Y-delta/delta-Y transformation problem

[yugeci]

Homework Statement

The question asks to find the current I going into the 2k resistor path using Y-delta or delta-Y transformations.

Homework Equations

Resistance in parallel

1 / R = 1 / R1 + 1 / R2 ..

Converting Delta to Y,

R1 = RaRb / (Ra + Rb + Rc)

Current divider formula

Ix = (Rt / Rx) * Is

The Attempt at a Solution

I converted the delta network of 3 k Ohm resistors into a Y-network of 1k Ohms each using the above equation. (3*3) / (3 + 3 + 3) = 1. Then I got the total resistance of the network as 1.5 k Ohm. Then I used the current divider formula and found I as 2.14 mA.

The only problem is I'm not sure if I found the total resistance of the network correctly. It's confusing. Can someone tell me if I did it right?

 

[szynkasz]

Right, it is $1.5\,k\Omega$. It is easy to check - delta-Y network is symmetrical so there is no current through $6\,k\Omega$ resistor and you can get rid of it.

 

[gneill]

It looks like you managed to find the correct equivalent resistance of the subnetwork, but I'm not sure how you went about it given that you started by transforming the outer Δ to a Y. The reason I say this is because the "new" Y won't have the same central connection as the existing one, nor will it be at the same potential by symmetry (since the existing one has legs 3-6-3, while the new one has legs 1-1-1 all in kΩ). Just looks like a lot of work from that point!

If you'd done the opposite and transformed the existing Y to a Δ, each new resistor would parallel one in the existing Δ, leaving a single Δ with a "wing" that can collapse to a single resistance and it's easy going from there.

 

[gneill]

Right, it is $1.5\,k\Omega$. It is easy to check - delta-Y network is symmetrical so there is no current through $6\,k\Omega$ resistor and you can get rid of it.

Ah, so it is! Well spotted!