Circuit Theory/Y Δ

Circuit after Δ Y transformation of top bridge .. can now use parallel/serial combinations rather than source injection to find Thevenin's resistance

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

Example23g.png

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.