applying nodal inspection technique to supernodes with dependent sources

Question

I was about to apply nodal analysis via inspection (to make it faster) to the above circuit, yet as we can see, it has a dependent source. So listing the equations for node(1 and 2) supernode a:

$$v_1(\frac{1}{2} + \frac{1}{6}) + v_2(\frac{1}{4}) -v_3(\frac{1}{6}) = 0$$

$$v_2(\frac{1}{4}) + v_3(\frac{1}{3} + \frac{1}{6}) - v_1(\frac{1}{6}) = 0$$

$$v_1-v_2 = 10V$$

Yet, it yields incorrect results (I tried solving it yet get wrong results). Are there any restrictions to nodal analysis by inspection when it has supernodes with dependent sources?

PS: correct answer are: \$v_1 = 3.043V\$, \$v_2 = -6.956V\$ and \$v_3= 0.6522V\$

Revision 2: instead of doing it the traditional way (like writing KCL equations), we can solve it via inspection. Are u familiar to it: The Inspection Method for Nodal Analysis and Nodal Analysis? I'm trying to apply solution by inspection here but it doesnt work the way it should be (notice my equation above).

Answer 1

Your supernode should enclose both sources. Your KCL equation is then:

\$\dfrac{v_1}{2} + \dfrac{v_2}{4} + \dfrac{v_3}{3} = 0\$

Then, you need three more equations (you have four unknowns):

\$v_1 - v_2 = 10V \$

\$v_3 - v_2 = 5\Omega \cdot i \$

\$i = \dfrac{v_1}{2\Omega} \$

Solve to get:

\$v_1 = 3.043V\$

\$v_2 = -6.956V\$

\$v_3 = 0.6522V\$

\$i = 1.522A\$

Answer 2

edit
If the numbers you've added in your PS are indeed correct then the assignment contains an error. "\$ 5 \text{ }i\$" has the dimension of current, if it should have been a voltage it should have said "\$ 5 \Omega \text{ }i\$". My answer assumes what the schematic says, not what may have been intended. I've added the answer for the corrected assignment below.
end of edit

Like I said in my answer to your other question, draw the chosen current directions so that you don't get confused and switch one of the directions halfway the calculation. I'll use the resistor values as current index, so the current though the 3 Ω resistor is \$I_{R3}\$.

I'm not sure what this supernode is, but I presume you want to apply KCL to a region instead of a single node. You'll have to clarify what this region is, because with the current equations there must be a mistake.

Anyway, we don't really need a supernode, you can set up a set of simultaneous equations as follows:

For the node \$v_1\$:

\$ I_{10V} = I_{R2} + I_{R6} \$

For the node \$v_2\$:

\$ I_{10V} + 5 I_{R2} + I_{R4} = 0 \$

For the node \$v_3\$:

\$ 5 I_{R2} + I_{R6} = I_{R3} \$

and

\$ v_1 = 2\Omega \cdot I_{R2} \$

\$ v_2 = 4\Omega \cdot I_{R4} \$

\$ v_3 = 3\Omega \cdot I_{R3} \$

\$ v_1 - v_3 = 6\Omega \cdot I_{R6} \$

We have 7 equations for 8 variables, so we need another independent equation:

\$ v_1 = v_2 + 10 V \$

This set of linear equations is easy to solve with some substitutions, and the result is:

\$ \begin{cases} v_1 = 0.989 V \\ v_2 = -9.01 V \\ v_3 = 5.27 V \\ I_{R2} = 0.495 A \\ I_{R3} = 1.76 A \\ I_{R4} = -2.25 A \\ I_{R6} = -0.714 A \\ I_{10V} = -0.220 A \end{cases} \$

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The solution for the corrected assignment:

\$ v_1 = v_2 + 10 V \$

\$ v_3 = v_2 + 5 \Omega \cdot \dfrac{v_1}{2 \Omega} = v_2 + 2.5 v_1 \$

and applying KCL to the ground node:

\$ \dfrac{v_1}{2 \Omega} + \dfrac{v_2}{4 \Omega} + \dfrac{v_3}{3 \Omega} = 0 \$

That gives us a set of 3 simultaneous linear equations in 3 unknowns, which indeed gives us

\$ \begin{cases} v_1 = 3.04348 V \\ v_2 = -6.95652 V \\ v_3 = 0.652174 V \end{cases}\$

Answer 3

This should work.

At the supernode consisting of \$v_1\$, \$v_2\$ and \$v_3\$ (the supernode encompasses both the independent and dependent voltage supplies):

$$\frac{v_1}{2}+\frac{v_1-v_3}{6}+\frac{v_2}{4}-\frac{v_1-v_3}{6}+\frac{v_3}{3} = 0$$

Conditional equations:

$$v_1-v_2 = 10V$$

$$v_3-v_2 = 5 \frac{v_1}{2}$$

Variables:

$$[v_1,v_2,v_3]$$

Results:

$$v1 = \frac{70}{23} = 3.043V$$ $$v2 = \frac{-160}{23} = -6.957V$$ $$v3 = \frac{15}{23} = 0.6522V$$