Solution of 3 to 14 Simultaneous Equations with
3 to 14 unknowns for AC 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.
Modified nodal analysis for the above circuit would require 6 equations. To simplify only 2 equations for v1 and v2 are
considered. From v1 and v2 the other node voltages are
readily calculated.
With V1 = 1V, R1 = 1, R2 = 2, R3 = 3, R4 = 4
jwL1 = 2j, jwL2 = 4j, 1/jwC1 = -3j.
The matrix equation for the node voltages v1 and v2 is:-
The image below shows the solution of the 2 complex equations
obtained by representing them as 4 real equations for the
4 variables v1r, v1i, v2r and v2i.
The blue text at the bottom of the image gives v1 and v2:-
v1 = 0.367868 - j0.34411 = 0.50372 Ð-43.088°
v2 = 0.12745 - j0.30819 = 0.3335 Ð-67.533°
Knowing v3 = 1.0, v4 and v5 can be calculated:-
v4 = 0.73593 - j0.18403 = 0.75859 Ð14.0398°
v5 = 0.21782 - j0.09037 = 0.23582 Ð-22.5328°
For the circuit above, the nodal equations for the node voltages
v1, v2 and v3 is shown below.
The image below shows how the applet can be used to solve the matrix equation.
The blue text at the bottom give the voltages v1, v2 and v3:-
v1 = 0.0418522 - j0.37104 = 0.373393 Ð-83.564°
v2 = 0.257918 - j0.08371 = 0.271162 Ð-17.918°
v3 = -0.0067873 - j0.14253 = 0.142691 Ð-92.726°
Knowing v4 = 1.0, v5 can be calculated:-
v5 = -0.0678713 - j0.0746586 = 0.100898 Ð-47.726°
The matrix equation for the voltages v1, v2, v3 and v4 in the
above circuit is shown below:-
Image below shows the solution of the matrix equation:-
v1, v2, v3 and v4 are shown at the bottom of the applet:-
v1 = 1.13551 - j0.13302 = 1.14327 Ð-6.6815°
v2 = 0.384571 - j0.044082 = 0.387089 Ð-6.53907°
v3 = 0.82472 + j0.111221 = 0.832186 Ð7.68052°
v4 = 1.60431 + j0.474729 = 1.67307 Ð16.4839°
knowing v5 = 1 and v8 = 2, v6, v7, v9 and v10 can be calculated.:-
v6 = 0.973894 - j0.080808 = 0.977241 Ð-4.74320°
v7 = 1.12604 + j0.127913 = 1.13328
Ð-6.48077°
v9 = 1.50146 + j0.308549 = 1.53284 Ð11.6126°
v10 = 1.12604 - j0.127913 = 1.13328 Ð-6.48077°
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Copyright © 2010 Cuthbert A. Nyack.