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Kirchoff's Laws

  1. May 16, 2007 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    for the first loop (2v battery)

    the effective resistance is 33/7
    the voltage is 2v

    so current=v/i
    =14/33 A

    Now loop 2 is the problem.
    i dont know how to add up the all the resistance.

    or am i using the wrong approach here.
  2. jcsd
  3. May 16, 2007 #2

    ok for the first lloop

    2=3(I1) + 4(I2) +3(I3)

    the second loop

    1=4(I4) +3(14)

    but its still wrong
  4. May 16, 2007 #3
    the problem im having iswith the I3 part of the second loop
    does current even flow there? or does i4 behave like i2 ?
  5. May 16, 2007 #4
    I2 and I4 are the same because there is no node between them. Try writing Kirchhoff's loop equations for the two loops indicated and the node equation for the blue node.
    Last edited: May 16, 2007
  6. May 16, 2007 #5
    ok for the first loop

    2=3(I1) + 3(I3)

    1=4(14) + 3(14)
  7. May 16, 2007 #6

    but dosend I4 go through the same direction as I1
    Last edited: May 16, 2007
  8. May 16, 2007 #7
    I don't know where that 14 came from. Just forget about what you wrote in your attempt and solve this by using only Kirchhoff's equations.

    The directions you choose are arbitrary. Only after you solved the equations you will know the real direction of each current.

    But for the voltage sources you must consider the direction of the voltage. Look again at the direction of the voltage from the 1V source and see what sign it should have in the equation you wrote.
  9. May 16, 2007 #8
    Yes, I2=I4
    Thats was what I meant
    But are my equations correct

    let me edit my 2nd equation

    1=4I2 +3(12-I3)
    Last edited: May 16, 2007
  10. May 16, 2007 #9
    Hint: what is the constraint on I3?

    Have you ever heard of mesh current analysis? You might be able to find some helpful examples in there.
  11. May 16, 2007 #10
    You defined I3 as the current that flows through the 3ohm resistor from the small loop. Why do you add I4 to it? I4=I2 is the current through the 4ohm resistor.
  12. May 16, 2007 #11
    I didnt add I4 to it

    1=4 I2 + 3(I2-I3)

    (12-13) would be the net current flowing into the 3 ohm resistor in the small loop
  13. May 16, 2007 #12
    or do you consider them separately.


    1 V = 4 I2 + 3 I3
  14. May 16, 2007 #13
    Then for the first loop

    2= 3 I1 + 3 i3
  15. May 16, 2007 #14
    oh sorry for the small loop i3 is negative
  16. May 16, 2007 #15
    Exactly! Now,you have 2 equations and 3 unknown currents. You need one more equation. That's where the node equation comes in.
  17. May 17, 2007 #16
    I3 doesn't matter because it is a combination of I1 and I2.
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