Stan przed przełączeniem

\begin{aligned} &\omega:=1 \\ &I:=4 \cdot e^{1 \mathrm{j} \cdot 90 \mathrm{deg}=4 \mathrm{i}} \\ &Z_{C}:=-1 \mathrm{j} \cdot \frac{1}{\omega \cdot C}=-5 \mathrm{i} \\ &Z_{L}:=1 \mathrm{j} \cdot \omega \cdot L=2 \mathrm{i} \\ &V_{A} \cdot\left(\frac{1}{Z_{C}}+\frac{1}{R_{2}+R_{0}+R_{3}}\right)=I \\ &V_{A}:=\frac{I}{\frac{1}{Z_{C}}+\frac{1}{R_{2}+R_{0}+R_{3}}}=17.041+7.101 \mathrm{i} \\ &V_{A}=18.462 \angle 22.62^{\circ} \\ &U_{C}=V_{A} \end{aligned} \begin{aligned} &u_{C}(t):=18.462 \cdot \sqrt{2} \cdot \sin (\omega \cdot t+22.62 \mathrm{deg}) \\ &u_{C .0}:=u_{C}(0)=10.042 \\ &i_{L .0}:=0 \end{aligned}

Stan ustalony po przełączeniu

\begin{aligned} &{\left[\begin{array}{c} \frac{1}{Z_{C}}+\frac{1}{Z_{L}}+\frac{1}{R_{2}+R_{0}}-\frac{1}{Z_{L}}-\frac{1}{R_{2}+R_{0}} \\ -\frac{1}{Z_{L}}-\frac{1}{R_{2}+R_{0}} & \frac{1}{Z_{L}}+\frac{1}{R_{2}+R_{0}}+\frac{1}{R_{3}} \end{array}\right] \cdot\left[\begin{array}{c} V_{A} \\ V_{B} \end{array}\right]=\left[\begin{array}{l} I \\ 0 \end{array}\right]} \\ &{\left[\begin{array}{l} V_{A} \\ V_{B} \end{array}\right]:=\left[\begin{array}{cc} \frac{1}{Z_{C}}+\frac{1}{Z_{L}}+\frac{1}{R_{2}+R_{0}}-\frac{1}{Z_{L}}-\frac{1}{R_{2}+R_{0}} \\ -\frac{1}{Z_{L}}-\frac{1}{R_{2}+R_{0}} \frac{1}{Z_{L}}+\frac{1}{R_{2}+R_{0}}+\frac{1}{R_{3}} \end{array}\right] \cdot\left[\begin{array}{l} I \\ 0 \end{array}\right]=\left[\begin{array}{c} -7.318+12.054 \mathrm{i} \\ 2.411+5.464 \mathrm{i} \end{array}\right]} \end{aligned} \begin{aligned} &I_{L}:=\frac{V_{A}-V_{B}}{Z_{L}}=3.295+4.864 \mathrm{i} \\ &I_{L}=5.876 \angle 55.886^{\circ} \\ &U_{C}:=V_{A}=-7.318+12.054 \mathrm{i} \\ &U_{C}=14.102 \angle 121.262^{\circ} \\ &u_{C}(t):=14.102 \cdot \sqrt{2} \cdot \sin (\omega \cdot t+121.262 \mathrm{deg}) \\ &u_{C . u}:=u_{C}(0)=17.048 \\ &i_{L}(t):=5.876 \cdot \sqrt{2} \cdot \sin (\omega \cdot t+55.886 \mathrm{deg}) \\ &i_{L . u}:=i_{L}(0)=6.88 \end{aligned}

Stan nieustalony po przełączeniu

\begin{aligned} &i_{L p}:=i_{L .0}-i_{L . u}=-6.88\\ &u_{C p}:=u_{C .0}-u_{C . u}=-7.005 \end{aligned} \left[\begin{array}{ll} \frac{1}{s \cdot L}+\frac{1}{R_{2}+R_{0}}+\frac{1}{\frac{1}{s \cdot C}} & \frac{\frac{-1}{1}}{s \cdot C} \\ \frac{\frac{-1}{1}}{s \cdot C} & \frac{1}{\frac{1}{s \cdot C}}+\frac{1}{R_{3}} \end{array}\right] \cdot\left[\begin{array}{l} V_{A} \\ V_{B} \end{array}\right]=\left[\begin{array}{l} \frac{-L \cdot i_{L p}}{s \cdot L}+\frac{\frac{u_{C p}}{s}}{\frac{1}{s \cdot C}} \\ -\frac{\frac{u_{C p}}{s}}{\frac{1}{s \cdot C}} \end{array}\right] \begin{aligned} &{\left[\begin{array}{l} V_{A} \\ V_{B} \end{array}\right]=\left[\begin{array}{l} \frac{-0.11505 \cdot s+31.533}{s^{2}+0.875 \cdot s+2.2917} \\ \frac{6.8904 \cdot s+3.2108}{s^{2}+0.875 \cdot s+2.2917} \end{array}\right]} \\ &u_{C}(s):=V_{A}-V_{B} \rightarrow \frac{-7.00545 \cdot s+28.3222}{s^{2}+0.875 \cdot s+2.2917} \end{aligned} \begin{aligned} &u_{C}(s):=-\frac{7.005 \cdot s+34.742}{s^{2}+0.875 \cdot s+2.292} \\ &a:=1 \\ &b:=0.875 \\ &c:=2.292 \\ &\Delta:=b^{2}-4 a \cdot c=-8.402 \\ &s_{1}:=\frac{-b+\sqrt{\Delta}}{2 a}=-0.438+1.449 \mathrm{i} \\ &s_{2}:=\frac{-b-\sqrt{\Delta}}{2 a}=-0.438-1.449 \mathrm{i} \end{aligned} \begin{aligned} &s:=-0.438+1.449 \mathrm{i} \\ &L:=7.005 \cdot s+34.742=31.674+10.15 \mathrm{i} \\ &M^{\prime}:=2 s+0.875=2.9 \mathrm{i} \\ &-\frac{31.674+10.15 \mathrm{i}}{2.9 \mathrm{i}} e^{-0.438 \cdot t+1.449 \mathrm{i} \cdot t} \\ &-\frac{31.674+10.15 \mathrm{i}}{2.9 \mathrm{i}} e^{-0.438 \cdot t} \cdot e^{1.449 \mathrm{i} \cdot t} \\ &-\frac{31.674+10.15 \mathrm{i}}{2.9 \mathrm{i}} e^{-0.438 \cdot t} \cdot(\cos (1.449 t)+1 \mathrm{j} \cdot \sin (1.449 t)) \end{aligned} \begin{aligned} &-2 e^{-0.438 \cdot t} \cdot\left(-10.922 \mathrm{i} \cdot 1 \mathrm{j} \cdot \sin (1.449 t)+\frac{10.15 \mathrm{i}}{2.9 \mathrm{i}} \cdot \cos (1.449 t)\right) \\ &-21.844 \cdot \sin (1.449 \cdot t) \cdot e^{-0.438 \cdot t}+7.0 \cdot \cos (1.449 \cdot t) \cdot e^{-0.438 \cdot t} \\ &u_{C}(t):=-21.844 \cdot \sin (1.449 \cdot t) \cdot e^{-0.438 \cdot t}-7.0 \cdot \cos (1.449 \cdot t) \cdot e^{-0.438 \cdot t} \\ &+14.102 \cdot \sqrt{2} \cdot \sin (\omega \cdot t+121.262 \operatorname{deg}) \\ &u_{C}(0)=10.048 \\ &u_{C .0}=10.042 \end{aligned}

Stany nieustalone

Teoria obwodów - Stany nieustalone - Przykład 3

Przykład 3