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So, in what looks like an attempt to explain mesh currents, my book has the following diagram:

enter image description here

Then it proceeds to convince us of the veracity of several equations. The first of them makes sense: \$i_d=i_1-i_c\$. The signs are preserved by the pattern in the use of current directions. Okay, I get that. Then it says something a bit harder to grasp: \$i_g=i_3-i_2\$ and \$i_f=i_1-i_3\$.

So my question becomes: What happens if we flip f and g? It seems that \$i_d=i_e\$ regardless of the chirality of the components, all of the numerically indexed currents are going to be the same. How, then, can these relationships be demonstrated?

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  • \$\begingroup\$ i_d and i_e will always be equal, but i_f and i_g will be swapped (i_f will be on the right, etc). Note that i_d = i_e = i_f + i_g. \$\endgroup\$
    – Shamtam
    Commented Nov 3, 2014 at 21:32
  • \$\begingroup\$ Summarising efox29 - If you swap the two components and do all calculations from the start all answer will be effectively as before. But the signs of some currents may change - but this is "artificial" and is entirely because of conventions related to the layout which are adopted for convenience when solving. \$\endgroup\$
    – Russell McMahon
    Commented Nov 3, 2014 at 23:02

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Because you (the book) have defined the direction what the branch currents then the mesh currents add/subtract so that you maintain the direction of the branch current. (That sounds a bit confusing).

Look at branch current ig for example. It passes through element g. It's direction is fixed - it goes down through element g.

Now look at mesh current i3. Does it also pass through element g ? It does. It goes down through element g. But what about mesh current i2 ? It also passes through element g however it goes up through element g. The direction of the branch current ig is already defined - down. So if we were to add up i3 and i2 together, their net sum must be in the direction of the branch current.

So ig = i3-i2.

If you had chosen the current ig to going up instead initially, then ig = i2 - i3.

It doesnt matter how you initally set up your directions, so long as you lock it in and dont change directions, the math will work itself out.

Hopefully that helped and give you enough to answer your own question.

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