Transformer, "EX6-V1/2".

Determine the currents and voltages of a circuit with a transformer of transformation ratio ' n '.
Method:
1) The time-domain voltage source Vs(t)=169.706COS(t) is transformed into a frequency-domain voltage phasor without the variable t : Vs=169.706.phase(0°)
2) C=166.67[mF] is transformed into a complex reactance and optionally put into modulus and phase form for frequency analysis, i.e. Xc=-6j. Knowing that j translates into phase(90°) and -j into phase(-90°), Xc can be put into modulus and phase form, i.e. Xc = -6.j = 6(-j) = 6.phase(-90°).
3) Apply the marker point rule:
if i1 and i2 both enter or leave marking points, i1/i2 = -n ; if i1 enters via marking point and i2 leaves marking point, i1/i2 = +n ; and if i1 leaves marking point and i2 enters via marking point, i1/i2 = +n .
Similarly, if V1 and V2 are both oriented towards the positive or negative poles of the coils, V2/V1 = +n, otherwise V2/V1 = -n.
4) V1 and V2 are both oriented towards the positive poles of the coils, so : V2/V1 = +n, V2/V1=1/2, i.e. V2 = 0.5V1 {1}
5) i1 enters through the marking point and exits through the foot of the first coil; i2 exits through the foot of the second coil. This also means that i2 enters through the marking point, so i1 and i2 both enter through the marking points, hence: i1/i2 = -n , i1/i2=-0.5 , so i2 = -2.i1 {2}
6) Mesh law gives Vs-Vr1-Vc-V1=0, i.e. (4 - 6j).i1 + V1=169.706.phase(0°) {3}
7) Similarly, the law of meshes gives V2+Vr2=0, i.e. V2 + R2.i2 =0.
Combining this equation with those {1} and {2} gives 0.5V1+R2(-2.i1)=0.
Replacing R2=20 gives -40.i1+0.5V1=0 {4}
8) Equations {3} and {4} form a system of equations whose unknowns are i1 and V1. Its resolution gives i1=2.015.phase(4.098°) and V1=161.214.phase(4.086°)
9) Equation {2}: i2 = -2.i1, i2=4.03.phase(-175.902°)
10) and equation {1}: V2=0.5V1, i.e. V2=80.607.phase(4.086°).
11) Check the ratio ' n ' using the values found for i1, i2 and V1, V2.