thevenin

Discussion in 'Homework Help' started by jstrike21, Oct 7, 2009.

  1. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    Just a quick question on starting this problem, is it ok to use mesh current on just the 1st two loops and solve for Ib, then substitute that into the dependent current source to find that current?
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Yes, you can solve the lefthand side and then take that results and use it to solve the righthand side. That is because there are no elements in the two lefthand loops that are dependent on the results on the righthand circuitry.

    hgmjr
     
  3. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    ok i started writing the mesh current out but i got to the dependent source and if im thinking correctly you cant use mesh current with that source there, right?
     
  4. hgmjr

    Moderator

    Jan 28, 2005
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    Perhaps it would be helpful if you posted your work so that others can see your process. If you take a wrong turn then someone can assist you in getting back on the right track.

    hgmjr
     
  5. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    can i say Ib: 100(Ib-.000571)+980Ib=0
     
  6. hgmjr

    Moderator

    Jan 28, 2005
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    My apologies, I just noticed that my earlier advice was in error. I just noticed that the voltage source in the left loop is a function of the voltage in the righthand loop. That means that you will not be able to compute the lefthand circuit without looking at the interaction between the two loops.

    hgmjr
     
  7. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    ok so now i really dont know how to get started on this one

    applying test voltages to find Rth after deactivating independent sources i think is the way to do it but im getting confused on that
     
    Last edited: Oct 7, 2009
  8. hgmjr

    Moderator

    Jan 28, 2005
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    Notice that the current source Is and R1 is the Norton's Equivalent of a voltage and series resistance. Try converting it back to the voltage and series resistance. That will make forming the equation for Ib very straightforward.

    hgmrj
     
  9. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    I am really confused about what you just said
     
  10. hgmjr

    Moderator

    Jan 28, 2005
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    Can you write the equation for Ib in the righthand loop?

    hgmjr
     
  11. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    Thats my problem, I don't know where to start
     
  12. hgmjr

    Moderator

    Jan 28, 2005
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    Can you form the equation for the voltage across R1?

    hgmjr
     
  13. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    wouldn't it just be .0571v?
     
  14. jstrike21

    Thread Starter Member

    Sep 24, 2009
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    sorry about the double post but I'm really stuck on this problem and I need some help on where to start with this.
     
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