# Ti-nSpire and Mesh Analysis

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1. Apr 6, 2014

### brickford5

I have just gotten my new Ti-nspire CAS calculator and seem to be stumped. I have created my two equations from my mesh analysis of a circuit.

Loop I2: I1(2+j2)+I2(8-j2)=12
Loop I1: I1(-j2)+I2(j-4)=0

I know I can solve this by hand but would like to learn to use my nspire. Any direction would be greatly appreciated, I feel I have tried everything I can think of and can't seem to get the result.

{{2+2i, -2i}, {8-2i, 12+i}} * {{I1, I2} = {{12, 0}} - I think this is right?

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Some things I've tried:

- Using dotP:
Creating the matrix of coefficients {{2+2i, -2i}, {8-2i, 12+i}}, taking the inverse (^-1) and dot-ting (dotP{}) it with [12,0] - I get 'Error Invalid data type'​

- Using cSolve:
Input: cSolve(x(2+i2)+y(8-2i)=12 and x(-i2)+y(i-4)=0,x,y​
Output: x=c6 and y=c7 and (then repeats equations)​

I reviewed numerous youtube videos and instructions but couldn't seem to find a solution the resulted in the right answer. I am trying to find I2 and did the hand calculations to get I2=1.76-1.06j

Thanks in advance to any input!

Last edited: Apr 6, 2014
2. May 4, 2014

### Greg Bernhardt

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

3. Jun 11, 2015

### Godzilla2456

Don't use matrix. Too tedious! Use the "Solve system of equations" feature.

For this basic circuit it would look like this! You access this feature by clicking menu, then 3 (algebra) then 7

4. Jun 11, 2015

### CalcNerd

You have an error in your equation from your first set of equations to your Matrix form (j-4) does not equal (12+J).

On using matrix features, have you tried your RREF (Reduced Row Echelon Form) the 2X3 matrix of your circuit ie the last column is your result. This command will reduce your matrix to provide an answer for I1 and I2 in the last column. Should work unless your Tinspire is based upon the Ti-84 (non-CAS model). The Ti-83/84 cannot handle complex matrix math. Not sure about the TinSpire NON-CAS calculator either.