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Using current divider rules

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the unknown current I1 and I2 in the circuit shown below. (Reference problem 7 on the PDF doc)


    2. Relevant equations
    V=iR
    i1=i(R4/(R12+R4))
    i2=i(R12/(R12+R4))


    3. The attempt at a solution
    I used V=ir where V=20v and R=2ohm so i=10A. Then plugging into i1 equation where R4=4 ohn, and R12=12 ohm. I get 2.5 A on i1 which makes sence because the resistance is higher and plugginf in i2 I get 7.5A again which makes sense because it has a lower resistance this also completes KCL rules. My question is did I do this right?
     

    Attached Files:

  2. jcsd
  3. Feb 1, 2013 #2
    Hello it would not allow me to ask a second question using the same PDF so I figured I would ask a second on this post sorry if this is wrong.

    1. The problem statement, all variables and given/known data
    Solve for the voltage Vx and ix?(Reference problem 8)


    2. Relevant equations
    v=ir
    0=i1+i2+3+...in
    0=V1+V2+V3+...Vn


    3. The attempt at a solution
    I suppose I am incorrect but this cirrcuit doesnt appear to be valid. When I do this 90V - 5A(8 ohm) = 50 volts. Then i10=50V/10 ohm = 5A well how can this be because due to KCL that would mean 5 amps and 5 out meaning no current would flow into the 4 ohm resistor. Ugh what have I done wrong.
     
  4. Feb 1, 2013 #3
    The problems are seperate from each other and both are on the PDF that was given
     
  5. Feb 1, 2013 #4

    gneill

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    Staff: Mentor

    The load that the 20V source sees is not just 2Ω, so your total current is not correct; the current that passes through the 2Ω resistor also must pass through the other resistors on its way back to the source. If you want to find the total current, you'll need to find the equivalent resistance presented by resistor network.

    attachment.php?attachmentid=55275&stc=1&d=1359763092.gif
     

    Attached Files:

  6. Feb 1, 2013 #5
    Ahhh ok makes sense so when I put the 12 ohm adn 4 ohm resistor in parrallel and then in series with the 2 ohm resistor I will get 5 ohms which equates to a total aperage of 4 A. Then by me above equation for i1=1A and i2=3A. Is that correct?
     
  7. Feb 1, 2013 #6

    gneill

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    Staff: Mentor

    Your result looks fine.
     
  8. Feb 1, 2013 #7
    Ok thanks man do you think you could help with my second question I posted I know I prolly wasnt suppose to do that but it wouldnt let post two seperate ones with the pdf.
     
  9. Feb 1, 2013 #8

    gneill

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    Staff: Mentor

    There's nothing wrong with the circuit or your analysis so far. What does that tell you about the potential drop across the 4Ω resistor?
     
  10. Feb 2, 2013 #9
    it would 0 correct
     
  11. Feb 2, 2013 #10
    oh heck woops I get it now so it would be 50-0=50 volts then v=iR so i2=25A going out which means. 0=5-5-25+ix so ix=25A and Vx = 50 volts is that correct?
     
  12. Feb 2, 2013 #11

    gneill

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    Staff: Mentor

    Yup. That's it.
     
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