Edupanda » Teoria obwodów »   Metody analizy obwodów elektrycznych   »   Metoda prądów oczkowych - prąd przemienny   » Przykład 1

Przykład 1

W obwodzie jak na rysunku obliczyć rozpływ prądów metodą prądów oczkowych. $$ \begin{aligned} &E_{1}=100 \cdot e^{1 \mathrm{j} \cdot 90 \mathrm{deg}}\\ &E_{2}=50 \\ &R_{1}=2, \quad Z_{L 1}=8 \mathrm{j}, \quad R_{2}=5 \\ &Z_{L 3}=5 \mathrm{j}, \quad Z_{C 4}=-2 \mathrm{j}, \quad Z_{C 5}=-1 \mathrm{j} \\ &R_{6}=10 \\ \end{aligned} $$

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Rozwiązanie

\begin{aligned} &I_{1}=I_{o 1} \quad+ \\ &I_{2}=I_{o 2} \\ &I_{3}=I_{o 3} \\ &I_{4}=I_{o 1}-I_{o 2} \\ &I_{5}=I_{o 2}-I_{o 3} \\ &E_{1}-I_{1} \cdot\left(R_{1}+Z_{L 1}\right)-I_{4} \cdot Z_{C 4}=0 \\ &I_{4} \cdot Z_{C 4}-I_{2} \cdot R_{2}-I_{5} \cdot Z_{C 5}=0 \\ &I_{5} \cdot Z_{C 5}-I_{3} \cdot\left(Z_{L 3}+R_{6}\right)+E_{2}=0 \end{aligned} \begin{aligned} &E_{1}-I_{1} \cdot\left(R_{1}+Z_{L 1}\right)-I_{4} \cdot Z_{C 4}=0 \\ &I_{4} \cdot Z_{C 4}-I_{2} \cdot R_{2}-I_{5} \cdot Z_{C 5}=0 \\ &I_{5} \cdot Z_{C 5}-I_{3} \cdot\left(Z_{L 3}+R_{6}\right)+E_{2}=0 \\ &E_{1}-I_{o 1} \cdot\left(R_{1}+Z_{L 1}\right)-\left(I_{o 1}-I_{o 2}\right) \cdot Z_{C 4}=0 \\ &\left(I_{o 1}-I_{o 2}\right) \cdot Z_{C 4}-I_{o 2} \cdot R_{2}-\left(I_{o 2}-I_{o 3}\right) \cdot Z_{C 5}=0 \\ &\left(I_{o 2}-I_{o 3}\right) \cdot Z_{C 5}-I_{o 3} \cdot\left(Z_{L 3}+R_{6}\right)+E_{2}=0 \end{aligned} \begin{aligned} &I_{o 2} \cdot Z_{C 4}+E_{1}-\left(Z_{L 1}+Z_{C 4}+R_{1}\right) \cdot I_{o 1}=0 \\ &I_{o 3} \cdot Z_{C 5}-\left(Z_{C 5}+Z_{C 4}+R_{2}\right) \cdot I_{o 2}+I_{o 1} \cdot Z_{C 4}=0 \\ &\left(-Z_{L 3}-\left(Z_{C 5}+R_{6}\right)\right) \cdot I_{o 3}+E_{2}+I_{o 2} \cdot Z_{C 5}=0 \\ &{\left[\begin{array}{ccc} Z_{L 1}+Z_{C 4}+R_{1} & -Z_{C 4} & 0 \\ -Z_{C 4} & Z_{C 5}+Z_{C 4}+R_{2} & -Z_{C 5} \\ 0 & -Z_{C 5} & Z_{L 3}+\left(Z_{C 5}+R_{6}\right) \end{array}\right] \cdot\left[\begin{array}{l} I_{o 1} \\ I_{o 2} \\ I_{o 3} \end{array}\right]=\left[\begin{array}{c} E_{1} \\ 0 \\ E_{2} \end{array}\right]} \\ &E_{1}:=100 \cdot e^{1 \mathrm{j} \cdot 90 \mathrm{deg}}=100 \mathrm{i} \\ &E_{2}:=50 \\ &R_{1}:=2 \\ &Z_{L 1}:=8 \mathrm{j} \\ &R_{6}:=10 \\ &R_{2}:=5 \\ &Z_{L 3}:=5 \mathrm{j} \\ &Z_{C 5}:=-1 \mathrm{j} \\ &Z_{C 4}:=-2 \mathrm{j} \end{aligned} $$ \left[\begin{array}{ccc} 2+6 \mathrm{i} & 2 \mathrm{i} & 0 \\ 2 \mathrm{i} & 5-3 \mathrm{i} & 1 \mathrm{i} \\ 0 & 1 \mathrm{i} & 10+4 \mathrm{i} \end{array}\right] \cdot\left[\begin{array}{c} I_{o 1} \\ I_{o 2} \\ I_{o 3} \end{array}\right]=\left[\begin{array}{c} 100 \mathrm{i} \\ 0 \\ 50 \end{array}\right] $$ $$ \begin{aligned} &I_{o 1}=-13.404-5.694 \mathrm{i} \\ &I_{o 2}=-4.093+3.676 \mathrm{i} \\ &I_{o 3}=-3.852+1.95 \mathrm{i} \end{aligned} $$ $$ \begin{aligned} &I_{1}:=I_{o 1}=-13.404-5.694 \mathrm{i} \\ &I_{2}:=I_{o 2}=-4.093+3.676 \mathrm{i} \\ &I_{3}:=I_{o 3}=-3.852+1.95 \mathrm{i} \\ &I_{4}:=I_{o 1}-I_{o 2}=-9.311-9.37 \mathrm{i} \\ &I_{5}:=I_{o 2}-I_{o 3}=-0.241+1.726 \mathrm{i} \end{aligned} $$ sprawdzenie $$ \begin{array}{ll} I_{1}-I_{2}-I_{4}=0 & O K \\ I_{2}-I_{3}-I_{5}=0 & \text { OK } \end{array} $$