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circuit

I am trying to determine the nodal voltages (namely, \$v_{1},\ v_{2},\ v_{3},\ \text{and}\ v_{4}\$).

I have determined that there are two supernodes: one between nodes \$v_1\$ and the reference node and one between \$v_2\$ and \$v_3\$.

Therefore, my equations are:

\$v_1 = 3\$

\$4 = \frac{v_2-v_1}{1} + \frac{v_3}{2} + \frac{v_3-v_4}{4}\$

\$0 = \frac{v_4}{3}+\frac{v_4-v_1}{2}+\frac{v_4-v_3}{4}\$

\$v_3-v_2=0.15(v_3-v_4)\$

Upon solving, I get a completely different answer than my textbook. My answer: \$(v_1,\ v_2,\ v_3,\ v_4)=\\\small (3,\ 1727/410,\ 928/205,\ 498/205)= \\\small (3,\ 4.212,\ 4.527,\ 2.429).\$

So my question is: is my system of equations correct and what are the nodal voltages for the circuit shown?

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  • \$\begingroup\$ Your schematic is nothing more than this. And you know that \$v_3 = v_2 + 0.15\cdot\left(v_3-v_4\right)\$ and can solve that for \$v_3\$ easily enough. (It's dependent, not independent.) So really only two nodal equations needed. And yes, I think you got those answers correct. \$\endgroup\$ Nov 6, 2023 at 1:07
  • \$\begingroup\$ Any idea why the solution according to my textbook is \$(v_1, v_2, v_3, v_4) = (3, -2.33, -1.91, 0.945)\$? \$\endgroup\$
    – kote
    Nov 6, 2023 at 3:03
  • \$\begingroup\$ Did you try a simulation? I didn't since the numbers worked out to yours. \$\endgroup\$ Nov 6, 2023 at 3:36
  • \$\begingroup\$ I trust the values you calculated, they seem correct to me. I think the book is wrong, but I couldn't tell you how. \$\endgroup\$ Nov 6, 2023 at 3:47
  • \$\begingroup\$ I just ran a circuit simulator and confirmed that these values are indeed correct. I must have redid the problem 5 times because my answer was not aligning with the textbook's answer. What an infuriating thing! Thanks for all the help. \$\endgroup\$
    – kote
    Nov 6, 2023 at 4:32

1 Answer 1

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Analysis based on the study of the network topology and pairs of independent nodes:

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