Chapter 3, Problem 74P

By inspection, obtain the mesh-current equations for the circuit in Fig. 3.118.

To determine

Write the mesh-current equations for the circuit in Figure 3.118.

Answer to Problem 74P

The matrix form representation of mesh-current equations for the given circuit is $\left[\begin{array}{cccc}\left(R_1+R_4+R_6\right)& -R_4& -R_6& 0\\ -R_4& \left(R_2+R_4+R_5\right)& 0& -R_5\\ -R_6& 0& \left(R_6+R_7+R_8\right)& -R_8\\ 0& -R_5& -R_8& \left(R_3+R_5+R_8\right)\end{array}\right]\left[\begin{array}{c}{i}_1\\ i_2\\ i_3\\ i_4\end{array}\right]=\left[\begin{array}{c}{V}_1\\ -V_2\\ V_3\\ V_2-V_4\end{array}\right]$.

Explanation of Solution

Given data:

Refer Figure 3.118 in the textbook for mesh analysis.

Calculation:

Apply Kirchhoff’s voltage law to loop 1 with current $i_1$ in Figure 3.118.

$\begin{array}{l}{i}_1{R}_1+\left(i_1-i_2\right)R_4+\left(i_1-i_3\right)R_6-V_1=0\\ i_1{R}_1+i_1{R}_4-i_2{R}_4+i_1{R}_6-i_3{R}_6-V_1=0\end{array}$

$\left(R_1+R_4+R_6\right)i_1-R_4{i}_2-R_6{i}_3=V_1$ (1)

Apply Kirchhoff’s voltage law to loop 2 with current $i_2$ in Figure 3.118.

$\begin{array}{l}{i}_2{R}_2+\left(i_2-i_4\right)R_5+V_2+\left(i_2-i_1\right)R_4=0\\ i_2{R}_2+i_2{R}_5-i_4{R}_5+V_2+i_2{R}_4-i_1{R}_4=0\end{array}$

$-R_4{i}_1+\left(R_2+R_5+R_4\right)i_2-R_5{i}_4=-V_2$ (2)

Apply Kirchhoff’s voltage law to loop 3 with current $i_3$ in Figure 3.118.

$\begin{array}{l}\left(i_3-i_1\right)R_6+\left(i_3-i_4\right)R_8-V_3+i_3{R}_7=0\\ i_3{R}_6-i_1{R}_6+i_3{R}_8-i_4{R}_8-V_3+i_3{R}_7=0\end{array}$

$-R_6{i}_1+\left(R_6+R_7+R_8\right)i_3-R_8{i}_4=V_3$ (3)

Apply Kirchhoff’s voltage law to loop 4 with current $i_4$ in Figure 3.118.

$\begin{array}{l}{i}_4{R}_3+V_4+\left(i_4-i_3\right)R_8-V_2+\left(i_4-i_2\right)R_5=0\\ i_4{R}_3+V_4+i_4{R}_8-i_3{R}_8-V_2+i_4{R}_5-i_2{R}_5=0\end{array}$

$-R_5{i}_2-R_8{i}_3+\left(R_3+R_5+R_8\right)i_4=V_2-V_4$ (4)

Represent the equations (1), (2), (3), and (4) are in matrix form as follows.

$\left[\begin{array}{cccc}\left(R_1+R_4+R_6\right)& -R_4& -R_6& 0\\ -R_4& \left(R_2+R_4+R_5\right)& 0& -R_5\\ -R_6& 0& \left(R_6+R_7+R_8\right)& -R_8\\ 0& -R_5& -R_8& \left(R_3+R_5+R_8\right)\end{array}\right]\left[\begin{array}{c}{i}_1\\ i_2\\ i_3\\ i_4\end{array}\right]=\left[\begin{array}{c}{V}_1\\ -V_2\\ V_3\\ V_2-V_4\end{array}\right]$

Conclusion:

Thus, the matrix form representation of mesh-current equations for the given circuit is $\left[\begin{array}{cccc}\left(R_1+R_4+R_6\right)& -R_4& -R_6& 0\\ -R_4& \left(R_2+R_4+R_5\right)& 0& -R_5\\ -R_6& 0& \left(R_6+R_7+R_8\right)& -R_8\\ 0& -R_5& -R_8& \left(R_3+R_5+R_8\right)\end{array}\right]\left[\begin{array}{c}{i}_1\\ i_2\\ i_3\\ i_4\end{array}\right]=\left[\begin{array}{c}{V}_1\\ -V_2\\ V_3\\ V_2-V_4\end{array}\right]$.