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Kirchoff's Rules

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine the current in each branch of the circuit shown in Figure P28.19 (attached)

    2. Relevant equations

    Kirchoff's Rules:
    [tex]\Sigma I_{in} = \Sigma I_{out}
    [tex]\Sigma \Delta V = 0[/tex]

    3. The attempt at a solution

    I have no idea where to begin. It's in the Kirchoff's rules section, so I'm assuming those will be used, but how I don't know how to apply them.

    Attached Files:

  2. jcsd
  3. Jul 7, 2008 #2


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    Staff: Mentor

    Begin at the beginning. Explain what the KCL (1st equation) means, and what the KVL (2nd equation) means.

    Then start with the KCL equations for the circuit. Call the bottom node ground, and write the 2 KCL equations for the two top nodes. There are three branches out of the top left node, and two branches out of the top right node.... That gives you two equations and two unknowns (the top two node voltages are the unknowns)...

    Show us some work, so that we can help you more. That's how it works here on the PF.
  4. Jul 7, 2008 #3


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    Staff: Mentor

    Oh, and you might try following the "Similar Threads" links at the bottom of the page, in case that helps to get you started writing the equations here in your own thread.
  5. Jul 7, 2008 #4


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    Homework Helper

    … give the unknowns names …

    Hi clairez93! :smile:

    Hint: start by calling the three unknown currents a b and c (with arrows).

    Then write out all the Kirchoff's equations you can think of … what do you get? :smile:
  6. Jul 7, 2008 #5
    Okay, so I have this a current I labeled a going upward along the vertical line at the far right with the 1 ohm resistor in it. Then b and c going downwards along the lines with the 8 ohm resistor and the 5 and 1 ohm resistor line. (b for the 5 & 1 ohm resistor line, and c for the one with 8)

    I'm thinking a = b+c according to Kirchoff's first?
  7. Jul 7, 2008 #6


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    Looks good! :smile:

    Now what about the voltage drops across the resistors?

    (going to bed now … :zzz:)
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