Circuit Theory/Y Δ

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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.

Wye-delta-2.svg


The equations are:

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}
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.