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A homework in Direct Circuit for beginning students

  1. May 23, 2014 #1
    1. The problem statement, all variables and given/known data

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    2. Relevant equations

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    3. The attempt at a solution

    Well, I am not familiar with the Latex system on this website, just because I am a newbie and I rarely access to this forum. I have a problem (may be simple for you guys) but I still need a solution from you, or at least, please give me the answer if you are boring with typing a lot of formulae.
    Thank you in advance
     

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  3. May 23, 2014 #2

    nsaspook

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  4. May 23, 2014 #3
    Thanks naspook
    I did have the solution that was solved by myself, and I think it's suitable for pupils, not a student but one day, my little brother gave a solution that was different to mine, and he believes that my solution is not right, So I need an answer from you to check.
     
  5. May 23, 2014 #4

    nsaspook

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    Why don't you give your answer and the steps you took to find your results first in the proper format for this forum.
     
  6. May 23, 2014 #5

    phinds

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    You need to read the forum rules. We help folks figure out how to solve problems, we do NOT just "give answers".
     
  7. May 23, 2014 #6
    Yeah, my steps: this DC circuit is a familiar one and we only need to know that with the given assumption as the ammeter is ideal, the potential difference between two points of [itex]\ R_5[/itex] is zero, then as a consequence [itex]\ {I_5}=0A[/itex]. And [itex]\ {I_1}=2A[/itex], [itex]\ {I_2}=2A[/itex],[itex]\ {I_3}={I_4}=1A[/itex], then from Kirchooff law, [itex]\ {I_A}=4A[/itex].
    Net resistance of the circuit is [itex]\ {R_{AB}}=6Ω[/itex]
     
  8. May 23, 2014 #7

    BvU

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    Yeah, well, you sure take big steps. Especially from I5 to I1, 2,3,4. Perhaps you want to explain the 2,2,1,1 in a little more detail ? And: what did your young brother find ?
     
  9. May 23, 2014 #8
    to BvU:
    Well, this problem can be solved simply by re-draw the circuit with the rule as: Consider all points that have the equal potential is one. So we can have a new and simple diagram, from it we get I1, I2, I3, I4, and of course, I5=0 as mentioned above. It's my steps

    My little brother: I think he did make a small fault when he re-draw the diagram
     
  10. May 23, 2014 #9

    BvU

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    Still: what did your young brother find ?

    I agree with I5. Also with the two 1. But don't you become suspicious at all when 30 V over 6 Ohm gives 4 A ?
     
  11. May 23, 2014 #10

    nsaspook

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    Points for trying but you need to review a few things. It's aways good to work backwards (what voltage is across each node with your current calculations) from your result to check for errors.
    http://www.usna.edu/Users/cs/vincent/suppnotes/EE301Topic06.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  12. May 23, 2014 #11
    Of course, nasaspook, we should review it, but I did knew this theory many years ago when I was in 9-grade class and I did give my answer.
    Could anyone give me your numerical answer, that is all I need
     
  13. May 23, 2014 #12
    Could anyone give the answer for me? I need it to check, this problem seems to be simple, but I need the answer from you
     
  14. May 24, 2014 #13

    BvU

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    Is not the way PF works. You have a choice to ignore answers to questions and hints to improve your work. You can not ask for 'the answer'.
     
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