a) Figure 9.39 shows the frequency-domain equivalent circuit. Note that the internal voltage of the source serves as the reference phasor, and that \pmb{V}_{1} and \pmb{V}_{2}, represent the terminal voltages of the transformer. In constructing the circuit in Fig. 9.39, we made the following calculations:
j\omega L_{1}= j\left(400\right)\left(9\right) = j 3600 \Omega,
j\omega L_{2}= j\left(400\right)\left(4\right) = j 1600 \Omega,
M = 0.5\sqrt{(9)(4)} = 3 H ,
j\omega M= j\left(400\right)\left(3\right) = j 1200 \Omega,
\frac{1}{j\omega C}=\frac{10^{6}}{j 400} =-j 2500\Omega.
b) The self-impedance of the primary circuit is
Z_{11} = 500 + j100 + 200 + j 3600 = 700 + j 3700 \Omega.
c) The self-impedance of the secondary circuit is
Z_{22} = 100 + j1600 + 800 – j 2500 = 900 – j 900 \Omega.
d) The impedance reflected into the primary winding is
Z_{r} =\left( \frac{1200}{|900 – j 900|}\right)^{2}\left(900 + j 900\right)
=\frac{8}{9}\left(900 + j 900\right) = 800 + j800 \Omega.
e) The scaling factor by which Z^{*}_{22} is reflected is \frac{8}{9}
f) The impedance seen looking into the primary terminals of the transformer is the impedance of the primary winding plus the reflected impedance; thus
Z_{ab} = 200 + j 3600 + 800 + j800 = 1000 + j4400 \Omega.
g) The Thévenin voltage will equal the open circuit value of \pmb{V}_{cd}. The open circuit value of \pmb{V}_{cd} will equal j1200 times the open circuit value of \pmb{I}_{1}.The open circuit value of \pmb{I}_{1} is
\pmb{I}_{1}=\frac{300 \angle0^{\circ}}{700 + j 3700}.
= 79.67 \angle -79.29^{\circ} mA.
Therefore
\pmb {V}_{Th}=j1200(79.67 \angle -79.29^{\circ}) \times 10^{-3}.
= 95.60 \angle 10.71^{\circ} V.
The Thévenin impedance will be equal to the impedance of the secondary winding plus the impedance reflected from the primary when the voltage source is replaced by a short-circuit.Thus
Z_{Th}=100 + j1600 + \left(\frac{1200}{|700 + j 3700|}\right)^{2} \left(700 – j 3700\right)
= 171.09 + j1224.26 Ω.
The Thévenin equivalent is shown in Fig. 9.40.
