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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.

schematic

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.

benjamin_ee
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1 Answers1

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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:

infinte

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

a concerned citizen
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  • 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? – benjamin_ee Aug 01 '22 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! ;-) – a concerned citizen Aug 02 '22 at 06:30
  • 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? – benjamin_ee Aug 02 '22 at 13:47
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    @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](https://i.stack.imgur.com/dPFTB.png). 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. – a concerned citizen Aug 02 '22 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? – benjamin_ee Aug 04 '22 at 01:29
  • @benjamin_ee That resistor is added to equalize the offset, and it's chosen as `R7 || R8`. Since the OP has been answered (or else, why would you have selected the answer?), it's best to ask new questions if you have additional info. That's because the comments section does not participate in a search (for future users). But first, please make sure that what you'er asking isn't already answered. – a concerned citizen Aug 04 '22 at 08:09