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This is a practice question from book: Engineering circuit analysis / William H. Hayt, Jr., Jack E. Kemmerly, Steven M. Durbin. — 8th ed. (pag. 91) .

Determine the nodal voltages in the circuit:

enter image description here

Ans: v1 = 3 V, v2 = −2.33 V, v3 = −1.91 V, v4 = 0.945 V.

And this is one of my attempts (The central node is the reference node): When I solve the matrix (I do it through software so that's not the problem) the answers for voltages are not the right ones. And I don't know what I'm doing wrong (Maybe I'm not stating the equations correctly).

enter image description here

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  • \$\begingroup\$ Spice doesn't even come close to the answer numbers you give (which I assume are supposed to be correct.) That 0.15 Vx component is a voltage-controlled voltage source, depending upon (V3-V4), yes? \$\endgroup\$
    – jonk
    Commented Jun 27, 2020 at 22:39
  • \$\begingroup\$ @jonk The ones on the "Ans: " quote? Those are exactly as they're in the book. Maybe I should have tested the book answer with a simulator. But I've done that with other exercises before just to notice that book answers are actually right. And yes to what you mention about the 0.15 Vx component. \$\endgroup\$
    – ricardovaras_99
    Commented Jun 27, 2020 at 23:23
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    \$\begingroup\$ Well, Spice says different things. And so do the equations I developed, which match exactly with Spice and do NOT match those numbers. See this. If you find a flaw in that, let me know. \$\endgroup\$
    – jonk
    Commented Jun 27, 2020 at 23:36

1 Answer 1

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Your equations are correct. The only error that you have is G1+G5 in the third line, third column of the matrix. It should be G4+G5.

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  • \$\begingroup\$ I didn't noticed, but then it'd be 0.75 instead of 1.25, and it still drops a different answer from that of the book. Gotta check with the simulator then, thanks. \$\endgroup\$
    – ricardovaras_99
    Commented Jun 27, 2020 at 23:27
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    \$\begingroup\$ I derived the results using another method and got the same answer with your equations. The answer of the book would be right if we switch the polarity of the voltage-controlled source 0.15vx. You can simply check it by comparing v3-v2 to 0.15x(v3-v4). \$\endgroup\$
    – Paul Ghobril
    Commented Jun 28, 2020 at 4:56

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