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The current in a circuit using nodal analysis

  1. Mar 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi!

    "Write an equation for I1."

    Here is the circuit:
    20140330_165223_zps9fa323e2.jpg

    That is the question. I suppose one could do this many ways but I've decided to do it with nodal analysis. The problem here is that we only got an ideal current source. I have actually no clue.

    So, I'd try to apply Kirchoff's Current Law to do this.

    2. Relevant equations
    KCL:
    ∑ Ii = 0


    3. The attempt at a solution

    I know that one is to pick a ground node somewhere. Let's pick the lower one for example.
    Then one is to identify all nodes, in this case there are just another one.
    Then what? Sorry but my teacher isn't very good at explaining this, at least not so I'd understand.

    Am I correct if I'm to assume that an ideal current source and a parallell resistance R could be simplified to just an ideal current source?

    If one is to apply KCL then it would maybe be something like:
    20140330_171448_zps3fc20c1b.jpg

    Am I on the right track? Some general advice to do this smooth is worth it's weight in gold! :biggrin:
    Thank you! :D


    The right answer for this question is: I1 = -R2I0/R1+R2
     
    Last edited: Mar 30, 2014
  2. jcsd
  3. Mar 30, 2014 #2

    CWatters

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    You seem to be on the right lines. Perhaps remember the voltage will be the same across all three devices as they are in parallel. Should give you some more equations.

    Sorry I might not be around to follow up.
     
  4. Mar 30, 2014 #3
    Ahh, thanks for replying anyway :) The current is so strange. My teacher said that it's good to imagine it like water and water tubes but I don't know how that would help me with the mathematical part. Lol.
     
  5. Mar 30, 2014 #4

    donpacino

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    Gold Member

    water through tubes is a good analogy. If 1 gallon of water passes through a pump (your current source) in 1 second and splits and goes through 2 pipes (R and R2), you know that the sum of the water that passes through the 2 pipes in that 1 second will be 1 gallon.
     
  6. Mar 30, 2014 #5
    Hi donpacino! :)
    I've got the idea of it clear however how do the calculus... o_O
     
  7. Mar 30, 2014 #6

    donpacino

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    haha. okay!!

    so first I can tell you your answer is not correct :(
    ------------------------
    The solution:

    first look at ohms law
    V=IR


    we know the voltage across all three components are the same. we can find that voltage by taking the current through the current source (Io) and multiplying it by the equivalent resistance of the two resistors.

    eq1: Req=____

    i'll let you solve that. hint: they are two resistors in parallel

    eq2: V=Io*Req

    so using ohms law the current through R is

    eq3: I1=V/R1

    so combining eq2 and eq3 we get

    I1=Io*Req/R1

    so generally we get an equation for current division. In any system comprised of parallel resistors and a current source, you can find the resistance through any of the paths by using the following equation

    Ix=Io*Req/Rx

    does that make sense?

    edit: in your case due to the direction of current your I1 will be inverted
    also if you really wanted to do it through nodal analysis, you would need 2 equations

    nodal analysis

    Io=V0/R+Vo/R2
    Vo/R=I1
     
  8. Mar 30, 2014 #7

    eq1: Req=____

    i'll let you solve that. hint: they are two resistors in parallel

    Hmm. Lets see. We have a ideal current source and the current is splitted up like the water tubes we mentioned earlier. If I ain't misstaken one should use current division rules, obtain that very current through that resistance - in this case R (which is where I is) and then use ohm's law on that very place?

    So if I0 is the total current and I1 is the sought current, then:

    I0 = RI/R2+R which is that very current. Now if I ain't completely lost we could take that whole formula and *bang* - use ohm's law? Or do I even want that? I'm so lost. hahha xD
    :D
     
  9. Mar 30, 2014 #8
    Seems like nodal analysis in this particular assignment is as wasteful as putting a V8 into a Smart car?
    hahah. It's basicall just one thing to do correct?
     
  10. Mar 30, 2014 #9
    HOWEVER since the current that is asked for is pointed at the opposite direction, it should therefore be negative correct? I just want be certain of what I do ;)
     
  11. Mar 30, 2014 #10

    donpacino

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    I think you have a fundamental misunderstanding of how this works.

    First you would write the Kcl
    I1=current through R
    I2=current through R2
    Io=source current

    Io+I1+I2=0

    then find an expression for Vo
    I1=-Vo/R
    I2=-Vo/R2
    Io=-Vo/R-Vo/R2

    then solve for Vo
    Vo=-Io/(1/R+1/R2)
    Vo=-Io*R1*R2/(R1+R2)

    then solve for I1
    I1=Vo/R
    I1=-Io*R2/(R1+R2)

    if you were to solve it using the equation i gave you
    1/Req=1/R+1/R2
    1/Req=(R+R2)/(R*R2)

    then plug it into the equation
    I1=-Io*R2/(R1+R2)
     
  12. Mar 31, 2014 #11
    Thank you! Yes it seems I have misunderstood. I have just started with this so it's a quite complex for me. However I think I know what I need to do to make it work.
     
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