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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

enter image description here

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  • \$\begingroup\$ You need to show your calculation so that folk don't waste time going over stuff you know how to do. \$\endgroup\$ – Andy aka Nov 21 at 18:57
  • \$\begingroup\$ Thats why I included a photo of my matrix setup. I setup the matrix and then put all of my data into the ti-89. \$\endgroup\$ – Etlewisg Nov 21 at 19:05
  • \$\begingroup\$ Have you searched the web? There seems to be many videos and how-to pages on exactly this topic. Here is a good one. This page covers cSolve which accomodates complex values. The TI-89 manual covers it as well of course. \$\endgroup\$ – relayman357 Nov 21 at 19:13
  • \$\begingroup\$ I found a video on youtube on how to input the numbers into a matrix and solve it as well as how to utilize the csolve function. I kept getting the wrong answer but I followed both instructions to the letter. \$\endgroup\$ – Etlewisg Nov 21 at 19:23
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    \$\begingroup\$ @relayman357 Thank you very much for such a detailed response. I see exactly where I went wrong. I appreciate you taking the time to help me. I will most certainly use the MathJax tool for future references. \$\endgroup\$ – Etlewisg Nov 22 at 20:30
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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.

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