在电路学中，星形连接和三角形连接的等效转换通常是超越中等教育的第一个电路知识。教科书通常使用断开一端建立电阻等式的推导方法。虽然这种方法是正确的，但初学者很难检测到它的正确性。本文通过纯数学推导了星形三角形连接的等效电阻转换公式。
满足星形联结
{ I a U c a R c a = U a b R a b I b U a b R a b = U b c R b c I c U b c R b c = U c a R c a \left\{ \begin{aligned} I_a \frac{U_{ca}}{R_{ca}}&=\frac{U_{ab}}{R_{ab}} \\ I_b \frac{U_{ab}}{R_{ab}}&=\frac{U_{bc}}{R_{bc}} \\ I_c \frac{U_{bc}}{R_{bc}}&=\frac{U_{ca}}{R_{ca}} \end{aligned} \right. ???????????????Ia RcaUcaIb+RabUabIc+RbcUbc=RabUab=RbcUbc=RcaUca
三角形联结满足
{ U a b = I a R a − I b R b U b c = I b R b − I c R c U c a = I c R c − I a R a \left\{ \begin{aligned} U_{ab}&=I_aR_a-I_bR_b \\ U_{bc}&=I_bR_b-I_cR_c \\ U_{ca}&=I_cR_c-I_aR_a \end{aligned} \right. ⎩⎪⎨⎪⎧UabUbcUca=IaRa−IbRb=IbRb−IcRc=IcRc−IaRa
若星形联结与三角形联结等效，则对任意的 I a , I b , I c , U a b , U b c , U c a I_a,I_b,I_c,U_{ab},U_{bc},U_{ca} Ia,Ib,Ic,Uab,Ubc,Uca，上两方程组都要成立。
首先消去 U a b , U b c , U c a U_{ab},U_{bc},U_{ca} Uab,Ubc,Uca。将三者代入到星形联结方程组中，化简得
{ I a + I b + I c = 0 I a ( R a + R c − R c a + R c R c a R b c ) + I b ( R c − R c a + ( R b + R c ) R c a R b c ) = 0 I a ( R a b + R b c ) R c + I a ( R c − R b c ) R c a + I b [ R b c ( R b + R c ) + R a b ( R b − R b c + R c ) + ( R b − R b c + R c ) R c a ] R a b = 0 \left\{ \begin{aligned} I_a+I_b+I_c&=0 \\ I_a\left(R_a+R_c-R_{ca}+\frac{R_c R_{ca}}{R_{bc}}\right)+I_b\left(R_c-R_{ca}+\frac{\left(R_b+R_c\right)R_{ca}}{R_{bc}}\right)&=0 \\ \frac{I_a\left(R_{ab}+R_{bc}\right)R_c+I_a\left(R_c-R_{bc}\right)R_{ca}+I_b\left[R_{bc}\left(R_b+R_c\right)+R_{ab}\left(R_b-R_{bc}+R_c\right)+\left(R_b-R_{bc}+R_c\right)R_{ca}\right]}{R_{ab}}&=0 \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧Ia+Ib+IcIa(Ra+Rc−Rca+RbcRcRca)+Ib(Rc−Rca+Rbc(Rb+Rc)Rca)RabIa(Rab+Rbc)Rc+Ia(Rc−Rbc)Rca+Ib[Rbc(Rb+Rc)+Rab(Rb−Rbc+Rc)+(Rb−Rbc+Rc)Rca]=0=0=0
上方程组对任意 I a , I b , I c I_a,I_b,I_c Ia,Ib,Ic 都要成立，只需
{ R a + R c − R c a + R c R c a R b c = 0 R c − R c a + ( R b + R c ) R c a R b c = 0 ( R a b + R b c ) R c + ( R c − R b c ) R c a = 0 R b c ( R b + R c ) + R a b ( R b − R b c + R c ) + ( R b − R b c + R c ) R c a = 0 \left\{ \begin{aligned} R_a+R_c-R_{ca}+\frac{R_c R_{ca}}{R_{bc}}&=0 \\ R_c-R_{ca}+\frac{\left(R_b+R_c\right)R_{ca}}{R_{bc}}&=0 \\ \left(R_{ab}+R_{bc}\right)R_c+\left(R_c-R_{bc}\right)R_{ca}&=0 \\ R_{bc}\left(R_b+R_c\right)+R_{ab}\left(R_b-R_{bc}+R_c\right)+\left(R_b-R_{bc}+R_c\right)R_{ca}&=0 \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎧Ra+Rc−Rca+RbcRcRcaRc−Rca+Rbc(Rb+Rc)Rca(Rab+Rbc)Rc+(Rc−Rbc)RcaRbc(Rb+Rc)+Rab(Rb−Rbc+Rc)+(Rb−Rbc+Rc)Rca=0=0=0=0
若星形联结电阻 R a b , R b c , R c d R_{ab},R_{bc},R_{cd} Rab,Rbc,Rcd 已知，则三角形联结电阻为
{ R a = R a b R c a R a b + R b c + R c a R b = R b c R a b R a b + R b c + R c a R c = R c a R b c R a b + R b c + R c a \left\{ \begin{aligned} R_a&=\frac{R_{ab}R_{ca}}{R_{ab}+R_{bc}+R_{ca}} \\ R_b&=\frac{R_{bc}R_{ab}}{R_{ab}+R_{bc}+R_{ca}} \\ R_c&=\frac{R_{ca}R_{bc}}{R_{ab}+R_{bc}+R_{ca}} \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧RaRbRc=Rab+Rbc+RcaRabRca=Rab+Rbc+RcaRbcRab=Rab+Rbc+RcaRcaRbc
若三角形联结电阻 R a , R b , R c R_a,R_b,R_c Ra,Rb,Rc 已知，则星形联结电阻为
{ R a b = R a R b + R b R c + R c R a R c R b c = R a R b + R b R c + R c R a R a R c a = R a R b + R b R c + R c R a R b \left\{ \begin{aligned} R_{ab}&=\frac{R_aR_b+R_bR_c+R_cR_a}{R_c} \\ R_{bc}&=\frac{R_aR_b+R_bR_c+R_cR_a}{R_a} \\ R_{ca}&=\frac{R_aR_b+R_bR_c+R_cR_a}{R_b} \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧RabRbcRca=RcRaRb+RbRc+RcRa=RaRaRb+RbRc+RcRa=RbRaRb+RbRc+RcRa