# thevenin

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

1. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
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?

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2. ### hgmjr Retired 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
104
0
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 Retired Moderator

Jan 28, 2005
9,029
219
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
104
0
can i say Ib: 100(Ib-.000571)+980Ib=0

6. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
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
104
0
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 Retired Moderator

Jan 28, 2005
9,029
219
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
104
0
I am really confused about what you just said

10. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
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Can you write the equation for Ib in the righthand loop?

hgmjr

11. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
Thats my problem, I don't know where to start

12. ### hgmjr Retired Moderator

Jan 28, 2005
9,029
219
Can you form the equation for the voltage across R1?

hgmjr

13. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
wouldn't it just be .0571v?

14. ### jstrike21 Thread Starter Member

Sep 24, 2009
104
0
sorry about the double post but I'm really stuck on this problem and I need some help on where to start with this.