Circuit Analysis Help-Delta to Wye

[IanPSwift]

I picked up an essentials of electrical and computer engineering book from the library the other day. It has been fascinating thus far, but I did come to one problem. The book offered this drill exercise, and I couldn't figure out how they solved it. All they want you to do is solve for total ohms. Any help?

Picture of circuit is attached.

 

[Resqueline]

That circuit does not seem to have much to do with delta & wye.
Anyway, the first "easy" thing is that the bridge consists of 4 equal resistors (R4-R7), so one can disregard the central/middle 12 ohm resistor. It never sees any "action".
What remains of the bridge is essentially two 24 ohm resistors in parallel (R4+R5//R6+R7), equaling one 12 ohm resistor.
This is then in series with the upper 12 ohm resistor (R3), making 24 ohms.
The next step is the parallel connection of these 24 ohms and the 12 ohm resistor (R2). R = 1 / (1/12 + 1/24)) = 1 / (3/24) = 1 / (1/8) = 8 ohms.
These 8 ohms are then connected in series with the first 12 ohm resistor (R1), resulting in 20 ohms.

 

[shrtrnd]

Resqueline answered the drawing question.
Look for information on 3-Phase 480VAC wiring (Electrical, not electronic).
For more information on the delta and wye configurations for power applications.
From what I remember the delta and wye are for increasing/decreasing the voltage to current ratio
depending on specific applications. (Google it for specifics)