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I don't understand how the current flows in the circuit below. How do I deal with the wire between nodes c and a, a and b, g and d. Why is the 5 Ω resistor shorted out? I am having a little trouble understanding what results from short circuiting.

enter image description here

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    \$\begingroup\$ "Why"? Because whoever gave you that homework thought it would be an interesting case to learn from. \$\endgroup\$
    – PlasmaHH
    Oct 26 '18 at 8:28
  • \$\begingroup\$ Hi and welcome. Yes, as @PlasmaHH wrote - the edge cases are great places to learn from. \$\endgroup\$
    – mike65535
    Oct 26 '18 at 12:08
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I believe that you're mistaken about "nodes a, b and c" and how you deal with the wires between them. They are actually all the same node because they are connected by a wire, and so they also have the same voltage.

This is how you should name your nodes:

Nodes

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Your convention is wrong. If you don't see any resistors across a line then you don't need to change the node name.

Here's the corrected circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, both ends of the 5-Ohm resistor has "a", so it's shorted out.

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There is no empty wires, just wires.

The convention in circuit drawing is that nodes connected by wires are at the same voltage, which means that the voltage at node a (Va) is the same as the voltage at node b (Vb) and c (Vc).

In turn, we know that the voltage across your 5 ohm resistor (Vc-Vb) = 0V, which means that it is in fact shorted out.

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