Simplifying a resistor network using pi-T (Y-delta) conversion

[JJBladester]

Homework Statement

Reduce the resistor network to a single resistor. Go step-by-step and indicate the series or parallel combinations being reduced.

Homework Equations

For series resistors: R_T=R_1+R_2+R_3+\cdot \cdot \cdot +R_N
For parallel resistors: R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdot \cdot \cdot +\frac{1}{R_N}}

∏-T Conversion:
R_1=\frac{R_BR_C}{R_A+R_B+R_C}
R_2=\frac{R_AR_C}{R_A+R_B+R_C}
R_3=\frac{R_AR_B}{R_A+R_B+R_C}

The Attempt at a Solution

The first thing I noticed is that R_A, R_B, and R_C are not in series and they're not in parallel. This led me to the ∏-T (Y-Δ) conversion. After the conversion, I was able to make further simplifications in steps (2) and (3).

In step (4), I get stuck because I don't know how to simplify the circuit given the way R₅ is hooked up. Any pointers?

 

[JJBladester]

My thinking, assuming I drew everything correctly, is that R₅ gets bypassed because a practically resistance-free path exists (the wire on the left-leg of the triangle above R₅).

Therefore the equivalent resistance would be:

R_eq=R₃+R₂||(R₁+R₄) = 10.57Ω + 58.92Ω ≈ 69.5Ω

Does this make sense?

 

[lewando]

Yes. You can also think of the wire as a 0-ohm resistor and use the parallel resistor formula.

 

[azizlwl]

I guess cable across Rc and R4
So mark all the points as zero voltage.
Now we have a parallel circuit of
1. Ra-150Ω
2. Series of Rb and parallel resistors of Rc and R4-129Ω

Equivalent resistance=69.35Ω

 

[JJBladester]

...you can also think of the wire as a 0-ohm resistor and use the parallel resistor formula.

This makes sense.

So, R_T=\frac{1}{\frac{1}{\sim 0 \Omega }+\frac{1}{220\Omega }}\approx 0 \Omega

Thanks lewando.