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

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:



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:


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.


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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