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Let the following circuit : enter image description here

Using KCL : \$ I_x = i + i', i' = I - 3 \$ and transforming the left-most branch that contains the 1 A current-source and the 1 ohm resistor, into a voltage current and the 1 ohm resistor in series with it thus the current \$ i \$ is going through that resistor, we get \$2 V = 1*i - 1 V \iff i = 3 A \$ so going back to the original circuit : \$ i_1 = 3 - 1 = 2 A \$ and since the 1 ohm resistor -the one on the right of the 2V source - is parallel with the other 1 ohm resistor on the left they have the current going through them 2 amps. so in conclusion : \$ I = 2 A \rightarrow i' = -1 A \rightarrow I_x = 2 A \$ Now using KVL to calculate \$V_x \$ : we have in the middle mesh ; \$ 3 V = V_x + 3*1 - 2 * 1 \iff V_x = 2 V \$ is this valid, my logic and everything.

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  • \$\begingroup\$ Yes, Ix is 2A but Vx is not 2V. It cannot be. \$\endgroup\$
    – G36
    Sep 16, 2023 at 13:09
  • \$\begingroup\$ @G36 -2V then ? tell me is my approach in the calculations of \$ V_x \$ is wrong \$\endgroup\$
    – HellBoy
    Sep 16, 2023 at 13:23
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    \$\begingroup\$ Yes, -2V is the right answer. As for the approach. Notice that we have 3V on the right and on the left we have 2V from the voltage source plus the voltage drop across 1 ohm's resistor (3V). Thus, we have 5V on the left and 3V on the right. Therefore Vx = 3V - 5V = -2V \$\endgroup\$
    – G36
    Sep 16, 2023 at 13:28
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    \$\begingroup\$ "Calculating the current produced by the voltage source" --- Voltage sources don't "produce" current. They don't determine the current. The load takes the current. The load determines the current. \$\endgroup\$ Sep 16, 2023 at 13:29
  • \$\begingroup\$ @DavideAndrea they do produce a current to keep a constant voltage, thats what i've read in another valid answer here in SE \$\endgroup\$
    – HellBoy
    Sep 16, 2023 at 13:30

3 Answers 3

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is this valid, my logic and everything.

Your logic For the determination of \$I_x\$ appears correct.
I find that the following approach has more clarity and is less error prone.
The elements \$R_3\$ and \$I_2\$ have been interchaged. Then \$R_3\$ is not included in the analysis of node a.

transforming the left-most branch that contains the 1 A current-source and the 1 ohm resistor, into a voltage source...

This transformation, while not incorrect, does not add value to the solution. Neither does the knowledge of \$i\$ and \$i'\$.

By inspection, notice in Figure 1, that there is one voltage-node indicated by the red wires, and 3 current-junctions. While KCL can be applied to each of the 3 luctions separately, KCL can be applied to all three junctions in the voltage-node with the same equation. $$ I_1+I_{R1}+I_x+I_{R2}+I_2=0 $$

Considering currents into the node as positive, and currents leaving the node as negative, the values are substituted and combined into a value for \$I_x\$ . $$ -1-2+I_x-2+3=0 $$

This approach allows the solution through inspection and a little mental arithmetic. Even writing as show allows quick use of a calculator for harder arithmetic.

schematic

simulate this circuit – Schematic created using CircuitLab

You had the right logic for the voltage \$V_x\$. Just missed the sign.
Figure 2 shows the reduced circuit for determination of \$V_2\$.

Since the current enters \$R_3\$ from the right, that end is labeled +. The left end labeled -.

Clear labeling helps preventing incorrect polarities on elements. So with the voltage \$V_{R3}=3V\$, KVL is applied as: $$ V_1+V_{R3}+V_x=V_2 $$ $$ 2+3+V_x=3 $$

schematic

simulate this circuit

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First, I will present a method that uses Mathematica to solve this problem. I know that this approach is not 'smart' but this method will work all the time, even when the circuit is (way) more complicated than this one. Also, this method will check your work.

Well, the circuit we want to analyze is given by:

schematic

simulate this circuit – Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} 0&=\text{I}_\text{d}+\text{I}_1+\text{I}_6\\ \\ \text{I}_7&=\text{I}_0+\text{I}_6\\ \\ \text{I}_2&=\text{I}_3+\text{I}_7\\ \\ \text{I}_8&=\text{I}_3+\text{I}_9\\ \\ \text{I}_4&=\text{I}_5+\text{I}_9 \end{alignat*} \end{cases}\tag1 $$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \begin{alignat*}{1} \text{I}_1&=\frac{\displaystyle\text{V}_\text{a}-0}{\displaystyle\text{R}_1}\\ \\ \text{I}_2&=\frac{\displaystyle\text{V}_\text{a}-0}{\displaystyle\text{R}_2}\\ \\ \text{I}_3&=\frac{\displaystyle\text{V}-\text{V}_\text{a}}{\displaystyle\text{R}_3}\\ \\ \text{I}_4&=\frac{\displaystyle\text{V}_\text{b}-0}{\displaystyle\text{R}_4}\\ \\ \text{I}_5&=\frac{\displaystyle\text{V}_\text{c}-\text{V}_\text{b}}{\displaystyle\text{R}_5} \end{alignat*} \end{cases}\tag2 $$

Using \$(2)\$ we can rewrite \$(1)\$ as follows:

$$ \begin{cases} \begin{alignat*}{1} 0&=\text{I}_\text{d}+\frac{\displaystyle\text{V}_\text{a}-0}{\displaystyle\text{R}_1}+\text{I}_6\\ \\ \text{I}_7&=\text{I}_0+\text{I}_6\\ \\ \frac{\displaystyle\text{V}_\text{a}-0}{\displaystyle\text{R}_2}&=\frac{\displaystyle\text{V}-\text{V}_\text{a}}{\displaystyle\text{R}_3}+\text{I}_7\\ \\ \text{I}_8&=\frac{\displaystyle\text{V}-\text{V}_\text{a}}{\displaystyle\text{R}_3}+\text{I}_9\\ \\ \frac{\displaystyle\text{V}_\text{b}-0}{\displaystyle\text{R}_4}&=\frac{\displaystyle\text{V}_\text{c}-\text{V}_\text{b}}{\displaystyle\text{R}_5}+\text{I}_9 \end{alignat*} \end{cases}\tag3 $$

Now, we can set up a Mathematica code to solve for all the voltages and currents:

In[1]:=Clear["Global`*"];
FullSimplify[
 Solve[{0 == Id + I1 + I6, I7 == I0 + I6, I2 == I3 + I7, 
   I8 == I3 + I9, I4 == I5 + I9, I1 == (Va - 0)/R1, I2 == (Va - 0)/R2,
    I3 == (V - Va)/R3, I4 == (Vb - 0)/R4, I5 == (Vc - Vb)/R5}, {I0, 
   I1, I2, I4, I5, I6, I7, I8, I9, V}]]

Out[1]={{I0 -> -I3 + Id + (1/R1 + 1/R2) Va, I1 -> Va/R1, I2 -> Va/R2, 
  I4 -> Vb/R4, I5 -> (-Vb + Vc)/R5, I6 -> -((Id R1 + Va)/R1), 
  I7 -> -I3 + Va/R2, I8 -> I3 + ((R4 + R5) Vb - R4 Vc)/(R4 R5), 
  I9 -> ((R4 + R5) Vb - R4 Vc)/(R4 R5), V -> I3 R3 + Va}}

Now, we can find:

$$\text{V}_x:=\text{V}_\text{b}-\text{V}=\text{V}_\text{b}-\left(\underbrace{\text{I}_3\text{R}_3+\text{V}_\text{a}}_{=\space\text{V}}\right)=3-\left(3\cdot1+2\right)=-2\space\text{V}\tag4$$

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The circuit can be simplified. I present the analysis of the partially modified circuit, based on the notions of network topology with the node voltage method:

enter image description here enter image description here enter image description here enter image description here enter image description here enter image description here

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  • \$\begingroup\$ such detailed answer by Franc even though its abit over my head currently, but infinite gratitude! \$\endgroup\$
    – HellBoy
    Sep 17, 2023 at 10:47

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