# Determining nodal voltages of a circuit with two supernodes

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?

• 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. Nov 6, 2023 at 1:07
• Any idea why the solution according to my textbook is $(v_1, v_2, v_3, v_4) = (3, -2.33, -1.91, 0.945)$?
– kote
Nov 6, 2023 at 3:03
• Did you try a simulation? I didn't since the numbers worked out to yours. Nov 6, 2023 at 3:36
• I trust the values you calculated, they seem correct to me. I think the book is wrong, but I couldn't tell you how. Nov 6, 2023 at 3:47
• 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.
– kote
Nov 6, 2023 at 4:32