What Kirchhoff Law should I use?

Discussion in 'Homework Help' started by mrvaledon, Aug 18, 2016.

  1. mrvaledon

    Thread Starter New Member

    Mar 1, 2016
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    Hey guys! I just want to solve a problem that have two resistance, one current source and one unknown current source. What Kirchhoff Law should I need to use? LVK or LCK?

    Here it is. I already make a example with others numbers and without the resistance of 8Ohms. I am confused with this! I tried several times. Need to find current and voltage on 8Ohms. I just want a hint or guide to solve this problem.

    Thanx!
     
    Last edited by a moderator: Aug 21, 2016
  2. BR-549

    Well-Known Member

    Sep 22, 2013
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    Show us.
     
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  3. mrvaledon

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    Mar 1, 2016
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    I already reply!
     
  4. DGElder

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    Apr 3, 2016
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    First solve for the voltage at the node on top, Vo, using KCL
     
  5. mrvaledon

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    Mar 1, 2016
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    What is the correct form to solve this circuit? I need to find voltage at resistance of 8ohms and current on dependent current source. I wanna some hints that can lead me to solve the circuit. Need to find the voltage first? Want some hint steps. I already tried several times with no correct results.
     
  6. WBahn

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    Since you've tried several times, pick which one you think was your best attempt (i.e., the one you think is the closest to being valid) and post that attempt. Then we can see what you did right and where you went wrong.
     
  7. WBahn

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  8. mrvaledon

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    Mar 1, 2016
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    I have not shown any effort here in the forum, but I've tried a lot of times. This circuit is a problem of the book and just want a guide where to start since it is the first time trying to solve a problem with dependent current using source current. The least I would expect would be a brief explanation on how to solve this type of circuit and the formulas and where to start. If it resolved by meshes, nodes or branches. So far all I have clear is that everything is in parallel and voltage is common. I ask only a guide to get the correct answer. For example: first you have to find voltage V0, but after that I have no idea if I need to solve it by meshes. I hope you understand. I write this post with the best intentions to learn.

    I hope you guys will understand.

     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Because this circuit have only two nodes you can try to use a nodal analysis. Treat the bottom node as a ground and find the voltage at the top node (Vo voltage). So, all you need is to write a one nodal equation and solve for Vo.
     
  10. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    I would start by noting the sum of the currents down the 3 legs is equal to the 6A source, and the current in two of the legs (actually all three once you expand it out)
    is in terms of Vo.

    That is just one unknown so it is immediately solvable.
     
  11. WBahn

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    Mar 31, 2012
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    So you expect us to be mind readers?

    That information is in the book. And it's probably been presented in class. And examples have been worked for you in both. Something hasn't clicked. So having people guess at what you do and don't know and just repeat the information that has already been given to you is unlikely to help very much. YOU need to struggle with it and make it your own. The best way for us to help is to see the tale of that struggle so that we can spot the exact place where you are going wrong and help you past that exact place and then do the same at the next place you go astray. But YOU need to do the work if YOU are going to learn the material.

    So SHOW us an attempt to apply KCL to the top node. Let us see what terms you come up with so that we have something to give feedback on.

    Or SHOW us an attempt to write the mesh equations. Let us see what equations you come up with so that we have something to give feedback on.
     
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  12. mrvaledon

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    Mar 1, 2016
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    Ok! Guys! I have this. I wrote:

    6A - i0 - i0/4 -8i0 = 0

    The second photo, are the equivalent of the circuit?
     
    Last edited: Aug 21, 2016
  13. WBahn

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    Please try to manage the size of your attachments better. First, take the time to orient the images so that people can read them easily. Remember, you are asking others for free help -- make it easy to help you if you want them to actually do so. Second, while 450 kB isn't too bad (at least compared to the 5 MB monsters we get), this can be rendered quite handily in less than a tenth that size:

    eq1.jpg

    Next, always, always, ALWAYS track your units. Most mistakes you make will mess up the units allowing you to catch and correct the mistake almost immediately.

    This is a shining case in point. Your very first equation has an error in it that messes up the units. Everything you did beyond that, whether two lines or two pages, is a pure waste of time.

    Your first line should have been

    <br />
6 \, A \; - \; i_o \; - \; \frac{i_o}{4} \; - \; \( 8 \, \Omega \) i_o \; = \; 0<br />

    In order to add things together, they must have the same units.

    The first three terms all have units of current, but the fourth term has units of voltage (resistance multiplied by current) and thus, you know right there that this equation is wrong.

    If you hadn't caught it at this stage and kept going, it would have showed it's ugly head eventually.

    <br />
6 \, A \; = \; (1 \; + \; \frac{1}{4} \; + \; 8 \, \Omega \) i_o \; = \; 0<br />

    Oops. How do you add 1 and 1/4 to 8 Ω? Can't! So something is guaranteed to be wrong!

    As you try to figure out what is going on, you would hopefully realize that the current in the right hand element is NOT even i_o.
     
    Last edited: Aug 21, 2016
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  14. mrvaledon

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    Mar 1, 2016
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    Why i0 = to 6. Th book says that V0 = 8 and i0 = 4
     
  15. WBahn

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    If the book doesn't at least supply units on things like that, then the instructor should find another book. Sadly, most textbook authors do a very poor job of properly tracking units through the examples worked in the text, but most at least tack the units onto the final results.

    Most textbooks are written by people that have little to no real world engineering experience. To them, units are a nuisance because the worst thing that can happen when you get a wrong answer due to failing to track units properly is that you lose some points on an assignment. In the real world, people can and do die. In the real world space probes that cost billions of dollars can and are slammed into planets instead of going into gentle orbits about them.

    Tracking units is perhaps the single most effective error detection tool available to the engineer. It is also free and easy to use. From my perspective, if an error is made and not caught and damage or injury results, it qualifies as criminal negligence. On occasion courts have agreed.
     
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  16. mrvaledon

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    Mar 1, 2016
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    This can be a possible answer?
     
  17. WBahn

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    You just aren't going to track your units, are you? Guess you just don't care too much about getting the right answer.

    You have i_o = 4.8 A.

    Then you have i_o = -1.2 A.

    Then you have i_o = 1.2 A.

    Do you REALLY need to ask if it is right?

    Look at your first equation. That's supposed to be KCL applied to the top node, right?

    There are four branches attached to that node, right?

    If current entering the node is positive and current leaving the node is negative, what is expression for the first (from the left) branch? For the second? For the third? For the fourth? Write them in terms of either i_o or in terms of v_o and the branch resistances.
     
  18. WBahn

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    Take a step back and start with the following:
    eq1.jpg

    In terms of the red quantities, what is KCL on the top node?

    Now express each of the red quantities in terms of Io or Vo.
     
  19. mrvaledon

    Thread Starter New Member

    Mar 1, 2016
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    is = i0 + i1 + i2
     
  20. mrvaledon

    Thread Starter New Member

    Mar 1, 2016
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    Ok! I got this equations and answers:

    For V0:
    8i0 = 6; therefore i0 = 6/8 = 0.75A in the resistance of 8 Ohm's

    For i0:
    6 = i0 + i0/4 = 6/1.25 = 4.8A

    I am assuming that the book is rounding the current on the 8Ohm's resistor to 1. And not rounding the current of i0 to 5.
     
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