Practice Problem 13.2 Magnetic Couple

 Determine the phasor currents $I_{1}, I_{2}$

loop 2:

$(j6-j4)I_{2}+(j3-j6)I_{1}=0$
$j2I_{2}-j3I_{1}=0$
$I_{2}=\frac{3}{2} I_{1}$

loop 1:
$100\angle60^\circ=(5+j2-2\times j3+j6)I_{1}-(j6-j3)I_{2}$
$100\angle60^\circ=(5+j2)I_{1}-j3I_{2}$

Substitusi:
$100\angle60^\circ=(5+j2)I_{1}-j4.5I_{1}$
$100\angle60^\circ=(5-j2,5)I_{1}$
$100\angle60^\circ=(5.59\angle-26.565^\circ)I_{1}$
$I_{1}=17.889\angle86.565^\circ$

$I_{2}=\frac{3}{2} I_{1}$
$I_{2}=26.833\angle86.565^\circ$