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Picture from khan academy

Hello, day 2 of learning electronics here.

Why is I3 = -I,

Why doesn't current come from the upper node, the one that comes after R3 and then splits into the current source?

Shouldn't it be I3 = -I plus whatever comes down the node?

Coder
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    Instruction note 2. says "Solve easy ones." I3 is the easy one because it's defined by the current source. "-I plus whatever comes down the node" is maybe a description of the current through R4, but it's not one of the mesh currents. – Theodore May 16 '23 at 19:19
  • For some reason I feel like emphasizing this is just one of many possible techniques for calculating currents and voltages in certain types of circuits. That's why the title says "Mesh Current Method" and not "How to Solve Circuits". – user253751 May 16 '23 at 21:49
  • Coder, if you are a programmer with an applied mathematics bend then you may enjoy learning how directed graph theory applies. Both KCL (nodal) and KVL (mesh) are brought under a single roof (column space, row space, and their associated null spaces, and the rank-nullity theorem.) This broader view, without alteration, applies to many domains you frankly wouldn't often remotely connect. For example probability often requires the sum of alternatives to be 1. Not unlike KCL's rule. See here for a small sample. – periblepsis May 17 '23 at 13:35

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There will be some current that enters or leaves the I/R3/R4 node in the R3 direction. Don't think of the loop currents as existing in isolation (even if you mathematically have to treat them as separate); they are generally just mathematical constructs that allow you to tidily represent the circuit as a system of equations.

In the case of I3, you do know what the currents are in the "right 3/4 of the loop" (the wires that are attached to the current source and nothing else) because the assumptions are that the current is confined to the wires and that the current source maintains the current through itself regardless of what the voltage across its terminals is. R4, however, is part of two loops and will be the sum of I2 and I3. What the total magnitude and sign of the current through R4 ends up being will depend on the values of the rest of the components in the rest of the circuit (which are not explicitly given).

vir
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Shouldn't it be I3 = -I plus whatever comes down the node?

The ideal current source defines the current that flows through itself. That's what a current source is and, it would be a terrible current source if the current it passed was alterable by other currents. I3 is defined by the current source; it's as simple as that.

After all, you wouldn't expect an ideal voltage source to alter its terminal voltage if it passes currents to other circuits would you?

Andy aka
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  • That explains it. How long is the current ideal, is the whole node equal to -I or is it only until the wire gets split (the point between R3, R4 and I)? – Coder May 17 '23 at 12:17
  • It remains at $-I$ only in that one wire because the current in R4 is added to by $I2$ @Coder – Andy aka May 17 '23 at 14:23
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The current I3 is equal to -I because that's how it is drawn in the diagram.

This is how the "mesh current method" (aka "loop current method") operates. You draw loops and voltages on your schematic; you end up working out the current in each loop and the voltage across each component. You work out the currents in a components by adding up the currents in all the loops that go through that component.

If you were thinking that it's wrong because the current through R4 would be I and it isn't, then don't worry about that. R4 is part of loops I2 and I3, so R4's current is not -I, it's actually I2-I. The current coming through R4 from the left is not counted in -I; it's counted in I2 instead.

user253751
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