First order transient RL circuit

Discussion in 'Homework Help' started by Artëm “Artjom” Dembski, Oct 8, 2015.

  1. Artëm “Artjom” Dembski

    Thread Starter New Member

    Sep 28, 2015
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    I know all the steps that are needed for solving the transient circuit.
    The small problem is that I just can't find the iL(0-).
    Someone ideas?
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    You can solve differential equations but can't solve a simple two-mesh DC problem?
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Индуктор должен быть полностью заряжен. Тоесть его можно просто заменить куском провада. Тоесть у вас простая схема, две батареи и три резистора. Найди ток в резисторе 4 Ома, это ток iL(0-).
     
  4. Artëm “Artjom” Dembski

    Thread Starter New Member

    Sep 28, 2015
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    If I make 2 loops, the first in the left part of the circuit and the second in the right part, I get 2.56A for I1 and 0.097A for I2.

    i1: -16 + 4*(i1+i2) + 6i1 = 0
    i2: -4*(i1+i2) - 12i2 - 5 = 0

    Are my equations wrong?
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Did you short out the inductor? An inductor is a short after a long time period with only DC sources in the circuit unless it oscillates. Once you do that, you have a DC circuit, that's it. If you know (and you should) how to solve DC circuits with two sources then you will be able to solve for IL(0-) quite easily.

    You can also check your own work by applying what you find with your first set of calculations to the circuit, then solving for all the missing quantities (such as the voltages across the resistors) then using the laws for voltage and current such as "the sum of all voltages in a series circuit equals zero" and "the sum of all currents entering a node equals zero". If you find one of these conditions that is not satisfied with your applied solutions, then you know something is still wrong.

    For example, say you find the current in a circuit through resistor R1 to be 1 amp, and the resistor is 10 ohms. If you do not get 10 volts across that resistor when you apply the solved-for current (1 amp), then you know something still isnt right. This has to work for every element in the circuit or something still isnt right.
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    These equations are wrong. Both of them.

    I suggest that you draw the current i1 and i2, then setup the equations.
     
  7. RBR1317

    Active Member

    Nov 13, 2010
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    Those equations have the right form, but the signs are wrong. Would need to see the loop currents drawn on the diagram in order to determine which signs are wrong. Nor do I understand why one would write two loop equations when a single node equation would suffice. Personally, I never use loop equations because it's too easy to get the signs wrong.
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    While I haven been taught both methods, I prefer loop equations. And yes, you are right, it would take only one node equation. With loops I just plug the equations into my calculator, and get the answers so it does not take too much longer to solve them.
     
  9. RBR1317

    Active Member

    Nov 13, 2010
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    I still dimly remember the time when I also preferred doing analysis using loop equations. Could not grok node equations when taking the linear circuits class, possibly because no one taught the right way to form node equations ("right" as in being 'easy' rather than 'awkward'). Then when my work became focused on signals and systems analysis where the signals are invariably voltages, I took a second look at node equations (because the variables are voltages), and found a method that gave me a very low error rate in forming the node equations. Have never looked back at loop equations, except sometimes to wonder at what I ever saw there.
     
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