Example 13.6 Mutual Inductance

Solve for and in Fig. 13.27 (the same circuit as for Practice Prob. 13.1) using the T-equivalent circuit for the linear transformer.

T-equivalent need ;$L_{a},L_{b},L_{c}$

$$L_{a}=L_{1}-M$$
$$L_{b}=L_{2}-M$$
$$L_{c}=M$$

So, coz current loop M become -M

$L_{a}=L_{1}-(-M)=j9$
$L_{b}=L_{2}-(-M)=j6$
$L_{c}=-M=-j$

-M can be represented by capacitor *)

KVL,

$I_{1}(4+j9-j)+I_{2}(-j)=60\angle90^\circ$
$I_{1}(4+j8)+(-j)I_{2}=60j$

$I_{1}(-j)+I_{2}(10+j6-j)=0$
$I_{1}(-j)+I_{2}(10+j5)=0$