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I need help figuring this problem. I'm reviewing for my final and I can't figure out how to write the nodal equations. What I have so far is:

$$6I_x = V_2- V_1 $$

KCL @ V_1: $$7A + \frac{V_1}{10} = 0 $$

$$V_1(\frac{1}{10}) = -7A$$

KCL @ V_2: $$\frac{V_2}{5} +\frac{V_2}{2} = 0 $$

$$ V_2(\frac{1}{5} + \frac{1}{2}) =0$$

I know that the answer will be the matrix:

$$ \begin{matrix} .4 & 1 & =& 0 \\ .1 & .7 & =& -7 \\ \end{matrix} $$

Circuit with depended voltage source

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Writing nodal equation at ground node, $$-7 =\frac{V_1}{10} + \frac{V_2}{5} + \frac{V_2}{2}$$ $$or,\ \mathbf{0.1V_1 + 0.7V_2 = -7}$$ Writing nodal equation at \$V_1\$ correctly will also produce the same equation. You actually missed the term \$0.7V_2\$ in it.

Now, $$V_2 = V_1+6I_x$$ but, $$I_x = -0.1V_1$$ then, $$V_2 = 0.4V_1$$ $$\mathbf{0.4V_1 - V_2 = 0}$$

Writing these equations in matrix format, $$ \left[ \begin{array}{cc} 0.4 & -1 \\ 0.1 & 0.7\end{array} \right]\left[ \begin{array}{c} V_1 \\ V_2\end{array} \right]=\left[ \begin{array}{c} 0 \\ -7\end{array} \right] $$

I am getting an extra -1. Correct me if I am wrong.

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  • \$\begingroup\$ I solved your last matrix equation and got the same result as me, so it seems correct. \$\endgroup\$ Jan 7 at 3:28
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Your KCL equations are wrong because you didn't consider the current through the current-dependent voltage source. You could consider it, and separately apply KCL at node 1 and node 2. Or, you could apply KCL at the supernode shown in your image. Below I show how to solve it with a supernode. I'll ignore the units.

Apply KCL at the supernode:

\$- 7 - \dfrac{V_1-0}{10} - \dfrac{V_2-0}{5} - \dfrac{V_2-0}{2} = 0 \tag*{} \$

\$\implies \left(- \dfrac{1}{10} \right) V_1 - \left(\dfrac{1}{5} + \dfrac{1}{2} \right) V_2 = 7 \tag*{}\$

\$\implies \left(- \dfrac{1}{10} \right) V_1 + \left(-\dfrac{7}{10}\right) V_2 = 7 \tag 1\$

Constraint equation for the supernode: express the voltage of the voltage source (\$6 \, I_x\$) in terms of nodal voltages:

\$6 \, I_x = V_2 - V_1 \tag 2\$

Constraint equation for the dependent source: express the controlling variable (\$I_x\$) in terms of nodal voltages:

\$I_x = \dfrac{0-V_1}{10} \implies I_x = -\dfrac{1}{10} V_1 \tag 3\$

Notice you have three equations ((1), (2), (3)) and three unknowns (\$V_1\$, \$V_2\$, \$I_x\$), therefore you can solve the system of linear equations.

However, the statement of your problem indicates to obtain a system of equations with only the unknowns \$V_1\$, \$V_2\$. So what you have to do is get rid the variable \$I_x\$, which will result in a system of two equations with two unknowns. To do that, simply substitute eq. (3) in eq. (2) (you could instead substitute (2) in (3)) to get:

\$6 \, \left(-\dfrac{1}{10} V_1\right) = V_2 - V_1 \tag*{}\$

\$\implies -\dfrac{6}{10} V_1 - V_2 + V_1 = 0 \tag*{}\$

\$\implies \left(-\dfrac{3}{5} +1 \right) V_1 - V_2 = 0 \tag*{}\$

\$\implies \left(\dfrac{2}{5}\right) V_1 + (- 1) V_2 = 0 \tag 4\$

Now, with equations (1) and (4) we have two equations and two unknowns, and only the node voltages as variables. We can write them as a single matrix equation:

\$ \begin{bmatrix} -\dfrac{1}{10} & -\dfrac{7}{10} \\ \dfrac{2}{5} & -1 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} 7 \\ 0 \end{bmatrix} \tag*{} \$

\$ \implies \begin{bmatrix} -0.1 & -0.7 \\ 0.4 & -1 \end{bmatrix} \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} 7 \\ 0 \end{bmatrix} \tag 5 \$

which is what you were asked to find.


Prove: To check if eq. (5) is correct, we can solve it for the nodal voltages (by any means, be it Cramer's rule, substitution, matrix inversion, with MATLAB, with a calculator, etc.) and compare it to a simulation. After solving eq. (5), we get:

\$ \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} = \begin{bmatrix} -\dfrac{350}{19} \\ -\dfrac{140}{19} \end{bmatrix} \approx \begin{bmatrix} -18.42 \\ -7.37 \end{bmatrix} \tag 6\$

Now we simulate the circuit. I used LTspice simply because it's free and fast. Here's the circuit in the simulator:

Circuit in LTspice

(The voltage source in series with the resistor of 10 Ω is to use its current as the controlling current of the dependent source. Notice its voltage is zero, so it doesn't affect the circuit). Here are the simulation results:

Results

If now you compare the highlighted text in the previous image with eq. (6), you'll see our answer checks.

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