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Kcl problem

  1. May 27, 2013 #1
    1. The problem statement, all variables and given/known data

    QQ??20130527135512.jpg

    2. Relevant equations

    KCL

    3. The attempt at a solution

    See picture. I got ix=i1. And I am stuck. Would someone please help me?

    Many thanks in advance.
     
  2. jcsd
  3. May 27, 2013 #2

    CWatters

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    Thats correct. The current through the 20kOhm must be 2i1

    How about writing some KVL equations.
     
  4. May 27, 2013 #3
    How can I write KVL without any resistors in some loops?

    For the leftmost loop: (v2-v1)/5000=0
    For the rightmost loop: (v4-v3)/20000=0
     
  5. May 27, 2013 #4

    CWatters

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    First off you can simplify things..

    V1 = V3
    V2 = V4

    Then write equations for the current through the 5k and 20k.

    Edit: oh heck I've given you the current through the 20K already. Just write an eqn for the current in the 5k.

    Then using those write equations for the voltage drop across the 5k and 20K.

    Then write your KVL equation. Try the outer loop.
     
    Last edited: May 27, 2013
  6. May 27, 2013 #5
    Since the current through 20K is 2i, and V1=V3 and V2=V4, the current through 20K can be expressed as (V4-V3)/20K=(V2-V1)/20K=2i

    Substitute that back to my node V1: 8i-i=0.004. So i=0.000571429A?
     
  7. May 27, 2013 #6
    Got it. Thanks!
     
  8. May 27, 2013 #7

    CWatters

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    For what it's worth the equation I got was..

    (4mA + i1)*5000 - 2i1*20,000 = 0

    which can be solved for i1
     
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