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Find all branch currents in the network shown.

  1. Dec 3, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    The problem and its solution are attached.

    2. Relevant equations
    V = RI
    R_eq = 1/R_1 + 1/R_2 + ... + 1/R_n (Parallel circuits)
    R_eq = R_1 + R_2 + ... + R_n (Series circuits)

    3. The attempt at a solution
    For the two lines of math below “Now referring to the reduced network of Fig. 3-14(b),”, could someone please explain to me why those specific numbers were chosen?

    I believe that the voltage across each resistor should be (2 Ω)(13.7 A) = 27.4 V and, I thought that I would have to do V/R = I for the 9.8 Ω and 2.0 Ω resistors such that I get (27.4 V) /(9.8 Ω) = 2.7959183673469387755102040816327 A and (27.4 V)/(2 Ω) = 13.7 A but, both are wrong, and, I don't understand what I am doing wrong.

    Any help in understanding what I am doing wrong as well as figuring out what is the correct way to approach the part that I am stuck on would be greatly appreciated!
     

    Attached Files:

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  2. jcsd
  3. Dec 3, 2012 #2

    gneill

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    Staff: Mentor

    The maths represent current dividers. There's a total of 13.7 A being split into two paths, one with resistance 9.8 Ω and the other 2.0 Ω. The sum of the resistances is 11.8 Ω. Thus one current will be 13.7(2/11.8) amps, and the other 13.7(9.8/11.8) amps.
    No, that's the potential across the 2 Ω resistor in series with the current supply, but it doesn't take into account the potential that will appear across the current supply itself. Current supplies produce whatever potential is required in order to maintain the flow of their specified current value.
     
  4. Dec 4, 2012 #3

    s3a

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    I get that now, thanks to you. :)

    Prior to this, I didn't know you could have a current source making a voltage; I thought you could only have a voltage inducing a current. (Just saying.)

    In an attempt to understand things better, I wanted to do some "reverse engineering" and found that the potential difference across the current source to be 4.64 V. Am I right?
     
  5. Dec 4, 2012 #4

    gneill

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    Staff: Mentor

     
  6. Dec 4, 2012 #5

    s3a

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    That makes sense.

    The following is what I did.:
    Let “cs” stand for “current source”.
    V_cs + [-(2)(13.7)] = -[(2)(11.38)]
    V_cs = 2(13.7) – (2)(11.38)
    V_cs = 4.64 V

    What's wrong with that?
     
  7. Dec 4, 2012 #6

    gneill

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    Staff: Mentor

    I don't follow what you're doing on the RHS of the first equation. The left had side seems to have units of volts, and I guess the 11.8 is the sum of the resistances of the two branches (Ohms), but what does the 2 stand for? Keep in mind that the branches are in parallel, so their resistances don't simply sum.
     
  8. Dec 4, 2012 #7

    s3a

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    I said 11.38 (A) and not 11.8 (Ω).

    The right side of the equation is the left-most 2Ω resistor's resistance multiplied by the current going through it (which is I_4). The other non-variable part of the equation is the voltage across the 2Ω resistor which is in series with the current source. The entire equation states that the sum of the voltages on the branch with the current source has the same voltage as the entire branch with the left-most 2Ω resistor.

    So, is the previous work I posted correct? (Tell me if there is more you want me to explain.)
     
  9. Dec 4, 2012 #8

    gneill

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    Staff: Mentor

    Okay, sorry I mixed up your resistance \ currents.

    Let's take the simplified circuit and add the current directions and polarities of the voltage drops on the resistors of interest:

    attachment.php?attachmentid=53631&stc=1&d=1354659151.gif

    Note the direction of the polarity of the potential drop across the 2Ω resistor in the center branch. So

    ##V_x - (13.7A)(2Ω) = (11.38A)(2Ω)##
     

    Attached Files:

    Last edited: Dec 4, 2012
  10. Dec 4, 2012 #9

    s3a

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    It's fine. I see my mistake now.

    By the way, that modification of the image I initially attached looks very aesthetically pleasing. :)

    Thanks a lot for everything!
     
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