circuit with dependent current/Laplace

Discussion in 'Homework Help' started by newzed, Jun 25, 2012.

  1. newzed

    Thread Starter New Member

    Jan 11, 2012
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    Hi guys!
    I have an exercise to ask you about, I attached the image with the assignment.
    As you can see there is a "box" that is repeatedly attached inside the circuit, once and then twice and i have to find v(t).
    I was wondering if I could just replace the "box" with a 2.5Ω resistor by the time the dependent current is depending on something that is inside the box as well but i don't think it's possible...
    What do you suggest?
    Thanks guys :) :)
     
  2. steveb

    Senior Member

    Jul 3, 2008
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    You are free to make an equivalent circuit for the "box". That's what I would do myself. You are correct that the equivalent resistor is now 2.5 ohms. However, you still need to keep the dependent current source. That's as simple as you can make it (EDIT: no it is not, see below). You need to be cautious because the factor of 7 on the dependent current source needs to be changed. I think you can figure out that this involves a factor of 2. So, which way does it go? Is the factor 7/2 or 14 ?

    EDIT: Thinking further, this is a special case because the dependent current source depends on an internal current. Hence, it seems you can make an equivalent resistor out of it. The resistance can be calculated by imagining a 1 V supply driving it, and then calculating the total current. Then the ratio V/I is the resistance.
     
    Last edited: Jun 25, 2012
  3. WBahn

    Moderator

    Mar 31, 2012
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    Whatever current is flowing in the control branch (the left hand side), the same amount will be flowing in the right hand side and seven times that will be flowing in the center path. So a total of 9 times the current will be flowing as would be for a 5Ω resistor for the same voltage. So what size resistor would make that happen?
     
  4. newzed

    Thread Starter New Member

    Jan 11, 2012
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    @steveb yeah I figured out that I could have made and equivalent circuit finding the equivalent Resistance like I was looking for the thevenin eq. circuit of the box.
    In this way my Rth would be = Vo/Io attaching a Voltage or a Current source to the circuit. The problem is that I don't know where I can attach it, I mean, i create a new branch in parallel with the other three with a voltage/current source and then i calculate the Rth?
    Furthermore I didn't get you second part of response about the factor of 2 involving the current...

    @WBahn I get that the total current is 9ix entering and leaving the box but that ix is not known by me so I'm sure I can replace it with just an equivalent resistance the thing is that I don't know how by the time I don't know where I have to calculate the equivalent resistance of the box
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Sometimes it takes a bit longer for the moment of enlightenment to come ...

    Suppose a current I flows into the box.

    You already have realized that

    I=9*ix

    The voltage drop V across all the elements in parallel is simply

    V=5*ix [can you explain why?]

    So the equivalent resistance R=V/I. What next?
     
  6. steveb

    Senior Member

    Jul 3, 2008
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    OK, sorry, I probably confused you because I first answered the question thinking you need to make a Norton equivalent circuit. In general, you would need to do this, and you could also convert this to a Thevenin equivalent.

    However, I spoke too quickly and then realized that you can go further and make this circuit an equivalent resistance only, without need for a current source or a voltage source. That's why I edited the post. This is a special case where the current source is dependent on the internal branch current in such a way that the box behaves as an ideal resistor.

    So, you only need a resistor in this case, which was your first line of thought, but the resistor value is not 2.5 Ohms because the current source modifies the behavior and changes the effective value.

    However, in the future, if you do need a Norton or Thevenin equivalent, just realize that you are making a 2 terminal device and the Norton circuit is a current source with resistor in parallel, while the Thevenin circuit is a voltage source with resistor in series. There should be no confusion about this.

    As far as the factor of 2, that was no longer relevant once I edited the post and pointed out that the final equivalent circuit would just be a resistor.
     
  7. WBahn

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    Mar 31, 2012
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  8. newzed

    Thread Starter New Member

    Jan 11, 2012
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    I love you guys, thanks for being so interested in this and helping me :)

    According to what @t_n_k said it should be easy to calculate the eq. resistance without applying any type of voltage/current source...
    The Voltage across any of the three branches is 5ix because they're in parallel and furthermore the current is the same across the two resistors by the time they have the same value!
    --> R= 5*ix / 9*ix = 5/9 ! Is it correct? So I can replace the creepy box by only this resistor! I was sure 2.5 was not correct anyway I wasn't able to figure out how to calculate the equivalent resistance because I didn't know how to deal with the attachment of a voltage/current source but now that you made me think about that, V is simply 5*ix! :)
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Congratulations!

    To be correct, V is simply 5Ω*ix. Saying 5*ix is saying that the voltage is a current that is five times as great as ix. Carry your units in your work, it could well save someone's life someday (I'm not kidding about that) and will definitely save you lots of points on homework and exams and get you better performance reviews and raises when you enter the work force.

    Hopefully you can see the value of us not just simply giving you the answer, even with an explanation attached. By forcing you to come to grips with it, you were forced to discover the key elements in ways that made sense to you.
     
  10. newzed

    Thread Starter New Member

    Jan 11, 2012
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    you're right, I would have put "ohms" it's just that i was too lazy to find the symbol because i'm using the quick reply thing :p
     
  11. newzed

    Thread Starter New Member

    Jan 11, 2012
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    Thanks again guys for helping me :D
     
  12. WBahn

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    I've got no problem with 5ohms*Ix. That is perfectly acceptable and how I do it when the Ω symbol isn't available. You'll find many of my early posts are that way because I didn't know about the alternate font.

    If you stick around, you will quickly discover that I am a self-admitted and completely unapologetic Units Nazi (that really should have been my user name). I cringe when I see other established members be sloppy, but I pick the hills I'm willing to fight on. In those cases I simply correct the units sloppiness if and when I quote their material. Folks that are here learning, especially if they appear to be students of one kind or another, I am pretty merciless in harping the units issue.
     
  13. newzed

    Thread Starter New Member

    Jan 11, 2012
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    Guys, I can't create another thread... Maybe I can post only one for a certain amount of time... I'd have another doubt on another exercise, I'll try to post it here :)

    It's about First Order Circuit with swtich
    It asks me to find a current that is not the one of the Inductor nor the capacitor's.

    When t<0
    the switch is closed and the inductor behaves as a short circuit while the capacitor as a open one...
    BUT having a short circuit where the inductor is, leads me to have the right part of the circuit eliminated and so the i i'm looking for is 0A (right?)
    On the other hand the current of the inductor is 1A

    For t>0
    the switch is opened and so the voltage source is gone and I have something that is even worse cause it's an RLC circuit (i guess)! I'm pretty sure this is not the way because second order circuits have not been treated yet so I still don't know how to calculate it!
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    UNITS!!

    Please put the proper units on all of the quantities in the schematic.

    I don't know why you couldn't start a new thread. If there was some kind of restriction on you (which I'm pretty sure there wasn't), it probably would have given you a message telling what the restriction was. The forum server may have hiccuped when you tried to post. I've lost two posts today that I submitted but that never made it to the thread I was responding to.


    It appears that you are only being asked to determine the current in the top '5' resistor (5Ω? 5kΩ? 5mΩ? 5 of Aunt Martha's hairpins?) immediately after the switch opens. Nothing more. Don't take on more than is asked for.

    Q1) What is the current in the inductor just before the switch is opened (magnitude and direction)?

    Q2) What is the voltage across the capacitor just before the switch is opened (magnitude and direction)?

    Q3) What is the current in the inductor just after the switch is opened (magnitude and direction)?

    Q4) What is the voltage across the capacitor just after the switch is opened (magnitude and direction)?

    Q5) After the switch is opened, what is the relationship between the current in the inductor and the current you are solving for?
     
  15. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    The current in the inductor can't change instantaneously. You are correct that the current in the inductor is 1A at t=0-. How would the value of current i at t=0+ ensure the inductor current is still 1A at that instant?
     
  16. newzed

    Thread Starter New Member

    Jan 11, 2012
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    The current iL(0-) is 1A so the current iL at t=0+ must be equal since it cannot change abruptly
    The voltage across the capacitor at t=0- is 10V right? So, also at t=0+

    at t=0+ the switch is already opened and the inductor is in series with my 5ohms resistor so i(0+) = 1 A as well?
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    Correct. What direction is it going? In general, currents and voltages have magnitudes and directions. Get in the happen of specifying the direction, as well.

    How do you figure it is 10V (and what direction?) before the switch opens? You said in your first post that the inductor is acting like a short and eliminating the circuit to the right of it. If the voltage across the cap is 10V and it looks like an open, how much current is flowing in the two resistors connected to it? Where is this current going from there?

    Yes, and for the right reasons.

    For extra practice, tabulate what ALL of the voltages and currents are at t=0- and t=0+.

    Then calculate all of the instantaneous powers both before and after as verify that they are consistent throughout the circuit.
     
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