# Need help finding the Thevenin Equivalent

Discussion in 'Homework Help' started by jut, Dec 20, 2007.

1. ### jut Thread Starter Senior Member

Aug 25, 2007
224
2
Dear All,

I've spent a couple hours on this problem but can't seem to crack it.

The goal is to find a Thevenin equivalent circuit to the left of terminals a,b.

Here are the equations that I've come up with, but have no solution when solving w/matlab.

KVL going CW around mesh on the left:
10 - 2i1 - v = 0

KCL at node c:
i1 - v/2 - i2 - 0.4v = 0

Nodal Analysis at node c, using node b as the reference node (not sure about this equation):
-(Vc - Vd)/2 - Vc/2 - (Vc - Va)/2 - 0.4v = 0

Va = v - 2*i2
Vd = 10

After simplifying each equation and putting them into a matrix, it does not yield a solution. Could someone please help verify the 3 equations?

2. ### leonidus Member

Sep 23, 2008
15
0
jut
U can solve this pb, as follows:
Consider I1 current flows in the left mesh having 10v supply.
Now, apply KVL in mesh1: 10-2(I1)-2(I1)=0 i.e. i1=I1=2.5A
So, v=2*(I1)=5v
Now, by KCL at node c with ref. to node b ,
(assume outgoing current to be positive)
-i1 +i1+0.4v+(Vc-Va)/2=0 ...(1)
Also, Vc=2*(I1)=v=5 ...(2)
Solve (1),(2), we get Va=9v i.e. Vth=Vab=9v
U will now get it.

3. ### veritas Active Member

Feb 7, 2008
167
0
Did you already solve for Rth? This involves removing the voltage/current sources.

This problem shouldn't require Matlab to solve: just solve one source at a time, and use superpositioning.

4. ### hitmen Active Member

Sep 21, 2008
159
0
Yeah. Rth = indept src killed. the 2 volt get short circuited so you are left with 2//2.

5. ### hitmen Active Member

Sep 21, 2008
159
0

KCL at node c:
i1 - v/2 - i2 - 0.4v = 0

There is only 1 current flowing thru i2 = 0.4v. you are substracting the same current twice. DOnt let the 0.4v current fool you. It is the same node! There is no pd difference between them. The same goes for your nodal equation!

Good luck. I am a student too & I am stuck with capacitors

6. ### neon9 Member

Oct 8, 2008
15
0
begin use the volts then R1XR2/R1+R2= A-B THEVENIN THEOREM SUGEST TO SIMPLiFY circuits by adding all voltages reducing all resistrance to equivalents r1∏r2
example +5v-10V = R1∏R2 R1=11K R2=10K then 5 -10= -5v this the source volts then 1/11k+1/10k/r1+r2=5.2381k to simplfy [-5v and 5.2381k is the thevenin equivalent] if more networks are added use the same principle. never try to use current when you are thevenising. if you use current then that is another method like node analysis and then some more. i hope this help.

7. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Take a look at this material on Thevenin's Theorem in the AAC ebook. There are some good worked examples for you to practice on.

hgmjr