# Solution of 3 to 14 Simultaneous Equations with 3 to 14 unknowns for DC circuit analysis.

The applet below can be used to find the solution of the kind of matrix equations (AX = B) which occur in circuit analysis.
The elements of the matrix A is shown in yellow and of the column matrix B in red. The element to be changed is shown in pink.

Step 1. Use the Neq scrollbar to set the number of equations.(3 to 14)
Step 2. Use the row and col scrollbars to move to the element to be entered. The element to be changed should be shown in pink. Only nonzero elements need to be entered.
Step 3. Use the c1, c2 and c3 scrollbars to set the value of the matrix element. The value is shown by the blue text at the top. Elements can be set with an accuracy of 6 figures. Negative numbers are displayed with 5 digits in yellow but are stored as 6 digits as shown by the blue text at top.
Step 4. Click the right arrow of the Lock scrollbar. The element to be changed should be updated and shown in cyan text. It is essential at this point to click the left arrow of the Lock scrollbar to return the text to pink.
Step 5. Repeat steps 2, 3 and 4 until all the matrix elements have been set.
Step 6. Click the right arrow of the soln scrollbar. The solution should be shown as the blue text at the bottom. The numbers are the solution to the variables in X and are listed in the same order as the column matrix X.

Using node voltage analysis, the matrix equation for the node voltages in the above circuit is shown below:-
The image below show the applet used to find the node voltages:-
The blue text at the bottom of the applet show the node voltages are:-
v1 = 0.398778V
v2 = -0.60121V
v3 = -4.503V
v4 = 5.86237V
v5 = 3.19033V
v6 = 8.0673V
v7 = -21.916V
v8 = -19.572V

Mesh current analysis applied to the above circuit gives the following equation for the mesh currents:-
Applet image showing solution of above equation is shown below:-
Blue text at bottom of applet gives the mesh currents:-
i1 = 1.98734A
i2 = 2.18228A
i3 = 1.92596A
i4 = 0.00576238A
i5 = 0.803457A
i6 = -0.80697A
i7 = -0.16837A
i8 = -0.099337A

Modified nodal analysis of the above circuit gives the following matrix equation. The number of variables is increased to 10 because the 2 voltage source currents are treated as unknowns. Matrix elements which must be added as a result are shown in green. Currents coming out of the positive terminal of the voltage sources are considered to be positive. However other conventions are possible and this would result in a change in sign for the voltage source currents obtained.
Image below shows the applet used to solve the matrix equation.
Blue text at bottom of applet show the node voltages and source currents:-
v1 = 0.104581V
v2 = -0.89540V
v3 = -5.2384V
v4 = 4.81643V
v5 = 0.836972V
v6 = 0.108401V
v7 = -2.8915V
v8 = -4.000V
i3 = 0.238288A
i4 = 0.865521A