# Circuit Theory/Simultaneous Equations/Example 3

The problem is find the unknown currents for the circuit from the table below. The resistance of each of the resistors is unknown, (even though they are all labeled 100 ohms).

### Label

This circuit comes to us labeled with this table of currents, but some of them are missing. There are four cases to figure out.
| Case | $i_{1}$ | $i_{2}$ | $i_{3}$ | $i_{4}$ | $i_{5}$ | $i_{6}$ | $i_{7}$ | $i_{9}$ |
| Case 1 | 1 | | | 2 | | 1 | | 3 |
| Case 2 | | 0 | 1 | 2 | | | 1 | |
| Case 3 | | | -1 | 1 | -1 | 1 | | |
| Case 4 | 4 | | 7 | | | 3 | | 9 |

### Loops

There are loops, but no power supply and all the resistor values are the same. There is not enough information to being analyzing the loops. The circuit came with the currents labeled and direction drawn. But no voltages Appears we are to figure this out with just the information in the table.

### Junctions

There are four nodes which can generate 3 equations. There are two trivial nodes with series devices having different currents labeled. One trivial node has currents going in opposite directions! This means have to write two trivial equations for a total of 5 equations.

### Count Eq

The first step is to list the variables and whether they are knowns or unknowns. The problem is there are four cases where the knowns and unknowns are different currents. This means there are really four problems here.
$i_{1}=?amps$ $i_{2}=?amps$ $i_{3}=?amps$ $i_{4}=?amp$ $i_{5}=?amps$ $i_{6}=?amps$ $i_{7}=?amps$ $i_{8}=?amp$ $i_{9}=?amps$ There are 9 currents that need to be found. 5 equations means that we need the values for 4 of the currents in order to find the other 9. Ok. Can do this problem.

### Terminal Eq

The resistors values are all unknowns. Writing 9 terminal equations will add 18 unknowns. So now the total would be 27 unknowns with 14 equations.

### Loop Eq

There are four loops that could result in four more equations. This gives us 18 equations to 27 unknowns. Given explicit values four of the currents, this reduces us to 23 unknowns. We are 5 unknowns from being able to compute everything!

### Junction Eq

$i_{1}+i_{2}=0$ $i_{6}-i_{7}=0$ $i_{2}-i_{3}-i_{4}-i_{5}-i_{6}=0$ $i_{8}+i_{4}-i_{9}=0$ $i_{1}+i_{3}-i_{8}=0$

### Solve the Equations
#### Algebra

The algebra solution is massive, messy, hard to check and doesn't inspire anyone. Algebra flows easily out of one's brain, lands on paper and makes sense in the moment to the author. But a year later, even the author would rather do the algebra again than run their mind through it. Other people look at this and are either intimidated or say to themselves, I would rather do this myself than try to figure that out.

#### Differential Equations

There are none in this problem, but three quarters of this course is going over circuits this complex with capacitors and inductors instead of resistors and looking at the differential equations.

#### Symbolic Computations
##### Wolfram Alpha

Could not get Wolfram Alpha to work. Tried 5 equations and 5 unknowns in symbol form:
Tried substituting for $i_{1}$ and $i_{6}$ to reduce the number of equations.
Still getting error messages. Giving up. The fact that wolfram alpha attempts to do this is amazing. Next tried "elvis pressley's eye color ... with trial "pro" version. Wolfram Alpha Pro said that Elvis sang a song called "Spanish Eyes."

##### MuPad

MuPAD had to be run once for each of the four cases.
The actual MuPAD syntax was typed once and then modified three times.
Each time only the variables to be solved for and the variables that were constant had to be changed.
This was not a lot of work ... compared to constructing and entering a 5 by 5 matrix and 1 by 5 matrix for each case like MatLab requires.

#### Numeric Solution

Only the first case is going to be set up and done with MatLab here. The algebra agrees with MuPAD. The authors trust that you can figure out how to do the other three cases in MatLab if you must.
It is the setting up the linear equations that is difficult, not the typing into MatLab. Steps to set up the solution:
* Substitute known values into the equations:
* $1+i_{2}=0$
* $1-i_{7}=0$
* $i_{2}-i_{3}-2-i_{5}-1=0$
* $i_{8}+2-3=0$
* $1+i_{3}-i_{8}=0$
* Organize the equations so that unknowns are in columns, numbers to the right of the equal sign:
* $i_{2}+0+0+0+0=-1$
* $0+0+0+i_{7}+0=1$
* $i_{2}-i_{3}-i_{5}+0+0=3$
* $0+0+0+0+i_{8}=1$
* $0+i_{3}+0+0-i_{8}=-1$
* Create two matrices, one square and the other a column with the numbers that are on the right of the equal sign
| ${\begin{array}{|c|c|c|c|c||c|}i_{2}&i_{3}&i_{5}&i_{7}&i_{8}&=\\\hline 1&0&0&0&0&-1\\0&0&0&1&0&1\\1&-1&-1&0&0&3\\0&0&0&0&1&1\\0&1&0&0&-1&-1\\\end{array}}$ | $\Rightarrow$ | ${\begin{bmatrix}1&0&0&0&0\\0&0&0&1&0\\1&-1&-1&0&0\\0&0&0&0&1\\0&1&0&0&-1\end{bmatrix}}$ | $\bullet$ | ${\begin{bmatrix}i_{2}\\i_{3}\\i_{5}\\i_{7}\\i_{8}\end{bmatrix}}$ | = | ${\begin{bmatrix}-1\\1\\3\\1\\-1\end{bmatrix}}$ |
* enter the matrices into MatLab or similar program
The first row was finished with MatLab above. The other rows come from MuPAD and the algebra.
| Case | $i_{1}$ | $i_{2}$ | $i_{3}$ | $i_{4}$ | $i_{5}$ | $i_{6}$ | $i_{7}$ | $i_{8}$ | $i_{9}$ |
| Case 1 | 1 | -1 | 0 | 2 | -4 | 1 | 1 | 1 | 3 |
| Case 2 | 0 | 0 | 1 | 2 | -4 | 1 | 1 | 1 | 3 |
| Case 3 | 0 | 0 | -1 | 1 | -1 | 1 | 1 | -1 | 0 |
| Case 4 | 4 | -4 | 7 | -2 | -12 | 3 | 3 | 11 | 9 |

### Simulate

This can not be simulated. There are no sources, and not enough information.

### Build Intuition

* This is an example of what forward circuit design is like. Given a vague requirement, start exploring possibilities.
* Current directions can be any direction. They capture part of the circuit topology and thus generate some constraints all by themselves.
* Symbolic algebra (MuPAD) becomes more useful in the forward engineering design process.