I've been having trouble trying to figure out how to properly get the equivalent resistance of this circuit (and ones similar) it is always an issue for me when there is a jumper wire that doesn't seem like it's only on one wire but multiple.
I've tried considering that the 4 ohm resistor is useless and working it out as if the 8 and 12 ohm are in series and parallel with the 5 ohm resistor.
The way the circuit is drawn suggests that all current flows are uniformly in the same direction across the drawing (right to left), which is incorrect; it was likely drawn this way to introduce a little misdirection.
The 4 Ohm and 12 Ohm are effectively in parallel, equivalent to 3 Ohms.
This equivalent 3 Ohm resistance is in series with the 8 Ohm, for an equivalent of 11 Ohms.
The equivalent 11 Ohm resistance is in parallel with the 5 Ohm, which calculates out to 3.4375 Ohms
The usual solution to this kind of problem uses a $Y-\Delta$ transformation
First, you redraw your circuit like so:
Next, you use the Y-\Delta transformation on the 5, 8, 4 Ohm "Delta" on the left; this gives you a new circuit that looks like this:
Now you just have to find the values of A, B and C; finally, you recognize that the circuit is now comprised of
$\mathbf{A}$, in series with a parallel combination of {$\mathbf{(B+12)}$ in parallel with $\mathbf{(C+0)}$}.
And that is something you should be able to do, using the equations in the link given. As an example, $R_A$ is given by
$$R_A = \frac{8*5}{8+5+4} = \frac{40}{17}$$
I will leave the rest as an exercise for you.
Note - since one of the resistors is zero, there is an easier way to do this particular problem, as shown in Anthony X's answer. But the above is more general, and works when all the resistors are non-zero.
