18 20 V i2 10 16 i7 14 V 15 i4 8 38 V i4 14 Figure 1: Electric Circuit This proj

Question

18 20 V i2 10 16 i7 14 V 15 i4 8 38 V i4 14 Figure 1: Electric Circuit This project refers to the electric circuit shown in the figure above. You are asked to do the following: 1. Write the circuit equations in matrix form using Kirchhoff's Laws, as explained in class. 2. Once you have the system matrix and the right hand side (RHS) vector, type them in a text file (a file with the .txt extension) using either Wordpad or Notepad. The first line of the file should contain a title that explains what is next in the file: something like 'size of the problem. The second line of the file should contain the actual size of the square matrix that corresponds to the circuit (an integer that is equal to the number of unknown currents in the circuit). Subsequent lines should contain the title of what is next plus the actual contents of what is next. The circuit matrix should be read in row order. The RHS vector should be read as a column vector

Explanation / Answer

   Before we start solving any circuit using Kirchoffs Laws we should be clear about our conventions. My conventions are:
   When we move from lower potential( negative terminal of a battery) to higher potential(positive terminal) we take positive voltage contribution.
   When we move in a loop, since current flows from higher potential to lower, if we move in the direction of current it is negative
   Sum of all voltages in a closed loop is zero
   Sum of all currents at a node is zero.
From the given circuit
Loop 1253:
18 i1 -16 i2 -10i3 +20 = 0
We can re write it as:
-9i1 + 8 i2+ 5 i3 + 0 i4 + 0 i5 + 0 i6 + 0i7 + 0i8 + 0i9 = 20 -(1)
Loop 4576:
16 i2 -14 + 8 i8 -6 i6 = 0
0 i1 + 8 i2 +0 i3 +0 i4 +0 i5 -3 i6 + 0 i7 + 4 i8 + 0 i9 = 7 -(2)

Loop 67109:

0 i1 + 0i2 + 0i3 + 0i4 + 6 i5+ 0 i6+ 0 i7 + 4 i8 – 7i9 = 19 -(3)

Loop 34698
0 i1 + 0 i2 + 10 i3 + 15 i4 + 12 i5 + 6 i6 +0 i7 + 0i8 + 0i9 = 0 -(4)

However these are insufficient as we have 9 currents to determine so we need atleast 5 more equations to determine all 9 currents:
Node 3:
i1 +i3 -i4 =0 -(5)
Node 5:
i1+i2+i7 = 0 -(6)
Node 4:
i2+ i6 -i3 = 0 -(7)
Node 6:
I6 + i8 -i5 = 0 -(8)
Node 9:
I5 + i9 -i4 = 0 -(9)

We have one more node 7, but that is extra information.

Now we can write above equations as matrix equation

( ( - 9 8 5 0 0 0 0 0 0@ 0 8 0 0 0-3 0 4 0@ 0 0 0 0 6 0 0 8 0@ 0 0 10 15 12 6 0 0 0@ 0 1-1 0 0 0 0 0@ 1 1 0 0 0 0 1 0 0 @ 0 1 -1 0 0 1 0 0 0@ 0 0 0 0-1 1 0 1 0@ 0 0 0-1 1 0 0 0 1))((i1@i2@i3@i4@i5@i6@i7@i8@i9))= ((10@7@19@0@0@0@0@0@0))
This is the solution
   Size of the circuit
9
Circuit Matrix
( - 9 8 5 0 0 0 0 0 0@ 0 8 0 0 0-3 0 4 0@ 0 0 0 0 6 0 0 8 0@ 0 0 10 15 12 6 0 0 0@ 0 1-1 0 0 0 0 0@ 1 1 0 0 0 0 1 0 0 @ 0 1 -1 0 0 1 0 0 0@ 0 0 0 0-1 1 0 1 0@ 0 0 0-1 1 0 0 0 1)
Right hand side vector:
((10@7@19@0@0@0@0@0@0))
   Now you can write a matlab program

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