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

1. Jul 8, 2012

1. The problem statement, all variables and given/known data

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

2. Relevant 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:
$$R_1=\frac{R_BR_C}{R_A+R_B+R_C}$$
$$R_2=\frac{R_AR_C}{R_A+R_B+R_C}$$
$$R_3=\frac{R_AR_B}{R_A+R_B+R_C}$$

3. The attempt at a solution

The first thing I noticed is that RA, RB, and RC 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?

2. Jul 8, 2012

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:

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

Does this make sense?

3. Jul 9, 2012

lewando

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

4. Jul 9, 2012

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

Last edited: Jul 9, 2012
5. Jul 9, 2012

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