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KCL Question

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine the current in each branch of the circuit in the below diagram.
    http://img341.imageshack.us/img341/7901/kclquestion.jpg [Broken]

    2. Relevant equations
    V=IR
    P=VI

    3. The attempt at a solution
    This is what i have done but it is incorrect.
    Vtotal=16V
    V=IR
    [tex]Io1=\frac{16.00}{8.00}[/tex]
    =2.00A

    [tex]Io2=\frac{16}{6}[/tex]
    =0.75A

    [tex]Io3=\frac{16}{1}+\frac{16}{3}[/tex]
    =21.3A

    P.S
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 6, 2010 #2

    berkeman

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    Staff: Mentor

    You are not writing the simultaneous KCL equations as far as I can tell. Combine series resistors where you can, and write the 1 KCL equation for this circuit (why just 1?). Then solve it.
     
    Last edited by a moderator: May 4, 2017
  4. Feb 6, 2010 #3
    Use mesh analysis and KVL
     
  5. Feb 6, 2010 #4

    berkeman

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    Staff: Mentor

    Is there a reason to pick KVL over KCL here? Just curious.
     
  6. Feb 6, 2010 #5
    After redoing the question this is what i have done:

    Three equations:

    Equation 1: Io3=Io1+1o2

    Equation 2: 8-(4)Io2-(6)Io3=0

    Equation 3: 4-(8)Io1-(6)Io2=0

    Substitute Equation 1 into Equation 2
    Equation 4
    8-(4)Io2-(6)(Io1+Io2)=0
    8=(10)Io2+(6)Io1

    Equation 4 - Equation 3

    8=(10)Io2+(6)Io1 - 4=(8)Io1+(6)Io2

    Io2=0.90Amps

    Substitute Io2 into Equation 3

    Io1=\frac{4-(6*0.90)}{8}

    =0.175Amps
    Subsitute Io1 and Io2 into Equation 1

    Io3=Io1+Io2
    =0.90+0.175
    =1.075Amps

    I checked with the book's answers and my solutions are incorrect. Can someone tell me where i have gone wrong.
     
  7. Feb 6, 2010 #6

    berkeman

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    Staff: Mentor

    I do not understand your equations. Are they meant to be KCL equations? The KCL only involves the voltage at that top node, and the 3 currents that flow out of that node.
     
  8. Feb 6, 2010 #7
    yes they are KCL equations
     
  9. Feb 7, 2010 #8
  10. Feb 7, 2010 #9
    ok thanks i know how to do it now.
     
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