Question 9.13: The parameters of a certain linear transformer are R1=200Ω,R...

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

$=\frac{8}{9}\left(900 + j 900\right) = 800 + j800 \Omega$.

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

= 171.09 + j1224.26 Ω.

The Thévenin equivalent is shown in Fig. 9.40.