Y-Δ变换

变换的基本思路是用${\displaystyle R'}$和${\displaystyle R''}$计算Y形电路端点的阻抗${\displaystyle R_{y}}$，其中${\displaystyle R'}$和${\displaystyle R''}$是Δ形电路中对应节点到邻接节点间的阻抗：

${\displaystyle R_{y}={\frac {R'R''}{\sum R_{\Delta }}}}$

其中${\displaystyle R_{\Delta }}$是Δ形电路的阻抗之和。具体公式如下：

${\displaystyle R_{1}={\frac {R_{a}R_{b}}{R_{a}+R_{b}+R_{c}}}}$

${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{a}+R_{b}+R_{c}}}}$

${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{a}+R_{b}+R_{c}}}}$

口訣为 Y形阻抗 = Δ形同側相邻阻抗乘积 / Δ形阻抗之和

变换的基本思路是计算Δ形电路的${\displaystyle R_{\Delta }}$：

${\displaystyle R_{\Delta }={\frac {R_{P}}{R_{\mathrm {opposite} }}}}$

其中${\displaystyle R_{P}=R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}$是Y形电路中的阻抗两两相乘之和，${\displaystyle R_{\mathrm {opposite} }}$是${\displaystyle R_{\Delta }}$所在支路对侧的端点在Y形电路中对应端点的阻抗。每一支路的阻抗计算公式为：

${\displaystyle R_{a}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{2}}}}$

${\displaystyle R_{b}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{3}}}}$

${\displaystyle R_{c}={\frac {R_{1}R_{2}+R_{2}R_{3}+R_{3}R_{1}}{R_{1}}}}$

口訣为 Δ形阻抗 = Y形阻抗两两相乘之和 / Y形对側端点阻抗

要将Δ形负载{${\displaystyle R_{a},R_{b},R_{c}}$}变换成Y形负载{${\displaystyle R_{1},R_{2},R_{3}}$}，需要比较二者对应节点的阻抗。要计算两种接法的阻抗，需要将电路中的一个节点断开。

Δ形电路中N₃断开後，N₁与N₂间的阻抗为

${\displaystyle {\begin{aligned}R_{\Delta }(N_{1},N_{2})&=R_{b}\parallel (R_{a}+R_{c})\\[8pt]&={\frac {1}{{\frac {1}{R_{b}}}+{\frac {1}{R_{a}+R_{c}}}}}\\[8pt]&={\frac {R_{b}(R_{a}+R_{c})}{R_{a}+R_{b}+R_{c}}}.\end{aligned}}}$

将{${\displaystyle R_{a},R_{b},R_{c}}$}之和用${\displaystyle R_{T}}$表示以简化方程：

${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$

得到

${\displaystyle R_{\Delta }(N_{1},N_{2})={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}}$

Y形电路中N₁与₂的对应阻抗为

${\displaystyle R_{Y}(N_{1},N_{2})=R_{1}+R_{2}}$

由以上两式得到：

${\displaystyle R_{1}+R_{2}={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}}$   (1)

同理，对於${\displaystyle R(N_{2},N_{3})}$与${\displaystyle R(N_{1},N_{3})}$，也分别有

${\displaystyle R_{2}+R_{3}={\frac {R_{c}(R_{a}+R_{b})}{R_{T}}}}$   (2)

${\displaystyle R_{1}+R_{3}={\frac {R_{a}(R_{b}+R_{c})}{R_{T}}}.}$   (3)

由此，{${\displaystyle R_{1},R_{2},R_{3}}$}的值可以由以上式子的线性组合（相加或相减）求出。

例如，将式(1)和式(3)相加，然後减去式(2)会得到

${\displaystyle R_{1}+R_{2}+R_{1}+R_{3}-R_{2}-R_{3}={\frac {R_{b}(R_{a}+R_{c})}{R_{T}}}+{\frac {R_{a}(R_{b}+R_{c})}{R_{T}}}-{\frac {R_{c}(R_{a}+R_{b})}{R_{T}}}}$

${\displaystyle 2R_{1}={\frac {2R_{b}R_{a}}{R_{T}}}}$

於是

${\displaystyle R_{1}={\frac {R_{b}R_{a}}{R_{T}}}.}$

其中 ${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$

求出所有的阻抗值如下：

${\displaystyle R_{1}={\frac {R_{b}R_{a}}{R_{T}}}}$ (4)

${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{T}}}}$ (5)

${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{T}}}}$ (6)

令

${\displaystyle R_{T}=R_{a}+R_{b}+R_{c}}$.

则Δ形电路到Y形电路的变换方程变为

${\displaystyle R_{1}={\frac {R_{a}R_{b}}{R_{T}}}}$   (1)

${\displaystyle R_{2}={\frac {R_{b}R_{c}}{R_{T}}}}$   (2)

${\displaystyle R_{3}={\frac {R_{a}R_{c}}{R_{T}}}.}$   (3)

将以上式子两两相乘得到

${\displaystyle R_{1}R_{2}={\frac {R_{a}R_{b}^{2}R_{c}}{R_{T}^{2}}}}$   (4)

${\displaystyle R_{1}R_{3}={\frac {R_{a}^{2}R_{b}R_{c}}{R_{T}^{2}}}}$   (5)

${\displaystyle R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}^{2}}{R_{T}^{2}}}}$   (6)

上式之和为

${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{a}R_{b}^{2}R_{c}+R_{a}^{2}R_{b}R_{c}+R_{a}R_{b}R_{c}^{2}}{R_{T}^{2}}}}$   (7)

将右侧式子中的公因式${\displaystyle R_{a}R_{b}R_{c}}$提出，约去分子中的${\displaystyle R_{T}}$和分母中的一个${\displaystyle R_{T}}$後得到

${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {(R_{a}R_{b}R_{c})(R_{a}+R_{b}+R_{c})}{R_{T}^{2}}}}$

${\displaystyle R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}}$ (8)

注意式(8)和式{(1),(2),(3)}的相似性，

将式(8)除以式(1)得到

${\displaystyle {\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}={\frac {R_{a}R_{b}R_{c}}{R_{T}}}{\frac {R_{T}}{R_{a}R_{b}}},}$

${\displaystyle {\frac {R_{1}R_{2}+R_{1}R_{3}+R_{2}R_{3}}{R_{1}}}=R_{c},}$

得到${\displaystyle R_{c}}$的表达式。同理，将式(8)除以${\displaystyle R_{2}}$或${\displaystyle R_{3}}$也能得到相应的表达式。