Simplifying a circuit with an empty wire (that may or may not be in parallel)

Question

Determine the value of the resistance R in the circuit shown below, given that Req = 9 ohms.

I called the top-left, top-right and middle-right nodes a, c and b respectively, and did a delta-Y transformation on the 24-ohm, 4-ohm and R resistor. I got that R3 and the 8-ohm resistor are in series, but now I'm completely stuck on trying to figure out how the empty wire in the middle of the circuit gets simplified.

If my understanding of parallel is correct then I think the wire is in parallel with the 30-ohm resistor because both ends connect to each other, so the equivalent resistance would be 0 and I could cancel out the 30-ohm resistor completely. But if that's true, then by the same reasoning wouldn't the wire also ultimately short everything above it including R? And if it's not in parallel with the 30-ohm resistor, how can I simplify anything?

Answer 1

your reasoning is correct that the 39 ohm resistor parallell with the wire can be replaced with just the wire. but there are no other parts that only parallel the wire. To elimitate a resistor entirely both ends must be connected to the same node.

from here-on in the solution can be found by making series and parallel combinations.

I re-drew the circuit without the 30 ohm

^{simulate this circuit – Schematic created using CircuitLab}

every path from A passes through a resistor...

it looks like the next simplification is to combine the 8 and the 24, and then combine that with the 4 and then that with the 12... etc.

Answer 2

Consider this T-shirt showing the crucial part of your circuit:

You can cancel the resistor only if there is some way of going from one end of the resistor to the other end through some wires, without going through any other resistor.

This is obviously true for the 30 Ω resistor. However, none of the resistors above (or any combination of those resistors) has a wire parallel in this way.

Once you have eliminated the 30 Ω resistor, you can find sufficient series/parallel combinations to simplify, as shown by Jasen's answer.