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Kirchhoff law problem

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data
    35mo3ty.jpg

    my work

    vcw4fa.jpg

    I know part A is incorrect because for my part B my quadratic contains an imaginary,which I doubt he would due. Some help would be awesome since it is due tomorrow, and will probably be on the exam as well.
     
  2. jcsd
  3. Oct 24, 2012 #2
    Part A looks correct to me. However, in part B, i guess you missed that there are two Ro resistors instead of 2Ro. Rewrite the equation.
     
  4. Oct 24, 2012 #3
    Hmm so what you are saying is.

    Req= R0 + (2ReqR0/Req+2R0)+R0 ?

    If that were the case couldn't I could the R0's and just have a single 2R0?

    Thanks for the help btw
     
  5. Oct 24, 2012 #4
    Yes, i mean that.
    Yes you can have single 2R_o, but you have included two 2R0 in the equation you posted.
     
  6. Oct 24, 2012 #5
    Excellent I got Req= 1+-√5, and I use the positive value

    Now in part c.

    I'm a bit confused as to what I use.

    I tried further condensing the picture I came up with in part A but that made no sense, then I reread what they are asking and I suppose I use the original picture and using kirchoffs rule around the loop from acbd as labeled in the picture. However looking at the picture acbd does not include i1 so should I include acebdf?

    Thank you
     
  7. Oct 24, 2012 #6

    gneill

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    Staff: Mentor

    You might want to think in terms of KCL rather than KVL. Effectively the current entering the junction divides into two paths and subsequently rejoins at another junction. You happen to know the resistances of the two paths, so you can work out how the current divides between the paths.
     
  8. Oct 24, 2012 #7
    I'm still rather confused, because I think I can solve for I1 using simply junction C.

    I know that R0 is going in junction C and R0 is going out of junction C

    But it doesnt add up unless I reverse the direction of one of the junctions.

    For example, R0+R0=2R0 Obeys KCL

    In which i1 would be 2R0.
     
  9. Oct 24, 2012 #8

    gneill

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    Staff: Mentor

    R0 is a resistance, not a current.

    The network after the junction has equivalent resistance Req, so the current io divides between the resistance 2Ro and Req.

    attachment.php?attachmentid=52286&stc=1&d=1351134423.gif
     

    Attached Files:

  10. Oct 24, 2012 #9
    So I0=Ix+i1

    That much I understand,

    What I don't understand is how do I use the relationship of Req and 2Ro to solve for I1.

    The way you drew the picture it seems that i1 is = to Req?
     
  11. Oct 24, 2012 #10

    gneill

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    Staff: Mentor

    Resistance is not current.

    You might want to look up "current divider". Like the voltage divider, it's one of the more handy circuits to know the math for.
     
  12. Oct 24, 2012 #11
    I looked one up online, and want to make sure I have it correct.

    It would be I0(Req)=I1

    Where I0 is = to R0
     
  13. Oct 24, 2012 #12

    gneill

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    Staff: Mentor

    No. I don't understand why you insist on equating current to resistance. They have different units. Current is in Amps, while resistance is in Ohms. For an equation to be correct, units must always agree.

    Two resistors in parallel must have the same potential (voltage) across them. If R1 and R2 are the resistances and the currents through them are I1 and I2 respectively, then by Ohm's law the relationship

    I1*R1 = I2*R2 ...(equating the voltage drops across each resistor due to their currents)

    must hold. If the total current is I0 = I1 + I2, where I0 is a given (known) value, then you have two equations in two unknowns I1 and I2 which you should be able to solve for.
     
    Last edited: Oct 25, 2012
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