Electric Circuit Analysis/Nodal Analysis/Answers Exercise 7: Answers KCL @ Node b: $i_{2}=i_{1}+i_{6}$ Thus by applying Ohms law to above equation we get. ${\begin{matrix}\ {\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{b}}{R_{1}}}-{\frac {V_{b}-V_{d}}{R_{6}}}=0\end{matrix}}$ Therefore ${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+{\frac {1}{R_{6}}})+V_{c}({\frac {1}{R_{2}}})+V_{d}({\frac {1}{R_{6}}})=0\end{matrix}}$ ............... (1) KCL @ Node c: $i_{3}=i_{2}+i_{4}$ Thus by applying Ohms law to above equation we get. ${\begin{matrix}\ {\frac {V_{s}-V_{c}}{R_{3}}}-{\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{c}-V_{d}}{R_{4}}}=0\end{matrix}}$ Therefore ${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{2}}})-V_{c}(({\frac {1}{R_{2}}}+{\frac {1}{R_{3}}}+{\frac {1}{R_{4}}})+V_{d}({\frac {1}{R_{4}}})={\frac {V_{s}}{R_{3}}}\end{matrix}}$ ............... (2) KCL @ Node d: $i_{5}=i_{4}+i_{6}$ Thus by applying Ohms law to above equation we get. ${\begin{matrix}\ {\frac {V_{s}-V_{c}}{R_{3}}}-{\frac {V_{c}-V_{b}}{R_{2}}}-{\frac {V_{c}-V_{d}}{R_{4}}}\end{matrix}}$ Therefore ${\begin{matrix}\ \ -V_{b}({\frac {1}{R_{3}}})-V_{c}({\frac {1}{R_{4}}})+V_{d}({\frac {1}{R_{3}}}+{\frac {1}{R_{4}}}-{\frac {1}{R_{5}}})=0\end{matrix}}$ ............... (3) $G_{1}={\frac {1}{R_{1}}}$ etc thus equations 1; 2 & 3 will be re-written as follows: ${\begin{matrix}\ V_{b}(-G_{1}-G_{2}-G_{6})+V_{c}(G_{2})+V_{d}(G_{6})&=&0....(1)\\\ \\\ V_{b}(-G_{2})+V_{c}(G_{2}+G_{3}+G_{4})+V_{d}(-G_{4})&=&(V_{s}\times G_{3})....(2)\\\ \\\ V_{b}(-G_{3})+V_{c}(-G_{4})+V_{d}(G_{3}+G_{4}+G_{5})&=&0....(3)\end{matrix}}$ Now we can create a matrix with the above equations as follows: ${\begin{bmatrix}(-G_{1}-G_{2}-G_{6})&(G_{2})&(G_{6})\\(-G_{2})&(G_{2}+G_{3}+G_{4})&(-G_{4})\\(-G_{4})&(G_{4})&(G_{3}+G_{4}+G_{5})\end{bmatrix}}.{\begin{bmatrix}V_{b}\\V_{c}\\V_{d}\end{bmatrix}}={\begin{bmatrix}0\\V_{s}\times G_{3}\\0\end{bmatrix}}$ The following matrix is the above with values substituted: $A.{\vec {X}}={\vec {Y}}$ → ${\begin{bmatrix}-0.056&0.05&0.001\\-0.05&0.0503&-0.0002\\-0.0001&-0.0002&0.000367\end{bmatrix}}.{\begin{bmatrix}V_{b}\\V_{c}\\V_{d}\end{bmatrix}}={\begin{bmatrix}0\\0.0009\\0\end{bmatrix}}$ 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 KCL 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: $detA=-9.8\times 10^{-8}$ $detA1=-1.7\times 10^{-8}$ $detA2=-1.8\times 10^{-8}$ $detA3=-1.5\times 10^{-8}$ Now we can use the solved determinants to arrive at solutions for Node voltages $V_{b};V_{c}andV_{d}$ as follows: 1. $V_{b}={\frac {detA1}{detA}}=0.17V$ 2. $V_{c}={\frac {detA2}{detA}}=0.19V$ 3. $V_{d}={\frac {detA3}{detA}}=0.15V$ Now we can apply Ohm's law to solve for the current through $R_{3}$ as follows: $I_{R_{3}}={\frac {V_{s}-V_{c}}{R_{3}}}={\frac {9V-0.19V}{10k\Omega }}=0.881mA$ 