Przykład 1

Metodą Thevenina obliczyć prąd \(I_x\) $$i_{1}(t)=5 \sin \left({\omega t}+45^{\circ}\right) \mathrm{A}, i_{2}(t)=10 \sin \left(\omega t\right) \mathrm{A}, \mathrm{R}=10 \Omega, \mathrm{X}_{\mathrm{c}}=5 \Omega$$

single-task-hero-img

Rozwiązanie

$$ I_{4} \cdot R-I_{3} \cdot R-I_{5} \cdot R=0 $$ $$ I_{4}-I_{3}-I_{5}=0 $$ $$A \quad I_{1}-I_{3}-I_{4}=0$$ $$I_{4}=I_{1}-I_{3}$$ $$B \quad I_{2}+I_{3}-I_{5}=0$$ $$I_{5}=I_{2}+I_{3}$$ $$I_{1}-I_{3}-I_{3}-\left(I_{2}+I_{3}\right)=0$$ $$-\left(3 \cdot I_{3}\right)+\left(I_{1}-I_{2}\right)=0$$ $$I_{3}=\frac{-I_{2}+I_{1}}{3}$$ \begin{aligned} &I_{1}:=\frac{5}{\sqrt{2}} \cdot e^{1 \mathrm{j} \cdot 45 \mathrm{deg}}=2.5+2.5 \mathrm{i} \\ &I_{2}:=\frac{10}{\sqrt{2}} \cdot e^{1 \mathrm{j} \cdot 0 \mathrm{deg}}=7.071 \\ &R:=10 \\ &Z_{C}:=-1 \mathrm{j} \cdot 5=-5 \mathrm{i} \\ &I_{3}:=\frac{-I_{2}+I_{1}}{3}=-1.524+0.833 \mathrm{i} \\ &U_{T}:=I_{3} \cdot R=-15.237+8.333 \mathrm{i} \end{aligned} \begin{aligned} &I_{4}:=I_{1}-I_{3}=4.024+1.667 \mathrm{i} \\ &I_{5}:=I_{2}+I_{3}=5.547+0.833 \mathrm{i} \\ &I_{4} \cdot R-U_{T}-I_{5} \cdot R=0 \quad+ \\ &U_{T}:=I_{4} \cdot R-I_{5} \cdot R=-15.237+8.333 \mathrm{i} \end{aligned} \begin{aligned} &Z_{A B}:=\frac{R \cdot(R+R)}{R+(R+R)}=6.667 \\ &I_{2} \cdot R-I_{1} \cdot R+I_{2} \cdot R=0 \\ &I=I_{1}+I_{2} \\ &Z_{A B}=\frac{U}{I} \\ &U=I_{1} \cdot R \\ &I_{2} \cdot R+I_{2} \cdot R=I_{1} \cdot R \\ &2 I_{2}=I_{1} \\ &I=2 I_{2}+I_{2}=3 I_{2} \\ &Z_{A B}=\frac{I_{1} \cdot R}{3 I_{2}} \\ &Z_{A B}:=\frac{2}{3} R=6.667 \end{aligned} \begin{aligned} &I_{x}:=\frac{U_{T}}{Z_{A B}+Z_{C}}=-2.063-0.297 \mathrm{i} \\ &I_{x}=2.084 \angle-2.999 \end{aligned}