# Help with Thevenin's Theorem!

Discussion in 'Homework Help' started by tonlika, Jun 25, 2011.

1. ### tonlika Thread Starter New Member

May 22, 2011
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I have to find the current across V2 using Thevenin's theorem.

Below in the attachments, draft2.jpg is the circuit and draft5.jpg is the Thevenin equivalent.

R2 and R3 are in series and give a 6 Ohm resistance.
R1 is shorted out. R4 is in parallel with R(23) so 1/Rt = 1/R(23) + 1/R4
1/6 + 1/6 = 2/6 which gives Rt the total resistance equal to 3 Ohm.

Then I write the nodal voltage equations:

Va/2 + (Va-V1)/4 = 0
Vb/6 + 4 + (Vb-V1)/2 = 0
(V1-Va)/4 - 4 + (V1-Vb)/2 = 0

which give:

3*Va - V1 = 0
4*Vb - 3*V1 = -24
-Va - 2*Vb + 3*V1 = 16

and solving the system/matrix for Va, Vb and V1 gives approximately:
Va = 1.1
Vb = -3.4
V1 = 3.4

The Thevenin voltage equivalent should be Va-Vb = 1.1 - (-3.4) = 4.5 Volts

So the current over V2 should be: I = (Vt + V2)/Rt = (4.5 + 8)/3 = 4.17 Amps

But when I do the Ltspice simulation the current is shown to be -6 A.

Can you please tell me what I'm doing wrong?

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Last edited: Jun 25, 2011
tonlika likes this.

May 22, 2011
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4. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Do you mean this version?
Now we have one supernode so the equation will look like this

For Vb

$\frac{Vb}{2}+\frac{Vb-V2}{6}+4 = 0$

And for V2 + Va

$\frac{Vb-V2}{6} = \frac{Va}{4}$

And

$Va = V2+12V$

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5. ### tonlika Thread Starter New Member

May 22, 2011
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Yes, this version.
Thank you so much! I got it!