A Le bloc de transformation 2 fils - 4 fils

 

Figure A.1: Schéma du bloc T

 Avec les notations de la figure A.1, on obtient:

$$\left\{ \begin{array} {lcr} 2R\cdot i_2+a-d & = & 0\ R\cdot... ...\ i_2 & = & \frac{Z}{R}\cdot\frac{e-a}{Z+Z_l}\end{array}\right.$$

D'où:

$$\left\{ \begin{array} {lcl} d & = & a+2Z\cdot\frac{e-a}{Z+Z_l}\ l & = & -Z_l\cdot\frac{e-a}{Z+Z_l}\end{array}\right.$$

Si Z=Z_l:

$$\left\{ \begin{array} {lcl} d & = & e\ l & = & \frac{a-e}{2}\end{array}\right.$$