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Inductors and circuit problem

  1. May 10, 2016 #1
    1. The problem statement, all variables and given/known data
    In the following figure, R1= 3.1 Ω, R2= 4.9 Ω, L1= 0.155 H, L2= 0.2 H and the ideal battery has ε = 5.6 V. Just after switch S is closed, at what rate is the current in inductor 1 changing?
    HW10Q16.png

    2. Relevant equations
    [itex]{ \varepsilon }_{ induced }\quad =\quad L\frac { di }{ dt } \\ V\quad =\quad IR[/itex]

    3. The attempt at a solution
    Don't know where to start. I know that the inductor will induce an emf in the opposite direction of the battery's. Right after the switch is closed, I guess it has two resistors in parallel, so the current can be calculated from Ohm's law. But that doesn't seem right since there will be a "resistance" from the inductors.
     
  2. jcsd
  3. May 10, 2016 #2

    Paul Colby

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    Can you write down a differential equation for ##i(t)##?
     
  4. May 10, 2016 #3

    Paul Colby

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    Actually, more basic question, what rules do you know about all circuits?
     
  5. May 10, 2016 #4
    Here is an attempt
    Loop 1 (left side), Loop 2
    [itex]\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad L\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -L\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0[/itex]
     
  6. May 10, 2016 #5

    Paul Colby

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    Looks close. I'm having a hard time understanding your definition of ##i_1## and ##i_2##? What happens to ##i_2## when it goes out of the inductor?
     
  7. May 10, 2016 #6
    I just have two currents because there are two loops. I guess in this case, i2 doesn't matter because the problem only asks for regarding the leftmost loop.

    So only the first equation.

    Edit: My bad, I think your confusion comes from the fact that I forgot to add subscripts to the L in the second equation
     
  8. May 10, 2016 #7

    Paul Colby

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    The reason I ask about ##i_2## is that your equations imply that ##i_2## just disappears. It's only in the second inductor. Doesn't that violate charge conservation?
     
  9. May 10, 2016 #8
    Here, I edited them
    [itex]\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad +\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0[/itex]
     
  10. May 10, 2016 #9

    Paul Colby

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    The thing about choosing loop currents as variables is they are automatically conserved because they form closed loops. If ##i_2## runs around the second loop shouldn't it be flowing through ##R_2## as well as ##L_2##?
     
  11. May 10, 2016 #10
    By that, equation #1 needs a i2R2 term as well correct?
     
  12. May 10, 2016 #11
    Conventional current flows from the positive of the terminal through inductor 1, then splits off into i1 downwards, and i2 rightwards. i2 then joins with i1 at the other junction. It doesn't make sense for i2 to flow upwards.
     
  13. May 10, 2016 #12

    Paul Colby

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    Well, ##R_2## is part of both loops so both ##i_1## and ##i_2## have to contribute to the voltage drop across ##R_2##. Be careful with the signs :)
     
  14. May 10, 2016 #13

    Paul Colby

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    I find it helpful to draw the two loop currents in their respective loops. Choose some loop direction for each loop and stick with it.
     
  15. May 10, 2016 #14
    okay, I have i1 for loop the left loop, and l2 for the right one
    [itex]\varepsilon \quad -\quad i_{ 1 }{ R }_{ 1 }\quad -\quad { L }_{ 1 }\frac { di_{ 1 } }{ dt } \quad -\quad i_{ 1 }{ R }_{ 2 }\quad =\quad 0\\ -{ L }_{ 2 }\frac { di_{ 2 } }{ dt } \quad -\quad i_{ 2 }{ R }_{ 2 }\quad =\quad 0[/itex]
     
  16. May 10, 2016 #15

    phyzguy

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    You're making it much too hard. The current through an inductor cannot change instantaneously. So immediately after the switch is closed, the current through the inductors is zero. So there is no voltage drop across the resistors, since there is no current.
    [ moderator's note: message abridged, too much help provided ]
     
  17. May 10, 2016 #16
    What about L2, doesn't that get factored in somehow too?
     
    Last edited by a moderator: May 11, 2016
  18. May 10, 2016 #17

    collinsmark

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    By the same logic that @phyzguy just mentioned, and also invoking Kirchhoff's current law, what is the current through [itex] R_2 [/itex]? Thus what is the voltage across [itex] R_2 [/itex]? Thus what is the voltage across [itex] L_2 [/itex]? :wink:
     
  19. May 10, 2016 #18

    Paul Colby

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    Both currents run through ##R_2## generating a voltage drop. The term in the second (and first) loop equations should be ##(i_2 - i_1)R_2##, right?
     
  20. May 11, 2016 #19
    Is it because of the resistance in the inductor to the change in current, that in time = 0, there is no current through it?
     
  21. May 11, 2016 #20

    collinsmark

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    When working with ideal, lumped parameter components, and assuming that there are no voltage/current sources that can generate an infinite amount of voltage/current (even for a very short period of time, such as an ideal impulse), then the following holds:
    • The current through an inductor cannot jump from one value to a different value instantaneously.
    • The voltage across a capacitor cannot jump from one value to a different value instantaneously.
    The current through an inductor can change gradually. That much is allowed. But it cannot jump from one value to another instantaneously. (The same is true for the voltage across a capacitor.)

    For example, if the moment before the switch is closed, an inductor has 0 A flowing through it, then the current flowing through it immediately after the switch is closed must be infinitesimally close to 0 A (in other words, 0 A).
     
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