Equivalent resistance and Dependent Sources

Discussion in 'Homework Help' started by newzed, Jan 26, 2012.

  1. newzed

    Thread Starter New Member

    Jan 11, 2012
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    Hi everyone! I'm asking your help in this exercise
    It's about finding the equivalent resistance...
    What should I do, connect a 1-V Voltage source to the terminals?
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes, connect 1V voltage source across A - B terminals and find current that is flow from this voltage source.
    Rab = 5Ω
     
  3. crutschow

    Expert

    Mar 14, 2008
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    You measure the open circuit voltage and the short circuit current of the circuit. The equivalent circuit resistance is V/I.
     
  4. m237b

    New Member

    Apr 8, 2013
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    @Jony130....would you be able to show in details how you get Rab = 5 Ohm in this circuit ?

    Thanks in advance.
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    I'm hoping he won't. It would have been better if he hadn't just given you the answer at all. It would have been much better if he had given you a range that the answer was in, such as "Rab is somewhere between 4.5Ω and 6Ω."

    Just giving you the answer or working the problem for you won't do you any good. Consider that you have examples worked in your text, examples worked in class, examples worked in the E-book here. If those haven't been sufficient to enable you to work a problem on your own, then seeing one more problem worked out for you isn't going to suddenly make the light click on; it will only let you get a good grade on the homework only to turn around and get reamed on the exam. You will only turn on that light by struggling with the problem yourself. So you need to make your best effort to work the problem and post your work. We can then see what you are doing right and where you are going wrong and guide you into finding your own mistakes and getting yourself back on the right path.
     
  6. m237b

    New Member

    Apr 8, 2013
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    I understand what are you saying...it's just I spent too much time on some similar but a bit more complex circuit and would like someone to provide some advice or instructions
     
  7. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    This one is particularly simple and can be solved almost by inspection.

    The current source current is twice the resistor current, and that current is:

    ix = Va-b / 30 ohms

    Isource = 2 * ix = 2 * (Va-b / 30 ohms) = Va-b / 15 ohms

    Since it follows ohms law the dependent current source can be replaced with a resistor.

    That leaves us with a 10 ohm, 15 ohm, and 30 ohm resistor in parallel.
     
    m237b likes this.
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Interesting concept. Knowledge of working complex circuits but not simplier ones. Is there a study out there that supports your supposition? Maybe your definition of more complex is different than mine.
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    And so you won't even TRY to solve a SIMPLER problem (not to mention that it is YOUR homework)?

    It sounds like this is one that you should try extra hard on and should solve it using several different techniques, particularly both nodal analysis and mesh analysis in addition to the ad-hoc method suggested by ErnieM.
     
  10. m237b

    New Member

    Apr 8, 2013
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    There is nothing mysterious here.....I tried more difficult circuit - failed, tried an easier one and failed to see how it was done too..so thought it may be ok
    to ask the folks....didn't think it may become a big deal...thanks for your reply
    anyway.
     
  11. m237b

    New Member

    Apr 8, 2013
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    I will try of course, I can see @ErnieM provided some tips on how to approach this.
     
  12. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    I wonder if the OP (newzed) is following this thread?
     
  13. m237b

    New Member

    Apr 8, 2013
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    Thanks ErnieM.....I didn't see how Isource was replaced by a resistor...so it looked unintuitive to see 5 Ohm as a result.

    Thanks again
     
  14. m237b

    New Member

    Apr 8, 2013
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    I guess he dropped a year ago....
     
  15. WBahn

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    Mar 31, 2012
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    Ah... I hadn't noticed that you weren't the OP. That changes things somewhat. While it is still better for you to post your best attempt so that we can see what you are doing and give much more directed assistance, it isn't as important that we avoid just giving the answer.

    So I owe Jony130 an apology for the mild rebuff I gave him earlier.

    If you'll post an attempt at either nodal analysis or mesh analysis, I will post a full walkthrough of the other.
     
  16. m237b

    New Member

    Apr 8, 2013
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    I've tried to use KCL and got stuck: -Ix-2Ix+I10=0 => -Vab/30-2*(Vab/30)+Vab/10=0
    I injected this equation into Wolframe (to make sure it's been solved correctly) and get result -> "True"...not sure how that result can be used further to solve the circuit.
     
  17. WBahn

    Moderator

    Mar 31, 2012
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    You need to indicate which way "I10" is flowing. Don't assume that people will make the same assumptions you do. From your equation it appears that you have defined it to be the current flowing downward through the resistor.

    Now, if the quantity "Vab/30Ω" turns out to be positive, which way would current be flowing through the 30Ω resistor? It would be flowing from A toward B, so right to left, which is in the direction opposite Ix. So

    Ix = -Vab/30Ω

    See where that leads you.

    Also, you need to get in the habit of tracking your units through your work from start to finish.
     
  18. m237b

    New Member

    Apr 8, 2013
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    I was under impression that I did mark the currents with accordance to the KCL. I follow the rule when all currents leaving the node are to be marked as "positives" so incoming are to be "negatives", hence the resulting quotation: -Ix-2Ix+I10=0 (obviously I10 is the one that leaving the node so going downwards from A to B.
    Agree o the units (just realized Advance editing mode provides more symbols)

    Now, do you say that the quotation I produced is not correct ? If it is correct then it leads to some sort of weird result if it is not, then I guess I have to rethink it again.
     
  19. WBahn

    Moderator

    Mar 31, 2012
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    What direction is Ix? It's going toward the node. But how do we know this? Because that is how Ix is defined by the arrow on the diagram.

    The same with the current in the dependent source. How do we know it is going toward the node? Because that is how it is defined by the arrow on the diagram.

    But what direction is I10 going? It's undefined. You added it (instead of subtracting it) in your equation because you are assuming it is flowing out of the node. But this is not the way to "define" things. If forces people reading your work to read your mind. So define it, then use it. Don't rely on people figuring out how you defined something based upon how you used it. What if that were where your mistake happened to be?
     
  20. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Then I misunderstood your posting. I thought you claimed to have a working knowledge of a more complex circuit and the simplier circuit was giving you a problem.

    Sorry about that.
     
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