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In this question we are required to find the nodal voltages and current in the circuit.

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Why is the 3A current not divided at node (d), that is, one going in 5 ohm resistor and other towards the source. The loop equation is given as:

enter image description here

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  • \$\begingroup\$ The 3 A current does divide at node d. However, the whole point of mesh analysis is to assume pseudo-currents (the word pseudo is just for your mental use so that you recognize it's not necessarily the same as final resulting device currents) in order to develop voltage drops around the loop which must sum to zero according to KVL. These pseudo currents add up (relative signs according to relative directions) in parts that share two loop currents, for example. I've not looked at your equations, though. Just saying something about your node d question, is all. \$\endgroup\$ Commented Nov 8, 2023 at 5:32
  • \$\begingroup\$ The 3 A loop current is a pseudo-current for that loop (it is also an actual current for the current source -- an accidental coincidence here) and you treat it as if it did not divide up. The simultaneous use of KVL on the three loops will work out in the end and will tell you how it actually divides up. But before that moment where you have the full answer, you treat them as though they run around the loop without dividing up. When you get i1 you will know how much of i3 actually goes through the 5 Ohm resistor and how much through the 12 V source. But you won't know until you get i1. \$\endgroup\$ Commented Nov 8, 2023 at 5:34

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schematic

simulate this circuit – Schematic created using CircuitLab

Consider an electrical network made up only of resistors, independent voltage and current generators. It turns out that the presence of current generators makes the direct application of the mesh-based analysis method impossible because the auxiliary variables, i.e. the co-tree currents aren't sufficient to express the voltages at the terminals of each current generator given that the voltage at the ends of these generators is independent of the current passing through them and therefore must be considered an unknown quantity. So the unknowns are: the auxiliary variables and the voltages at the terminals of the current generators. To this increase in the number of the unknowns, corresponds an equal increase in the number of equations, as each current generator introduces a constraint between the currents and therefore a new equation.

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    \$\begingroup\$ For material copied from another source we need to see a link to the original source or at least a citation for it. Can you add that, please? \$\endgroup\$ Commented Nov 8, 2023 at 15:48
  • \$\begingroup\$ The two graphs, i.e. the circuit provided by the person who published the post and the oriented graph with independent meshes, were created by me with Windows Paint , as well as calculations. The theory, however, can be traced back to the text from the nineties, "Fondamenti di Elettrotecnica" Volume one, Second edition; authors: "G. Martinelli, M. Salerno" Editrice Siderea. \$\endgroup\$ Commented Nov 10, 2023 at 9:00
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Let's analyze the circuit using nodal analysis. Here's how we can break down the problem:

  1. Identify the Nodes

Node 1: The junction point where the 12V source, 5-ohm resistor, and 2-ohm resistor meet.
Node 2: The junction point where the 6V source, 1-ohm resistor, and 2-ohm resistor meet.
Node 3: The junction point where the 3A source, 5-ohm resistor, and 6-ohm resistor meet.
Reference Node: We'll choose the bottom node as our reference node (ground).

  1. Define Node Voltages

Let V1 be the voltage at Node 1 with respect to the reference node.
Let V2 be the voltage at Node 2 with respect to the reference node.
Let V3 be the voltage at Node 3 with respect to the reference node.

  1. Apply Kirchhoff's Current Law (KCL) at each Node

Node 1:
(V1 - 12) / 5 + (V1 - V2) / 2 = 0
Node 2:
(V2 - V1) / 2 + (V2 - 6) / 1 = 0
Node 3:
(V3 - V1) / 5 + (V3 - V2) / 6 = 3

  1. Solve the System of Equations

We now have three equations with three unknowns (V1, V2, V3). Solving this system of equations will give us the nodal voltages.

Simplifying the Equations

Equation 1: 2V1 - 24 + 5V1 - 5V2 = 0 => 7V1 - 5V2 = 24
Equation 2: 3V2 - 3V1 + 6V2 - 36 = 0 => -3V1 + 9V2 = 36
Equation 3: 6V3 - 6V1 + 5V3 - 5V2 = 90 => -6V1 - 5V2 + 11V3 = 90
Solving for V1, V2, and V3

You can use methods like substitution, elimination, or matrix methods to solve this system. Here's a possible approach using elimination:

Multiply Equation 1 by 3 and Equation 2 by 7:
21V1 - 15V2 = 72
-21V1 + 63V2 = 252
Add the two equations to eliminate V1:
48V2 = 324 => V2 = 6.75V
Substitute V2 back into Equation 1:
7V1 - 5(6.75) = 24 => V1 = 9.75V
Substitute V1 and V2 into Equation 3:
-6(9.75) - 5(6.75) + 11V3 = 90 => V3 = 15.75V
5. Calculate the Currents

Now that we have the nodal voltages, we can calculate the currents using Ohm's Law:

i1: (V1 - 12) / 5 = (9.75 - 12) / 5 = -0.45A (Current flows opposite to the assumed direction)
i2: (V2 - V1) / 2 = (6.75 - 9.75) / 2 = -1.5A (Current flows opposite to the assumed direction)
i3: (V3 - V1) / 5 = (15.75 - 9.75) / 5 = 1.2A

Summary

V1 = 9.75V
V2 = 6.75V
V3 = 15.75V
i1 = -0.45A
i2 = -1.5A
i3 = 1.2A

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