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• Notes

$\begin{array}{l}{\text { A linear transformer couples a load consisting }} \\ {\text { of a } 360 \Omega \text { resistor in series with a } 0.25 \mathrm{H}} \\ {\text { inductor to a sinusoidal voltage source, as }} \\ {\text { shown. The voltage source has an internal }}\end{array}$

$\begin{array}{l}{\text { impedance of } 184+j 0 \Omega \text { and a maximum }} \\ {\text { of } 245.20 \mathrm{V} \text { , and it is operating at } 800 \mathrm{rad} / \mathrm{s} \text { . }}\end{array}$

$\begin{array}{l}{\text { The transformer parameters are } R_{1}=100 \Omega} \\ {L_{1}=0.5 \mathrm{H}, R_{2}=40 \Omega, L_{2}=0.125 \mathrm{H}, \text { and }} \\ {k=0.4 . \text { Calculate (a) the reflected impedance; }} \\ {\text { (b) the primary current; and (c) the secondary current. }}\end{array}$

الخطوات

(1) نحول $Z \leftarrow c, l$

نحول المصادر $\text { phasor } \leftarrow$

(2) قيمة $M=K \sqrt{L_{1} L_{2}}$

(3) قيمة V من $\text { Mutual } \text { Inductance }$

(4) تطبق قواعد $\text { linear tr }$

$R=360 , L=0.25 \quad Z=R+J \omega L=360+J0.25*800=360+J200$

$Z_{1}=R_{1}+J \omega L_{1}=100+J800*0.5=100+J400$

$Z_{2}=R_{2}+J \omega L_{2}=40+J*800*0.125=40+J100$

$V_{s}=245.2 \ \angle 0$

$Z_{r e f}=\frac{(wm)^2}{\left|Z_{22}\right|^{2}}(Z_{22})^* \rightarrow \Omega$

$Z_{22}=40+100J+360+J200=(400+J300) \Omega$

$M=K \sqrt{L_1 L_{2}}=0.4 \sqrt{0.5 *0.125}=0.1 H$

$Z_{ ref }=\frac{(0.1* 800)^{2}}{(400)^{2}+(300)^{2}}(400 -J300)$

$Z_ { ref }=(10.24-J7.68) \Omega$

Prim

$-\overline{V}_{1}+\left(Z_{3}+R_{1}+J \omega L_{1}\right) I_{1}-J \omega M I_{2}=0$

$-245.2 \ \angle 0+[184+100+400J]+I_1-J(800)(0.1)I_2=0 \longrightarrow (1)$

Sec

$- J \omega M I_1+(J \omega L_{2}+R_{2}+Z_{2})I_2=0$

$-J(+800) *(0.1) *I_1+(400+300J)I_2=0 \longrightarrow (2)$

$I_{1}=\frac{400+300J}{-J(80)} I_{2}=[-3.75+5]I_2 \longrightarrow (2')$

$I_{2}=0.08 \ \angle 177^{\circ}$    Sec Current

$I_{2}=0.5 \angle-56.118^{\circ}$    Prim Current