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,
$I_{1}(4+j9-j)+I_{2}(-j)=60\angle90^\circ$
$I_{1}(4+j8)+(-j)I_{2}=60j$
$I_{1}(-j)+I_{2}(10+j6-j)=0$
$I_{1}(-j)+I_{2}(10+j5)=0$
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