calculatorengineeringhomeworkmesh
Good Afternoon,
I am having calculator difficulties in regards to inputing my mesh equations into my TI-89. The answer I am receiving is not the answer that was presented in the solution provided by my TA. I also double checked that my mode was in degrees.
I uploaded the original solution and problem. The second link is to an image of my matrix.
Thank you!
Edit: Working link to my matrix: https://imgur.com/5j0g3Ce
The key matrix works by "strobing": apply power to one line briefly, see if it's shorted to another one. Normally I'd expect to see pullups or pulldowns to avoid floating lines, but I can't see them; they might be inside the chip.
You're on the right track with your transistors to convert 1.5V to 3.3V. You can get logic level converter ICs to do this all in a nice package. Given that you don't have a scope, you might just as well get the propellor to do the data reading, if you can write an app that just reads pins, looks for changes, and sends timestamped results to a PC for visualisation. Connect this to the A-Z lines. Power on the calculator. You should find that it's strobing some of the lines, and some of the other lines will become connected when keys are pressed.
Many of us have quite low brain capacity. For example your equations have too much variables at least for me. I can't at a glance see, is somewhere an error. But here's one proper solution. The facts are collected to equations with smaller number of variables:
The equations are
All terms are moved to the left side to get zeros to the right sides.
The solution is found by MS Excel Solver:
The orange area has the modified variables. The blue area has the left sides of the three equations. The yellow area is the goal - make this zero. It's the sum of the squares of the left sides.
The result: V0 = 48V
Best Answer
Compare each of your loop equations to what I show below.
$$\text{Bottom left loop: }\text{ }-(-6j) + (i_1-i_3)(-\frac{2}{3}j) + (i_1-i_2)=0 $$ $$\text{Upper left loop: }\text{ }5i_1+(i_3-i_4)(4j)+(i_3-i_1)(-\frac{2}{3}j)=0 $$ $$\text{Upper right loop: }\text{ }(i_4-i_2)(-\frac{1}{2}j)+(i_4-i_3)(4j)=0 $$ $$\text{Bottom right loop: }\text{ }9+(i_2-i_1)+(i_2-i_4)(-\frac{1}{2}j)=0 $$
Now, collecting terms and writing in matrix form, $$Ax=b$$ $$\begin{bmatrix}(1-\frac{2}{3}j)&-1&\frac{2}{3}j&0\\(5+\frac{2}{3}j)&0&3\frac{1}{3}j&-4j\\0&\frac{1}{2}j&-4j&3\frac{1}{2}j\\-1&(1-\frac{1}{2}j)&0&\frac{1}{2}j\end{bmatrix}\begin{bmatrix}i_1\\i_2\\i_3\\i_4\end{bmatrix}=\begin{bmatrix}-6j\\0\\0\\-9\\\end{bmatrix}$$
Then solve however you want (e.g. in your TI-89). I used this nice online complex capable simultaneous equation solver. The results are,
$$\begin{bmatrix}i_1\\i_2\\i_3\\i_4\end{bmatrix}=\begin{bmatrix}-1.8-1.2j\\-7-0.033j\\-9.05+6.608j\\-9.34+7.556j\end{bmatrix}\text{amps}$$
And, thanks to that handy 1Ω resistor we can easily find Vx,
$$V_x=(i_1-i_2)(1Ω)=5.33\angle-12.65⁰\text{ V}$$
The above is a phasor which we can convert to,
$$V_x(t)=5.33 cos(2000t-12.65⁰)\text{ V}$$
p.s. I recommend you take the time to enter your work in MathJax for 2 reasons. It is much easier for people to read (vs. a camera snap) and it gives you a chance to slow down and check your work. This neat tool lets you draw symbols and it tells you the corresponding Latex code.