and thanks for posting on this thread what i overall search for is the iL(t) and the Uc(t), i think i have figured out how things are with the iL(t) with mostly the help of t_n_k and some other members of this site but i am still confused about the Uc. Assuming that the numbers of i,u in the steady states are right we proceed in removing that curent source and we get a series rlc circuit with 14 ohm r, is that right ?
    if yes wont the Uc be equal with the i*r+UL ?
    as i said i am confused :/

    And t_n_k i think this problem is taken from alexander-sadiku book for basic circuits but i can check it out later to make sure if you want
     
  15. MrAl

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    Jun 17, 2014
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    Hello,

    So you mean after some time you are opening the switch and want to know the voltage across the cap and across the inductor? If so then yes it forms a series circuit after that.
     
  16. JoyAm

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    Aug 21, 2014
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    Then why do i fail to get the correct answer for Uc(t) ?
     
  17. MrAl

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    Jun 17, 2014
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    Hello,

    If the current source is there to pump in the initial conditions, then what you have is a basic series circuit, but the inductor will have an initial current of 3 amps and the cap will have an initial voltage of 12 volts.

    Lets start at the beginning to make sure i understand what you want to do.

    First, 3 amps is applied to the network (switch closed) and then we wait a very long time. At first the inductor does not conduct, but the cap does, so the total voltage at the top of the network will be 30 volts close to t=0.
    With that, the cap starts to charge up and the inductor starts to conduct current.
    Some time later, the inductor takes all of the current and since it is 3 amps its resistor (R2) drops 3*4=12 volts and the inductor voltage is zero. That puts 12v at the top of the network so the cap charges gradually to +12v and then stays there.
    After a long long time, the cap voltage remains at 12v and the inductor current remains at 3 amps and the circuit would stay this way forever if we did not 'open' the switch.

    Once we 'open' the switch, we have a series RLC circuit, but it's not *just* a series RLC circuit it's a series RLC circuit with initial conditions of:
    vC=+12v at the top of the cap,
    iL= 3 Amps (conventional current flow flowing from top to bottom)

    So you do have an RLC series circuit, but the initial conditions must be applied in order to solve for what happens after the switch is opened.

    IS this what you did, and if so, can you show your work so far?

    Just to note, depending on the damping of the network there may be one or more oscillation cycles before it settles down, but this circuit is probably overdamped so that wont happen with the resistor values shown.

    Also, since the current in the inductor is flowing from top to bottom the initial current is negative. So the capacitor discharges after the switch is 'opened'.
     
    Last edited: Aug 24, 2014
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  18. t_n_k

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    Mar 6, 2009
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    There is clearly some confusion. Going back to JoyAm's original post the attached problem formulation clearly shows the 3A current source being connected at t=0 seconds. Presumably the problem statement (never fully disclosed by the OP) then requires one to solve for the inductor current and the capacitor voltage as functions of time beyond t=0.

    JoyAm states the two expected solutions as being

    i_{\small{L}}(t)=3(1-e^{-5t}) \ amps

    and

    u_{\small{C}}(t)=12(1-e^{-5t}) \ volts

    Both of these are correct for the aforementioned scenario.

    It seems JoyAm has resolved the means of finding the inductor current equation but is unable to do so for the capacitor voltage equation. It seems to me the roadblock is the insistence that the equivalent circuit shown in the second attachment of JoyAm's original post is the correct basis of the capacitor voltage derivation - which it is not.

    One must go back to the original circuit to resolve the matter.

    There are a couple of approaches, one of which I will develop as follows.

    The current source terminal voltage will be

    v_s(t)=R_2 i_{\small{L}}(t)+L \frac{di_{\small{L}}(t)}{dt} \ volts

    Where R2=4Ω

    With the result for the inductor current known, it is possible to deduce the source terminal voltage as

    v_s(t)=12+18e^{-5t} \ V

    It is also possible to find the capacitor branch current (from Kirchoff's Current Law) as

    i_{\small{C}}(t)=3 e^{-5 t} \ A

    From this we may find the capacitor voltage as the difference between the source terminal voltage and the capacitor branch series resistor (R1) voltage drop, or

    u_{\small{C}}(t)=v_s(t)-R_1 i_{ \small{C}}(t) \ volts

    Where R1=10Ω

    Substitution of the various terms leads to the (correct) answer given earlier.
     
    Last edited: Aug 24, 2014
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  19. JoyAm

    Thread Starter Member

    Aug 21, 2014
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    @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 ?
     
  20. MrAl

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    Jun 17, 2014
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    Hi again,

    Oh so you were trying to solve this by eliminating the source? What told you to start that way?

    If you equate both sides of the circuit to the current source, you can form two equations, one for current and one for voltage, where you pretend to know the voltage at the top of the network and call it 'v':

    C*dv/dt+iL=3
    (v-vC)/10+(v-L*di/dt)/4=3

    dv/dt is the cap voltage time derivative, and
    di/dt is the inductor current time derivative,
    vC is the cap voltage,
    iL is the inductor current.

    We also pretend we know the current through the inductor is iL, so we have for v:
    v=iL*4+L*di/dt

    Now substituting this v into the second equation far above and solving for dv/dt and di/dt we end up with two coupled ODE's. You can then solve them how you like.

    Note that what helps solve this is knowing two basic things:
    1. The inductor current iL is also equal to (v-vL)/4 which is simply the difference between the voltage at the top of the network (v) minus the inductor voltage (vL=L*di/dt) divided by the resistance in series with the inductor (4).
    2. The cap current iC is also equal to (v-vC)/10 which is the difference between the voltage at the top of the network (v) minus the cap voltage (vC) divided by the resistance in series with the cap (10).

    The assumed solution for vC is correct BTW.

    Using Laplace Transforms we could put the two side networks in parallel, solve for V(s) at the top of the network, then solve for Vc(s), then find the Inverse Laplace Transform to get Vc(t).
     
    Last edited: Aug 25, 2014
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