How to solve for multiple unknowns using substitution?

Question

$R_1$, $R_2$, $R_3$, $R_4$, $R_5$ and $V_6$ suppose to be 'known' values.

$$\frac{V_{n_1}}{R_1} + \frac{V_{n_1}-V_{n_3}}{R_2} + i_6 = 0$$

$$ \frac{V_{n_2}-V_{n_3}}{R_4} + \frac{V_{n_2}-V_{n_4}}{R_3} - i_6 = 0 $$

$$ \frac{V_{n_3}-V_{n_1}}{R_2} + \frac{V_{n_3}-V_{n_2}}{R_4} + \frac{V_{n_3}}{R_5} = 0 $$

$$ V_{n_4} = V_6 $$

$$ V_{n_1}-V_{n_2} = 2V_6 $$

I could not see a quick way to solve it using substitution, because if you want to solve for $V_{n_1}$ for example, you have $V_{n_2}$ and $V_{n_3}$ as unknowns. Could anyone help me with this?

Could I do this?: Substitute $ V_{n_1} = 2V_6 + V_{n_2} $ in $\frac{V_{n_1}}{R_1} + \frac{V_{n_1}-V_{n_3}}{R_2} + i_6 = 0$ to get:

$$\frac{2V_6 + V_{n_2}}{R_1} + \frac{2V_6 + V_{n_2}-V_{n_3}}{R_2} + i_6 = 0$$

And add $ \frac{V_{n_2}-V_{n_3}}{R_4} + \frac{V_{n_2}-V_{6}}{R_3} - i_6 = 0 $ (after this substitution: $ V_{n_4} = V_6 $) to it to get:

$$\frac{2V_6 + V_{n_2}}{R_1} + \frac{2V_6 + V_{n_2}-V_{n_3}}{R_2} + \frac{V_{n_2}-V_{6}}{R_3}+ \frac{V_{n_2}-V_{n_3}}{R_4} = 0$$

Now I can substitute $ V_{n_1} = 2V_6 + V_{n_2} $ in $ \frac{V_{n_3}-V_{n_1}}{R_2} + \frac{V_{n_3}-V_{n_2}}{R_4} + \frac{V_{n_3}}{R_5} = 0 $ to get:

$$ \frac{V_{n_3}-2V_6 - V_{n_2}}{R_2} + \frac{V_{n_3}-V_{n_2}}{R_4} + \frac{V_{n_3}}{R_5} = 0 $$

And add the last two equations to get:

$$ \frac{2V_6 + V_{n_2}}{R_1} + \frac{V_{n_2}-V_{6}}{R_3} + \frac{V_{n_3}}{R_5} = 0 $$

But now, I still have two unknowns.

$ \frac{V_{n_3}-2V_6 - V_{n_2}}{R_2} + \frac{V_{n_3}-V_{n_2}}{R_4} + \frac{V_{n_3}}{R_5} = 0 $

$ V_{n_3} (\frac{1}{R_5} + \frac{1}{R_4} + \frac{1}{R_2}) = \frac{2V_6}{R_2} + V_{n_2} (\frac{1}{R_2} + \frac{1}{R_4}) $

$ V_{n_3} = \frac{\frac{2V_6}{R_2} + V_{n_2} (\frac{1}{R_2} + \frac{1}{R_4})}{(\frac{1}{R_5} + \frac{1}{R_4} + \frac{1}{R_2})} $

$ \frac{2V_6 + V_{n_2}}{R_1} + \frac{V_{n_2}-V_{6}}{R_3} + \frac{\frac{\frac{2V_6}{R_2} + V_{n_2} (\frac{1}{R_2} + \frac{1}{R_4})}{(\frac{1}{R_5} + \frac{1}{R_4} + \frac{1}{R_2})}}{R_5} = 0 $

$ \frac{2V_6 + V_{n_2}}{R_1} + \frac{V_{n_2}-V_{6}}{R_3} + \frac{\frac{2V_6}{R_2} + V_{n_2} (\frac{1}{R_2} + \frac{1}{R_4})}{(\frac{1}{1} + \frac{R5}{R_4} + \frac{R5}{R_2})} = 0 $

$ \frac{(1 + \frac{R5}{R_4} + \frac{R5}{R_2})(2V_6 + V_{n_2})}{R_1} + \frac{(1 + \frac{R5}{R_4} + \frac{R5}{R_2})(V_{n_2}-V_{6})}{R_3} + \frac{2V_6}{R_2} + V_{n_2} (\frac{1}{R_2} + \frac{1}{R_4}) = 0 $

As you can see, this takes a lot of steps to solve and does not lead to a simple and quick solution....

Answer 1

You can use the third equation to eliminate $V_{n_4}$ The second gives $V_{n_1}=2V_6+V_{n_2}$. Use that to eliminate $V_{n_1}$ from the first two, which leaves two equations in two unknowns. In general, you just use one equation to eliminate one unknown from the rest and continue until you are down to one equation in one unknown. If $i_6$ is not already known, you are one equation short as you have five unknowns.

Answer 2

Let's simplifly the notation putting, kor $k=1,\ldots, 6$, $V_{n_k}=V_k$, $\frac{1}{R_k}=G_k$, $i_6=i$ and $V_6=v$. In this way, your system becomes $$\left\{ \begin{align} V_1G_1+(V_1-V_3)G_2&=-i\\ (V_2-V_3)G_4+(V_2-V_4)G_3&=i\\ (V_3-V_1)G_2+(V_3-V_2)G_4+V_3G_5&=0\\ V_4&=v\\ V_1-V_2&=2v \end{align}\right. $$ or $$\left\{ \begin{align} (G_1+G2)V_1-G_2V_3+i&=0 & (E_1)\\ (G_3+G_4)V_2-G_4V_3-G_3V_4-i&=0 & (E_2)\\ -G_2V_1-G_4V_2+(G_2+G_4+G_5)V_3&=0 & (E_3)\\ V_4&=v & (E_4)\\ V_1-V_2&=2v & (E_5) \end{align}\right. $$ So you have five equation in the five unkwnown $V_1,V_2,V_3, V_4, i$.

Now follow these steps:

1. $E_1+E_2+G_3E_4\to E_2$: substitute the second equation with the sum of the first and the second equation and the third multiplied by $G_3$;
2. $E_2+E_3\to E_3$: substitute the third equation with the sum of the second and the third equation;
3. $\frac{1}{G_5}E_3\to E_3$: substitute the third equation with the third equation divided by $G_5$;
4. $E_5\leftrightarrow E_1$: exchange the first equation and the fifth equation;
5. $(G_2+G_4)E_3+E_2\to E_2$: substitute the second equation with the sum of the second equation and the third multiplied by $G_2+G_4$;
6. $-E_1\leftrightarrow E_2$: exchange the second equation and the first equation multiplied by $-1$;
7. $[(G_2+G_4)G_3/G_5+(G_3+G_4)]E_2-E_1\to E_1$: substitute the first equation with the difference of the second equation multiplied by $(G_2+G_4)G_3/G_5+(G_3+G_4)$ and the first equation.

After the above steps the system becomes $$\left\{ \begin{align} g_{11}V_1&=gv & (E_1)\\ V_2-V_1&=-2v & (E_2)\\ \frac{G_1}{G_5}V_1+\frac{G_3}{G_5}V_2+V_3&=\frac{G_3}{G_5}v & (E_3)\\ V_4&=v & (E_4)\\ (G_1+G_2)V_1-G_2V_3+i&=0 & (E_5) \end{align}\right. $$ where $$\begin{align} g_{11}&=(G_2+G_4)\frac{G_1+G_3}{G_5}+G_1+G_2+G_3+G_4\\ g&=2\left[(G_2+G_4)\frac{G_3}{G_5}+G_3+G_4\right]+G_3+(G_3+G_4)\frac{G_3}{G_5} \end{align} $$ Finally we will find $$\small \left\{ \begin{align} V_1&=v\frac{g}{g_{11}}\\ V_2&=V_1-2v=v\left(\frac{g}{g_{11}}-2\right)\\ V_3&=\frac{G_3}{G_5}v-\frac{G_1}{G_5}V_1-\frac{G_3}{G_5}V_2=v\left[\frac{G_3}{G_5}-\frac{G_1}{G_5}\frac{g}{g_{11}}-\frac{G_3}{G_5}\left(\frac{g}{g_{11}}-2\right)\right]\\ V_4&=v\\ i&=-(G_1+G_2)V_1+G_2V_3=v\left\{-(G_1+G_2)\frac{g}{g_{11}}+G_2\left[\frac{G_3}{G_5}-\frac{G_1}{G_5}\frac{g}{g_{11}}-\frac{G_3}{G_5}\left(\frac{g}{g_{11}}-2\right)\right]\right\} \end{align}\right. $$