Electric Circuit Analysis/Mesh Analysis/Answers

Exercise 8: Answers Model Answer KVL arround abca loop: $I_{1}*R_{1}+(I_{1}-I_{3})*R_{2}+(I_{1}-I_{2})*R_{3}=-V_{s}$ Therefore $I_{1}(R_{1}+R_{2}+R_{3})-I_{2}(R_{3})-I_{3}(R_{2})=-V_{s}$ ............... (1) KVL arround acda loop: $(I_{2}-I_{1})*R_{3}+(I_{2}-I_{3})*R_{4}+I_{2}*R_{5}=V_{s}$ Therefore $-I_{1}(R_{3})+I_{2}(R_{3}+R_{4}+R_{5})-I_{3}(R_{4})=V_{s}$ ............... (2) KVL arround bcdb loop: $I_{3}*R_{6}+(I_{3}-I_{2})*R_{4}+(I_{3}-I_{1})*R_{2}=0$ Therefore $-I_{1}(R_{2})-I_{2}(R_{4})+I_{3}(R_{2}+R_{4}+R_{6})=0$ ............... (3) Now we can create a matrix with the above equations as follows: ${\begin{bmatrix}(R_{1}+R_{2}+R_{3})&(-R_{3})&(-R_{2})\\(-R_{3})&(R_{3}+R_{4}+R_{5})&(-R_{4})\\(-R_{2})&(-R_{4})&(R_{2}+R_{4}+R_{6})\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-V_{s}\\V_{s}\\0\end{bmatrix}}$ The following matrix is the above with values substituted: $A.{\vec {X}}={\vec {Y}}$ → ${\begin{bmatrix}10220&-10000&-20\\-10000&30000&-5000\\-20&-5000&6020\end{bmatrix}}.{\begin{bmatrix}I_{1}\\I_{2}\\I_{3}\end{bmatrix}}={\begin{bmatrix}-9\\9\\0\end{bmatrix}}$ Now that we have arranged equations 1; 2 & 3 into a matrix we need to get Determinants of the General matrix, and Determinants of alterations of the general matrix as follows: Solving determinants of: Matrix A  : General matrix A from KVL equations Matrix A1 : Genral Matrix A with Column 1 substituted by ${\vec {Y}}$. Matrix A2 : Genral Matrix A with Column 2 substituted by ${\vec {Y}}$. Matrix A3 : Genral Matrix A with Column 3 substituted by ${\vec {Y}}$. As follows: $detA=9.86\times 10^{11}$ $detA1=-857700000$ $detA2=11016000$ $detA3=6300000$ Now we can use the solved determinants to arrive at solutions for Mesh Currents $I_{1};I_{2}andI_{3}$ as follows: 1. $I_{1}={\frac {detA1}{detA}}=-0.00086968A$ 2. $I_{2}={\frac {detA2}{detA}}=0.00001117A$ 3. $I_{3}={\frac {detA3}{detA}}=0.000006388A$ Now we can solve for the current through $R_{3}$ as follows: $I_{R_{3}}=I_{1}-I_{2}=-0.881mA$ The negative sign means that $I_{R_{3}}$ is flowing in the direction of $I_{2}$.