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LR circuit in parallel, need someone to check answer

  1. Mar 11, 2016 #1
    1. The problem statement, all variables and given/known data
    S6owVy9.png
    L is a perfect inductor with inductance L. The switch is turned on at t=0 and the currents i1 and i2 run trhough the circuit as indicated by the figure.
    2. Relevant equations
    Given this circuit I need to find i1 and i2 for t=0 and t => infinity.

    3. The attempt at a solution
    For i1 i figure it is simply an LR circuit so we can use the following:

    \begin{equation}
    I(t) = \frac{V}{R}(1 - e^{-\frac{Rt}{L}})
    \end{equation}

    Which becomes:

    \begin{equation}
    i_{1}(t) = \frac{\mathcal{E}}{R_0 + R_1}(1 - e^{-\frac{(R_0 + R_1)t}{L}})
    \end{equation}

    for t=0 i1 = 0 and for t => infinity i1 => E/(R0 + R1)

    i2 is constant and is:

    \begin{equation}
    i_{2} = \frac{\mathcal{E}}{R_0 + R_2}
    \end{equation}

    Is this correct?
     
    Last edited: Mar 11, 2016
  2. jcsd
  3. Mar 11, 2016 #2

    SteamKing

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    You've still got R1 and R2 in parallel, and both of these are in series with R0.

    In the short term, this circuit is going to act like L = 0, and I don't think that's what you have.
     
  4. Mar 11, 2016 #3
    Would you be so kind as to elaborate a bit on that? What kind of implication does this have? I'm a bit lost now.
     
  5. Mar 11, 2016 #4

    SteamKing

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    For the short term, pretend the inductor is not in the circuit and analyze it just with the three resistors present in order to find i1 and i2.
     
  6. Mar 11, 2016 #5

    SammyS

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    You have not given a complete description of the problem.

    Is the switch open for a long time, then it's closed at t=0, or is it closed for a long time, then opened at t=0 ?

    In either case, i2 is definitely not constant.
     
  7. Mar 11, 2016 #6
    I updated the question to reflect the full description. Sorry must have forgot some text when I copied it over from latex.

    The switch is *CONNECTED* exactly at t=0, so it isn't connected at any time before.
     
  8. Mar 11, 2016 #7

    SammyS

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    I think you mean the switch is open for a long time, then closed at t=0, thus completing the circuit through ##\ \mathscr{E} \ ## .
     
  9. Mar 11, 2016 #8
    yes
     
  10. Mar 11, 2016 #9

    cnh1995

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    Think about inductor's behavior at t=0 and at t=∞. You can analyze the circuit by removing the inductor for t=0 and replacing it by a wire for t=∞.
     
  11. Mar 11, 2016 #10

    SammyS

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    @Erik P ,
    Are you familiar with Thevenin's Theorem ?
     
  12. Mar 12, 2016 #11
    No, don't recall having had about that. Is the first part i1 correct or incorrect? I could see a possible solution for i2 as being the following, assuming i1 is correct:
    \begin{equation}
    V_2 = i_2(R_0 + R_2)
    \end{equation}

    \begin{equation}
    V_1 = \mathcal{E}(e^{-\frac{(R_0 + R_1)t}{L}})
    \end{equation}

    \begin{equation}
    V_1 = V_2
    \end{equation}

    \begin{equation}
    \mathcal{E}(e^{-\frac{(R_0 + R_1)t}{L}}) = i_2(R_0 + R_2)
    \end{equation}

    \begin{equation}
    i_2 = \frac{\mathcal{E}}{R_0 + R_2}(e^{-\frac{(R_0 + R_1)t}{L}})
    \end{equation}

    For t = 0, i2 is:

    \begin{equation}
    \frac{\mathcal{E}}{R_0 + R_2}
    \end{equation}

    for t => infinity, i2 => 0

    This circuitry stuff really isn't a strong point for me :(
     
  13. Mar 12, 2016 #12

    ehild

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    If you want the currents as function of time, apply Kirchhoff rules, but take into account that the potential change along an inductor is UL=L dI/dt.

    upload_2016-3-12_14-21-23.png
    Write up the change of potential in both loops, along the red arrows. In the right loop, it is
    -LdI1/dt -R1 I1 + R2 I2 =0.
    What is the equation for the left loop?
    You can eliminate I2, and you get a differential equation for I1 as function of time.
    In case you need the currents at t=0 and t--->infinity only, think about the inductor. When you get induced voltage across it?
    What is I1 through the inductor when the switch gets closed? What is the voltage across the inductor after very long time?
     
    Last edited: Mar 12, 2016
  14. Mar 12, 2016 #13

    SteamKing

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    The basic problem with your calculations is you are treating the inductor like it was an open circuit, similar to a capacitor, when it is not.

    After the current passes thru R0, it is going to split in the parallel branches, with one part going thru R1 and the rest thru R2. You must figure out how the current divides in these two branches, based on R1 and R2.

    The presence of the inductor affects only the transient behavior of the current just after the switch is closed. Once these transients settle down, the inductor no longer has any effect, at least for a circuit having DC current.
     
  15. Mar 12, 2016 #14
    I found the solution for the long term, the transient part I'm a little in doubt as to.
    For the long term here is the solution:
    \begin{equation}
    V_1 = i_1R_1
    \end{equation}

    \begin{equation}
    V_2 = i_2R_2
    \end{equation}

    \begin{equation}
    V_1 = V_2 \Rightarrow i_1R_1 = i_2R_2 \Rightarrow i_2 = \frac{i_1R_1}{R_2}
    \end{equation}

    \begin{equation}
    V_0 = (i_1 + i_2)R_0 = (i_1 + \frac{i_1R_1}{R_2})R_0
    \end{equation}

    \begin{equation}
    V_0 + V_1 = (i_1 + \frac{i_1R_1}{R_2})R_0 + i_1R_1 = \mathcal{E}
    \end{equation}

    \begin{equation}
    \mathcal{E} = i_1R_0 + i_1\frac{R_1R_0}{R_2} + i_1R_1
    \end{equation}

    \begin{equation}
    i_1 = \mathcal{E}\frac{R_2}{R_1R_2 + R_0R_1 + R_0R_2}
    \end{equation}

    likewise for i2:

    \begin{equation}
    i_2 = \mathcal{E}\frac{R_1}{R_1R_2 + R_0R_1 + R_0R_2}
    \end{equation}

    For t=0 the inductor stops current flow so i1 = 0, and i_2 becomes E/(R0 + R2) as i posted in one of the previous replies. Is this reasoning correct?
     
  16. Mar 12, 2016 #15

    ehild

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    It is correct.
     
  17. Mar 12, 2016 #16
    Thanks a lot, could you perhaps help me with another quick question?

    I'm supposed to show, by using Kirschhoff's law, that at the time t >= 0 the following applies:

    \begin{equation}
    \frac{di_1}{dt} + \frac{\tilde{R}}{L}i_1 = \frac{\tilde{\mathcal{E}}}{L}
    \end{equation}

    and i'm also supposed to find E~ and R~, however I can't seem to find any information on the ~ over E and R, what does it mean? Is there a specific law I should be looking at, etc.
     
    Last edited: Mar 12, 2016
  18. Mar 12, 2016 #17

    ehild

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    Can you show the original problem text?
    Remember the voltage across an inductor is UL=L dI/dT. It is still true that the voltages along a closed loop add up to zero.
    You might look at my Post #16.
     
  19. Mar 12, 2016 #18

    gneill

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    That's where Thevenin's Theorem comes in. You should have covered Thevenin and Norton equivalent circuits in circuit analysis prior to the introduction of inductor and capacitor components.

    You want to remove the inductor from the circuit and replace the remaining voltage source and resistor network with a Thevenin equivalent circuit. It will consist of a voltage source and a single resistor. Those are your "E~" and "R~". Then when the inductor is re-inserted you have a simple series RL circuit to analyze.
     
  20. Mar 12, 2016 #19
    That is more or less the original text, I translated it from Danish to English.

    EDIT:
    It would seem the part about R0 = R1 = R2 = R is for the last problem not this one.
     
    Last edited: Mar 12, 2016
  21. Mar 12, 2016 #20

    SammyS

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    What does That refer to?

    In the OP, your statement of the problem is not clearly stated. It's split between 1. The problem statement, and 2. Relevant equations .

    Then, in Post #20, you add to the problem statement.

    So I repeat ehild's request for a complete statement of the problem.
    Where does this last part come from? I see it mentioned nowhere else. Did I miss it?

    So I repeat ehild's request for a complete statement of the problem.
     
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