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Current in a Resistor network (need help 2 parts of part b)

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the resistor network shown in the figure below, where R1= 5[tex]\Omega[/tex] and R2= 7[tex]\Omega[/tex] .
    26_51alt.gif

    (a) Find the equivalent resistance between points a and b
    Req=([1/6 +1/5]+7)+12+6=(9.73)-1+18-1=6.32[tex]\Omega[/tex]

    (b) If the potential drop between a and b is 12 V, find the current in each resistor.
    I12[tex]\Omega[/tex]=I6[tex]\Omega[/tex]upper=12/18=2/3A
    I6[tex]\Omega[/tex]lower=.56A
    I5[tex]\Omega[/tex]=?
    I7[tex]\Omega[/tex]=?
    2. Relevant equations
    Req=V/(Inet)
    I=V/R; resistor in parallel I/2
    3. The attempt at a solution
    For I6[tex]\Omega[/tex]lower I think I just did V/11[tex]\Omega[/tex] and rounded.
    Now for I5[tex]\Omega[/tex] shouldn't it be .53A or the same .56A they both got marked wrong though...
    And for I7[tex]\Omega[/tex] shouldn't it be .56A+I5[tex]\Omega[/tex]
    wrong answers for I7[tex]\Omega[/tex]:.92A,.615A, 1.09A,.67A I have one more attempt. I want to make it count.
    wrong answers for I5[tex]\Omega[/tex]:.56A, 1.23A,.53A 2 more tries.
    which I think is more important.

    Also the total I would be 1.899A?
     
  2. jcsd
  3. Oct 26, 2009 #2

    rock.freak667

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    If the voltage across ab is 12 V, then isn't the voltage in 6 and R1 resistors also 12V?
     
  4. Oct 26, 2009 #3

    Delphi51

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    Check the current for the lower branch again. The 6 and 5 in parallel make 2.727 ohms, right? So the lower branch has resistance 7 + 2.727 and the current should be 12/9.727 = 1.23 A. That's the current through R2. You haven't said what I5 or I7 mean so I don't know what else you are having trouble with.
     
  5. Oct 27, 2009 #4
    I5 would mean the current threw the 5 ohms resistor and I7 the current threw the 7ohms resistor
     
  6. Oct 27, 2009 #5
    and for the 6ohms resistor the answer was .56A
     
  7. Oct 27, 2009 #6
    Ah I got it thanks the current threw the bottom wire was the key. So that would make I5=.67A and I7=I6lower+I5= 1.23A
     
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