Solving 4 complex equations with your Ti-84
Most complex solver programs, written for the Ti-84, can only solve equations with a maximum of 3 variables. This Ti-84 Basic program however is unique in that it can solve up to 4 complex linear equations (w, x, y, z). This is very useful for electrical engineering students to calculate complex voltages or currents in networks. Of course, it can also solve 2 and 3 linear equations. At the end of this page, an extended example including network theory is provided. Of course, equations with real numbers (DC networks) can also be solved.
Example : solving 4 linear complex equations.
Suppose you have the next complex equations,
which hold for an electric network
(2i)Ia + (5)Ib + (9)Ic - (5i-12)Id = 3
(7)Ia + (3-i)Ib + (5+6i)Ic + (8)Id = 12-3i
(13)Ia - (2i-2)Ib + (3+6i)Ic + (10)Id = 3i
(0)Ia +(6+3i)Ib + (14)Ic - (3i-17)Id = 4
The Ti program solv234.8xp solves:
In the complex equations of the Ti84 program, we choose
w=Ia, x=Ib, y=Ic, z=Id, so in the first equation M=2i, A=5, B=9 and C=-(5i-12)=+(12-5i) and D=3
Be aware that all signs are positive in the Ti-equations, therefore we take C=(12-5i). For the other 3 equations, we use the same method as explained. Solving these equations "by hand" would be an incredible task.
To check if the answer is correct, we fill in the answer in equation 2
(7)Ia + (3-i)Ib +(5+6i)Ic + (8)Id should be 12-3i for the Ti program
Ia=w, Ib=x, Ic=y, Id=z. After running the program, the solutions are stored in w, x, y, z
Therefore : (7)w + (3-i)x +(5+6i)y + (8)z should be 12-3i.
After this check, we can trust the program that it gives the right answers.
Example : solving 2 linear complex equations.
A complete example of solving 2 complex equations is now given. Suppose the network with 2 AC voltage (60 Hz) sources with different voltage level and 90 degrees of phase shift as well as components such as inductance resistance and capacitor. Current Ia and Ib must be calculated. Simulation with Multisim gives Ia=2.351A (rms) and Ib=0.532104 A (rms).
The electric circuit is transferred to a complex scheme with complex impedances and sources.
According to Kirchhoff's Second Law, the sum of voltages in closed loops 1 and 2 are equal to zero, resulting in two equations.
1: +110-Ia*(50+75.398j) –(Ia-Ib)(-53.052j+40)-19052j=0
2: +110-Ia*(50+75.398j) –(Ib)(188.49j+60)=0
or reduced in the way that they can be used in the Ti-84 program
eq1: (-90-22.345j)*Ia +(40-53.052j)*Ib=190.52j-110
eq2 : (-50-75.398j)*Ia +(-60-188.49j)*Ib=-110
In the Ti-program, we choose 2 equ in which x=Ia and y=Ib
Be aware of the minus signs in -90-75.398j, For the minus sign of -90 you have to use the (-) key.
Fill in the complex number for a, b, p, c, d, q from equation 1 and 2
The Ti-84 solves this problem within seconds .
The results are given in the a+bi mode. To compare the RMS values for Ia en Ib with Multisim we change the mode of the Ti in re^(θi), amplitude r with angle in rad e-notation. The results of x and y are stored in the calculator in memory x and y. So if we recall x and y than x and y are shown in re^(θi) mode.
We can conclude that the results agree very well with those obtained with Multisim.
At the end Ia-Ib is calculated with x-y=Ia-Ib=2,48 Arms
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