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Y-Δ transformation 星三角变换

. 1 min read

Y-Δ transformation

Given (a) $R_1,R_2,R_3$ find$R_{12},R_{23},R_{31}$ is equivalent to (a)

Which means that If each circuit is placed in a black box, we can't tell by external measurements whether the box contains Y or Δ connection.

Method 1 Hard way

given $V_x' = V_x$, current should all to be the same.

So that $i_x' = i_x$.
$$
i_1=(V_1-V_0)/R_1 \\
i_2=(V_2-V_0)/R_2 \\
i_3=(V_3-V_0)/R_3
$$

$$
i_1'=(V_1'-V_2')/R_12+(V_1'-V_3')/R_31
\\ i_2'=(V_2'-V_1')/R_12+(V_2'-V_3')/R_23
\\ i_3'=(V_3'-V_1')/R_31+(V_3'-V_2')/R_23
$$

$$
i_x=i_x'
$$

Method 2 Easy way

等效电阻

$$
R_{eq_{12}} = R_1+R_2 = R12∥(R_{31}+R_{23})\\R_{eq_{23}} = R_2+R_3 = R23 ∥(R_{31}+R_{12})\\
R_{eq_{31}} = R_3+R_1 = R31 ∥ (R_{12}+R_{23})\\
$$

Solve the function we will get
$$
R_1=\frac{R_{12}R_{31}}{R_{12}+R_{23}+R_{31}}\\
R_2=\frac{R_{23}R_{12}}{R_{12}+R_{23}+R_{31}}\\
R_1=\frac{R_{31}R_{23}}{R_{12}+R_{23}+R_{31}}\\
$$

$$
R_{12}=\frac{R_{12}+R_{23}+R_{31}}{R_3 }\\
R_{23}=\frac{R_{12}+R_{23}+R_{31}}{R_1 }\\
R_{31}=\frac{R_{12}+R_{23}+R_{31}}{R_2 }\\
$$

A little conclusion



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