The bridge circuit example was solved with source injection, but could have been solved with a Y Δ transformation.

The 5,2 and 3 ohm resistors form a Δ that could be transformed into a Y.

The equations are:

- $R_{1}={\frac {R_{b}R_{c}}{R_{a}+R_{b}+R_{c}}},R_{2}={\frac {R_{a}R_{c}}{R_{a}+R_{b}+R_{c}}},R_{3}={\frac {R_{a}R_{b}}{R_{a}+R_{b}+R_{c}}}$
- $R_{1}={\frac {5*2}{3+5+2}}=1,R_{2}={\frac {3*2}{3+5+2}}=0.6,R_{3}={\frac {3*5}{3+5+2}}=1.5$

Now Thevenin's resistance can be solved by parallel and serial combinations:

- $R_{th}={\frac {1}{{\frac {1}{8}}+{\frac {1}{10.6}}}}+1.5=6.0591$

Which is the same value found through source injection.