Rozwiązanie
\begin{aligned}
&I_{1}=\frac{V_{A}-V_{B}}{Z_{L}}\\
&I_{2}=\frac{V_{A}}{R_{2}+Z_{C}}\\
&I_{4}=\frac{V_{B}}{R_{4}}\\
&\begin{array}{ll}
I-I_{1}-I_{2}=0 & I-\frac{V_{A}-V_{B}}{Z_{L}}-\frac{V_{A}}{R_{2}+Z_{C}}=0 \\
I_{1}-I_{3}-I_{4}=0 & \frac{V_{A}-V_{B}}{Z_{L}}-\frac{V_{B}+U_{S}}{R_{3}}-\frac{V_{B}}{R_{4}}=0
\end{array}\\
&V_{A} \cdot\left(\frac{1}{Z_{C}+R_{2}}+\frac{1}{Z_{L}}\right)+V_{B} \cdot\left(-\frac{1}{Z_{L}}\right)=I\\
&V_{B} \cdot\left(\frac{1}{R_{3}}+\frac{1}{Z_{L}}+\frac{1}{R_{4}}\right)+V_{A} \cdot\left(-\frac{1}{Z_{L}}\right)=\frac{-U_{s}}{R_{3}}\\
&\left[\begin{array}{cc}
\frac{1}{Z_{C}+R_{2}}+\frac{1}{Z_{L}} & -\frac{1}{Z_{L}} \\
-\frac{1}{Z_{L}} & \frac{1}{R_{3}}+\frac{1}{Z_{L}}+\frac{1}{R_{4}}
\end{array}\right] \cdot\left[\begin{array}{c}
V_{A} \\
V_{B}
\end{array}\right]=\left[\begin{array}{c}
I \\
-U_{s} \\
R_{3}
\end{array}\right]\\
&\left[\begin{array}{cc}
\frac{1}{Z_{C}+R_{2}}+\frac{1}{Z_{L}} & -\frac{1}{Z_{L}} \\
-\frac{1}{Z_{L}} & \frac{1}{R_{3}}+\frac{1}{Z_{L}}+\frac{1}{R_{4}}
\end{array}\right] \cdot\left[\begin{array}{c}
V_{A} \\
V_{B}
\end{array}\right]=\left[\begin{array}{c}
I \\
\frac{-a \cdot U_{L}}{R_{3}}
\end{array}\right]=\left[\begin{array}{c}
I \\
\frac{-a \cdot\left(V_{A}-V_{B}\right)}{R_{3}}
\end{array}\right]\\
&I_{1}=\frac{V_{A}-V_{B}}{Z_{L}} \quad U_{L}=I_{1} \cdot Z_{L}=V_{A}-V_{B}
\end{aligned}
\begin{aligned}
&\left[\begin{array}{cc}
\frac{1}{Z_{C}+R_{2}}+\frac{1}{Z_{L}} & -\frac{1}{Z_{L}} \\
-\frac{1}{Z_{L}}+\frac{a}{R_{3}} & \frac{1}{R_{3}}+\frac{1}{Z_{L}}+\frac{1}{R_{4}}-\frac{a}{R_{3}}
\end{array}\right] \cdot\left[\begin{array}{l}
V_{A} \\
V_{B}
\end{array}\right]=\left[\begin{array}{l}
I \\
0
\end{array}\right]\\
&\omega=1\\
&I=5 \cdot e^{1 j \cdot 30 d e g}=4.33+2.5 \mathrm{i}\\
&U_{s}=a \cdot U_{L}\\
&Z_{L}:=1 \mathrm{j} \cdot \omega \cdot L=8 \mathrm{i} \quad Z_{C}:=-1 \mathrm{j} \cdot \frac{1}{\omega \cdot C}=-5 \mathrm{i}\\
&\left[\begin{array}{c}
V_{A} \\
V_{B}
\end{array}\right]:=\left[\begin{array}{cc}
\frac{1}{Z_{C}+R_{2}}+\frac{1}{Z_{L}} & -\frac{1}{Z_{L}} \\
-\frac{1}{Z_{L}}+\frac{a}{R_{3}} & \frac{1}{R_{3}}+\frac{1}{Z_{L}}+\frac{1}{R_{4}}-\frac{a}{R_{3}}
\end{array}\right]^{-1} \cdot\left[\begin{array}{l}
I \\
0
\end{array}\right]=\left[\begin{array}{c}
32.771+21.671 \mathrm{i} \\
3.886-8.535 \mathrm{i}
\end{array}\right]\\
&I_{1}=\frac{V_{A}-V_{B}}{Z_{L}}=3.776-3.611 \mathrm{i}\\
&I_{2}=\frac{V_{A}}{R_{2}+Z_{C}}=0.554+6.111 \mathrm{i}\\
&I_{3}=\frac{V_{B}+a \cdot U_{L}}{R_{3}}=1.833+0.657 \mathrm{i}\\
&I_{4}=\frac{V_{B}}{R_{4}}=1.943-4.267 \mathrm{i}
\end{aligned}
\begin{aligned}
&U_{S}:=a \cdot U_{L}=14.443+15.103 \mathrm{i} \\
&I-\frac{V_{A}-V_{B}}{Z_{L}}-\frac{V_{A}}{R_{2}+Z_{C}}=0 \\
&\frac{V_{A}-V_{B}}{Z_{L}}-\frac{V_{B}+U_{S}}{R_{3}}-\frac{V_{B}}{R_{4}}=0 \\
&I_{2} \cdot R_{2}+I_{2} \cdot Z_{C}-I_{1} \cdot Z_{L}+a \cdot I_{1} \cdot Z_{L}-I_{3} \cdot R_{3}=0
\end{aligned}
Jeżeli masz jakieś pytania, uwagi lub wydaje Ci się, że znalazłeś błąd w tym rozwiązaniu, napisz proszę do nas wiadomość na kontakt@edupanda.pl lub skontaktuj się z nami przez nasz profil na FB: