# How would you find the equivalent resistance of these circuit diagrams given a wire with no resistor?

I understand how to combine resistors to find their equivalent resistance, but I am completely oblivious as to how to remove resistors that are shorted. For example, in the first image, there is a wire that has zero resistors on it. Would that mean that the 2 4 ohm resistors between the ends of the empty wire are completely ignored?

Same with the second image. What happens to the resistance when there is a wire connecting two junctions that are parallel with each other?

Would that mean that the 2 4 ohm resistors between the ends of the empty wire are completely ignored?

No. It means the two 4 ohm resistors are in parallel.

Same with the second image. What happens to the resistance when there is a wire connecting two junctions that are parallel with each other?

The wire makes the top and bottom pairs of resistors in parallel.

Hope this helps

These layout diagrams are designed to confuse you, especially the first.
For the second diagram, the 60 ohm resistor is in parallel with the 40 ohm resistor across ab. The 20 and 30 ohm resistors are in parallel and then in series with the 4 ohm resistor across ab (That is in parallel with the first 2).

The two 4 ohm resistors in the first diagram are not shorted, as there is another connection between the two. Try redrawing the layout.
Clue: Those two 4 ohm resistors are actually in parallel.

I have re-drawn the first circuit to make it obvious where the optical trickery is going on and it can be seen that no resistors are actually short circuited and there are no bridges, so it can all be solved simply using the rules for adding resistors in parallel (//) and series (+).

The total resistance of the whole circuit is:

((R1 // R2) + R3) // ((R4 // R5) + R6),

where $$(R_a // R_b) = \frac{R_a \times R_b}{R_a + R_b}$$,

is the usual formula for adding two resistors in parallel.

Rich has already given a full explanation of the second circuit in his answer.