thevenin voltage

Using nodal analysis i labeled the node closes to 2 v node v2 next v1 and the far right one vt. since the circuit is open there is no current running through the 1 ohm resistor to far right. The current value on far right says 1/2

kcl v2: (v2-2/2) + (v2-v1/2) = 1
kcl v1: (v1-v2/2) + (v1/4) +(v1-vt/2) = 0
kcl vt : (vt-v1/2) = 1/2

2((v2-2/2) + (v2-v1/2) = 1)
v2 - 2 + v2 -v1 = 2
2v2 -v1 = 2
v2 = v1/2 + 1

vt-v1 = 2
vt = 1 + v1

(4(v1-v2/2) + (v1/4) +(v1-vt/2) = 0)4
2v1 - 2v2 + v1 + 2v1 - 2vt = 0
 5v1 - 2v2 - 2vt = 0

5v1 - 2(v1/2 +1) - 2(v1 +1) = 0
5v1 - v1 - 2 - 2v1 - 2 = 0
2v1 = 4
v1 = 2

vt = 1 +2
vt = 3 V

However, the answer key says that it's 4 V. Am I doing something wrong her?
voltage.jpg
kuntilanakAsked:
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programmer78Commented:
4 v is correct
DSC02235.JPG
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kuntilanakAuthor Commented:
why doesn't my solution work? your circuit looks quite different...hmmmm
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programmer78Commented:
I haven't understood your solution, but my circuit is exactly the same as your (it is just looks more simple), sorry, I've designated resisters with bars (Russian GOST requires such a designation and haven't paid attention to this feature).

So first of all, I've considered voltage source as a current source, calculating I1. Then I used Kirchgof theorem about currents and calculated voltage drop across each resistor. Knowing these voltages you can easily calculate the Thevenin's voltage. 1 Ohm resistor doesn't affect it cause the circuit is opened.
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aburrCommented:
"Using nodal analysis i labeled the node closes to 2 v node v2 next v1 and the far right one vt. since the circuit is open there is no current running through the 1 ohm resistor to far right. The current value on far right says 1/2"
-
a good and clear start. It will be a while (hours) before I can look at the details
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kuntilanakAuthor Commented:
>>So first of all, I've considered voltage source as a current source, calculating I1. Then I used Kirchgof >>theorem about currents and calculated voltage drop across each resistor. Knowing these voltages you >>can easily calculate the Thevenin's voltage. 1 Ohm resistor doesn't affect it cause the circuit is opened.

I understand your solution now... but I like my solution better... mind giving feedback on it?
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programmer78Commented:
Would be better, if you put a picture with refdes-s (R1, R2, etc..) and if you mark nodes with latin letters (like A,B,C,D) so you was able to refer to the segments of the network. And, please, rewrite your solution, using new picture. Otherwise it is very difficult to understand, what you meant and where the mistake is.
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kuntilanakAuthor Commented:
I labeled the nodes I hope it made more sense... it's just KCL at those nodes... I think you'll get what I mean
voltage.jpg
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kuntilanakAuthor Commented:
I just want to know whether the three equations below is correct, as the steps below it are just substitution to these three equations to solve for Vt.

kcl v2: (v2-2/2) + (v2-v1/2) = 1
kcl v1: (v1-v2/2) + (v1/4) +(v1-vt/2) = 0
kcl vt : (vt-v1/2) = 1/2

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programmer78Commented:
1st equation should be: (v2 - 2)/2 + (v2-v1)/2 = 1;
2nd equation should be: (v1 - v2)/2 + (v1/4) + (v1 - vt)/2 = 0
3rd equation should be: (vt - v1)/2 = 1/2
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kuntilanakAuthor Commented:
yes... that's actually what I meant, just problems with parantheses...but then it results in 3 as the final result
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programmer78Commented:
(v2 - 2)/2 + (v2-v1)/2 = 1;
(v1 - v2)/2 + (v1/4) + (v1 - vt)/2 = 0
(vt - v1)/2 = 1/2

*2: v2 - 2 + v2 - v1 = 2;
*4: 2v1 - 2v2 + v1 + 2v1 - 2vt = 0;
*2: vt - v1 = 1

v1 = 2v2 - 4;
2*(2v2 - 4) - 2v2 + 2v2 - 4 + 2*(2v2 - 4) - 2vt = 0;
vt = 1 + v1;

4v2 - 8 - 2v2 + 2v2 - 4 + 4v2 - 8 - 2*(1+v1) = 0;

4v2 - 2v2 + 2v2 + 4v2  -  8 - 4 - 8 - 2*(1+v1) = 0;

8v2 - 20 - 2 - 2v1 = 0

8v2 - 22 - 2*(2v2 - 4) = 0

8v2 - 22 - 4v2 + 8 = 0

4v2 - 14 = 0;

v2 = 3.5 v.

v1 = 2v2 - 4 = 2*3.5 - 4 = 7  - 4 = 3 volts;

vt = 1 + v1 = 4 volts;

Seems you've mistaken in arithmetics.

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