# Situation with controlled source in Mesh Analysis. Situations without solution?

I was analyzing the circuit below, and by using Mesh Analysis (this I am currently practicing), I obtained that the voltage at the $$\6k\Omega\$$ resistor (R6) is $$\9V\$$. I was able to work this out and then checked with the simulator.

simulate this circuit – Schematic created using CircuitLab

Before that, when annotating the circuit, the $$\R2 = 4k\Omega\$$ resistor I put wrong, leaving $$\3k\Omega\$$. In this situation, considering the loop currents $$\i_A\$$ and $$\i_B\$$, for the two loops from left to right, I obtained:

$$-2V_x + 2ki_A + 3k(i_A-i_B) = 0 \Longrightarrow -2\left(3k(i_A-i_B)\right) + 2ki_A + 3k(i_A-i_B) = 0 \Longrightarrow -ki_A+3ki_B=0$$ $$-3k(i_A-i_B)-3+6ki_B = 0 \Longrightarrow -3ki_A+9ki_B = 3 \Longrightarrow -ki_A+3ki_B=1$$

Which consists of a system without a solution. In this situation, what is the explanation for not being able to find a solution for the loop currents and get the voltage at resistor R3? In the simulator, the result gave -250MV, which is clearly a "response" of the simulator for something you have no answer for.

• No solution means, such a circuit cannot exist in real and satisfy KVL at same time. Aug 2 at 9:18

You're so very close. Try solving this symbolically. You've already used the mesh equations, but you used values for resistors. Instead, use this:

$$\begin{cases} 2R_2(I_1-I_2)&=R_1I_1+R_2(I_1-I_2) \\ R_2(I_1-I_2)+3&=R_3I_2 \end{cases} \\ \Rightarrow \\ \begin{cases} I_1&=\dfrac{-3R_2}{R_1R_3+R_2(R_1-R_3)} \\ I_2&=\dfrac{3(R_1-R_2)}{R_1R_3+R_2(R_1-R_3)} \end{cases}$$

Now substitute the values for $$\R_1=2\$$ and $$\R_3=6\$$ (kΩ, Ω, mΩ, doesn't matter, they're all made in Taiwan) and you get:

$$\begin{cases} I_1&=\dfrac{-3R_2}{12-4R_2} \\ I_2&=\dfrac{6-3R_2}{12-4R_2} \tag{1} \end{cases}$$

If you use $$\R_2\$$ as a variable, like $$\x\$$, and plot it, you'll see this:

At $$\R_2=3\$$, the solution blows up: $$\12-4\cdot 3=0\$$.

• This seems comical to me. Among the infinite possible numbers for me to make a mistake, I accidentally chose the only possible one whose circuit has no solution. Well, as a final conclusion, can I understand that my system has no solution because according to those parameters involved, Kirchhoff's Laws (basic to a circuit) are not satisfied? Aug 1 at 23:14
• @benjamin_ee There are an infinite number of solutions for $R_1=2,\; R_3=6$, except when $R_2=3$. As you've seen, yourself, for $R_2=4$ there is a very nice solution. Try imposing values for any other two resistors, while making the third a variable, and you'll see that for each resistor there is a case where it fails. it also depends on the numbers. E.g. for $R_1=2,\; R_2=4$, $I_1$ fails when $R_3=4$. In fact, there are an infinite number of combinations for failures, too, but some will be negative (and some will work). So, I wouldn't call it a a mistake, I'd say treasure! ;-) Aug 2 at 6:30
• Perhaps this question is somewhat philosophical, and I understand that a controlled source is not something easily conceivable, but... in theory can I say that it is not possible to build such a circuit if I somehow had the construction of these ideal resistors/sources? Aug 2 at 13:47
• @benjamin_ee There's a bit of a contradiction in there: "build in theory". Either you have ideal components, in which case you also get infinity, or you don't. It's not that you can't build such a circuit, it's that it will not show what you expect. If you think about it, it makes sense: close to saturation, that "VCVS" will change its operating point, it will no longer be 2*V(a), it will be lesser and lesser, until it hits a dead end, when it's a fixed gain for any input. So all the currents will be a function of that limitation. Aug 2 at 15:23
• Seeing your simulation, I tried to reproduce it but ran into a number of problems. Checking, I saw that you used an R9 = 5k Ohm, why is this necessary? And how do I know it is 5k? Aug 4 at 1:29