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Homework Help: Current in a branch

  1. Jul 14, 2006 #1
    I was wondering,when we have a voltage source connected to a node in series with a resistance connected to the ground like this: ,when determining the current on the branches through node method is the current on that branch 5/1k or v1+5/1k?

    http://i75.photobucket.com/albums/i281/esmeco/branchcurrent.jpg [Broken]

    and a resistance co
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Jul 14, 2006 #2
    The current accross the resistor is:

    [tex] \frac{V_1-V_g}{1k} | V_g = 0 [/tex]

    Ground is a node voltage also, you just define it as [itex] 0V[/itex]

    On a side note, if someone reads this thread...how do I make the "|" larger in LaTeX?
    Last edited: Jul 14, 2006
  4. Jul 14, 2006 #3


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    [tex] \left| \frac{V_1-V_g}{1k} \, \right| \, V_g = 0 [/tex]

    hmm...was trying the \left and \right command which works for brackets \left( content \right). I only got it to work with a leading \left| command to go with the \right|.


    [tex] \frac{V_1-V_g}{1k} \, \Bigg| \, V_g = 0 [/tex]

    Test3, finally got it :smile:

    [tex]| \big| \Big| \bigg| \Bigg|[/tex]
    Last edited: Jul 14, 2006
  5. Jul 14, 2006 #4


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    I must be brain dead (well, it is Friday afternoon, and it's been a tough week). What equation are you guys trying to write (a power?), and what is the "|" character supposed to be? Sorry if I'm being totally lame. Maybe I should go home.... :uhh:
  6. Jul 14, 2006 #5
    So,the current that passes through the branch where is the voltage source in series with a resistance is (5v-0v)/1k?
  7. Jul 14, 2006 #6


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    Nope, not necessarily. If you erase the little stub at the lower left that is labelled I1, and the little stub off the lower right that goes nowhere, then the loop current is 5V/(R1+R2). But if you inject some extra I1 current in that lower left stub (without showing it in a standard way, BTW) or something-something at the lower right stub, then you need to take that into account....
  8. Jul 14, 2006 #7
    :rofl: :smile:

    thanks man!

    Yeah I was trying the \left and \right commands, but wasn't getting it either. Well cool. I didn't know about the big commands. Thanks again :)
  9. Jul 14, 2006 #8
    [tex] f(x,y,z,t) = t(x+y+z) [/tex]

    [tex] f(x,y,z,t) \, \Bigg| \, \begin{array}{c} x=y=z=1 & t =4 \end{array} \,\,\, = 2(1+1+1) [/tex]

    I believe it means evaluated at, or where...
  10. Jul 14, 2006 #9

    And to clarify this above, it would be:
    [tex] I_{R1} = \frac{V_1-V_g}{1k}=\frac{V_1 -(0)}{1k} = \frac{5}{1k} [/tex]
  11. Jul 14, 2006 #10
    But the equation for the node on the left(since the other node is ground) is:

    1 + (v1-0)/1k + 5/1k=0

    Or is:

    1 + (v1-0)/1k + (v1-5)/1k=0

    Edit:Thanks for the reply!
  12. Jul 14, 2006 #11


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    Ohhhhh! Evaluated at! Now I get what you were trying to do. But the evaluation condition would normally be in smaller font, down at the bottom of the long vertical line, I believe. At least that's how I've seen it before.

    I feel much better now. Have a good weekend, all.
  13. Jul 14, 2006 #12
    :smile: Yeah. I don't know how to do that in LaTeX though :frown:

    You too man, have a good one!
  14. Jul 14, 2006 #13


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    Another test-session:

    [tex] f(x,y,z,t) \, \Big|_{\substack {x=y=z=1 \\ t =4}} \, = 2(1+1+1) [/tex]

    _{\substack{content}} does the trick. Not sure if it could be placed further down, so that the equals sign is placed more to the left.
    Last edited: Jul 14, 2006
  15. Jul 14, 2006 #14
    Where is the 1 from!?

    I don't know what this means. Label the node you are talking about, and post that circuit.

    For sake of clarity. Lets just assume that your circuit is a single loop. Thus those little branches at the bottom are not connected to anything, so they therefore have no current running through them.

    so your circuit looks something like this:
    Code (Text):

      |                     |
      |                     |
    ( + )                  |
    ( - )                   |--|||
      |                     |
    Now if you want to use nodal analysis, you have use a super node. So you label TWO nodes. Both are on the left hand side. One you have already labeled V1, the other lets call V0 (and this is in the bottom left hand corner).

    The super node expression is:
    [tex] \frac{V_0-V_g}{R_2}+\frac{V_1-V_g}{R_1}=0[/tex]

    You write the expression for the voltage:
    [tex] V_1-V_0 = 5 [/tex]

    And also note that Vg is defined as 0V so:

    [tex] V_g = 0 [/tex]

    [tex] \frac{V_0}{R_2}+\frac{V_1}{R_1}=0[/tex]
    [tex] V_1-V_0 = 5 [/tex]

    Does that make sense?
  16. Jul 14, 2006 #15
    dude you are the man at LaTeX! :smile:

    To be really picky :biggrin: do you know how to left-justify the expression at the evaluated-at symbol?
    Last edited: Jul 14, 2006
  17. Jul 14, 2006 #16


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    I'm just clicking on various examples in the following thread :wink:


    [tex] f(x,y,z,t) \, \Big|_{\begin{subarray}{l} x=y=z=1 \\ t =4 \end{subarray}} \, = 2(1+1+1) [/tex]
    Last edited: Jul 14, 2006
  18. Jul 14, 2006 #17
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