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Kirchoff's Laws

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Hello there, hope you are well.

    I am having trouble solving this problem involving Kirchoff's laws.

    I have attached a copy of the circuit and problem to this post in jpeg format.

    Thanks again in advance for any help.

    Mike


    2. Relevant equations



    3. The attempt at a solution
    Well I merged to the two 20 ohms into one 10ohm and said that the total voltage in any loop equals zero.

    so 60=20Ia + 10Ib

    and 20=20Ib + 10Ia

    I have tried solving this many times and keep getting a negative current for Ib. My original equation must be wrong.

    Thanks alot for any help or input,

    Mike
     

    Attached Files:

  2. jcsd
  3. Jan 14, 2010 #2

    tiny-tim

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    Hi Mike! :smile:
    Which loop has Ia, and which has Ib, and in which direction? :confused:

    And what about Ic ???

    (and where do those 20s on the RHS come from?)
     
  4. Jan 14, 2010 #3
    I think Ive got it now.

    I was using loop analysis so thats why there is only two currents.

    The 20s came from 60=10Ia + 10(Ia+Ib) which expands to give 20Ia + 10Ib

    You can work out the currents using simultaneous equations, giving a Ia=3.333A and Ib=-0.6666A

    I of the load resistor, Il=Ia+Ib = 2.6667A

    So Vl=2.6667*10 = 26.667V

    Does this look about right? Thanks alot for your help

    Mike
     
    Last edited: Jan 14, 2010
  5. Jan 14, 2010 #4

    tiny-tim

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    ohhh! you should always state an intermediate step like that :rolleyes:

    (apart from anything else, you stand a good chance of making a mistake with ±s if you don't write this stuff out clearly)

    Yes, the answer looks ok now. :smile:
     
  6. Jan 14, 2010 #5
    Sorry about that, you are absolutely right... that's most probably why I was getting it wrong in the first place!

    Thanks for your help and take care,

    Mike
     
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