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• Notes

$\begin{array}{l}{\text { The load impedance connected to the secondary }} \\ {\text { winding of the ideal transformer in Fig consists of }} \\ {\text { a } 237.5 \mathrm{m} \Omega \text { resistor in series with } \mathrm{a} 125 \mu \text { H inductor. }} \\ {\text { If the sinsoidal vilage source }\left(v_{g}\right) \text { is gennationg }} \\ {\text { the voltage } 2500 \cos 400 \mathrm{t} \mathrm{V} \text { . find the steady-state }} \\ {\text { expressions for: (a) } i_{1} :(\mathrm{b}) v_{1 :}(\mathrm{c}) i_{2} : \text { and }(\mathrm{d}) v_{2}}\end{array}$

$R_{1}=0.25 \quad L_{1}=5 \mathrm{m} H \Rightarrow Z_{1}=R_{1}+J \omega l_1=0.25+J[400*5*10^{-3}]=(0.25+J2) \Omega$

$R_{2}=237.5 * 10^{-3} \quad L_{2}=125 * 10^{-6} \Rightarrow Z_2=R_2+J \omega l_2=(0.2375+J0.05) \Omega$

$\overline{V}_{1}=\overline{V}_{2} a=10 \overline{V}_{2} \longrightarrow (1)$

KVL prim:

$-2500 \ \angle0+(0.25+2J) \overline{I}_1+\overline{V}_1=0 \longrightarrow (2)$

KVL sec:

$-\overline{V}_{2}+i_{2}[0.2375+J0.05] = 0 \longrightarrow (3)$

$i_{2}=a i_{1}=10 i_{1} \longrightarrow (4)$

$i_1=100 \ \angle-16.26 \ A$

$i_{2}=1000 \ \angle-16.26$

$v_{1}=2427.06 \ \angle -4.37^{\circ}$

$v_{2}=242.7 \ \angle-4.37^{\circ}$