Linear Algebra and the C Language/a04p


Analysis of an electrical circuit

Studying the circuit in a clockwise direction, we can write:


At node A:

      Input = Output
    I2 + I3 = I1

At node B :

     Input = Output
        I4 = I3 + I5

At node C :

     Input = Output
   I6 + I5 = I4

At node D :

    Input = Output
       I1 = I6 + I2

Then:

a) -I1  +I2  +I3                 = 0
             -I3  +I4  -I5       = 0
                  -I4  +I5  +I6  = 0
   +I1  -I2                 -I6  = 0

Left part of the circuit:

              -90 +I1 R1 + I2 R2 = 0

b)                +I1 R1 + I2 R2 = +90

Central part of the circuit:

   -I2 R2 + I3 R3 + I4 R4 + I6 R6 = 0

c) -I2 R2 + I3 R3 + I4 R4 + I6 R6 = 0

Right part of the circuit:

             90 - I5 R5 - I4 R4 = 0

d)                -I4 R4 -I5 R5 = -90

External part of the circuit:

   -90 + I1 R1 + I3 R3 + 90 - I5 R5 + I6 R6 = 0

e)           +I1 R1 + I3 R3 - I5 R5 + I6 R6 = 0

Then:

a) -I1     +I2     +I3                             =  +0
                   -I3     +I4      -I5            =  +0
                           -I4      +I5    +I6     =  +0
   +I1     -I2                             -I6     =  +0
b) +I1 R1  +I2 R2                                  = +90
c)         -I2 R2  +I3 R3  +I4 R4          +I6 R6  =  +0
d)                         -I4 R4  -I5 R5          = -90
e) +I1 R1          +I3 R3          -I5 R5  +I6 R6  =  +0

With R1 = 15, R2 = 60, R3 = 15, R4 = 15, R5 = 60, R6 = 15

a) -I1     +I2     +I3                             =  +0
                   -I3     +I4      -I5            =  +0
                           -I4      +I5    +I6     =  +0
   +I1     -I2                             -I6     =  +0
b) +I1 15  +I2 60                                  = +90
c)         -I2 60  +I3 15  +I4 15          +I6 15  =  +0
d)                         -I4 15  -I5 60          = -90
e) +I1 15          +I3 15          -I5 60  +I6 15  =  +0

Let's rectify the system:

    I1   I2   I3   I4   I5   I6
   
a)  -1   +1   +1   +0   +0   +0     +0
    +0   +0   -1   +1   -1   +0     +0
    +0   +0   +0   -1   +1   +1     +0
    +1   -1   +0   +0   +0   -1     +0
b) +15  +60   +0   +0   +0   +0    +90
c)  +0  -60  +15  +15   +0  +15     +0
d)  +0   +0   +0  -15  -60   +0    -90
e) +15   +0  +15   +0  -60  +15     +0

The code in C language:

double ab[RA*(CA+Cb)]={
//   I1    I2    I3    I4    I5    I6    
     -1,   +1,   +1,   +0,   +0,   +0,   +0, +0,   +0,
     +0,   +0,   -1,   +1,   -1,   +0,   +0, +0,   +0,
     +0,   +0,   +0,   -1,   +1,   +1,   +0, +0,   +0,
     +1,   -1,   +0,   +0,   +0,   -1,   +0, +0,   +0,
    +15,  +60,   +0,   +0,   +0,   +0,   +0, +0,  +90,
     +0,  -60,  +15,  +15,   +0,  +15,   +0, +0,   +0,
     +0,   +0,   +0,  -15,  -60,   +0,   +0, +0,  -90,
    +15,   +0,  +15,   +0,  -60,  +15,   +0, +0,   +0
};

The solution is given by solving the system:

 I1     I2     I3     I4     I5     I6 
+1.00  -0.00  -0.00  +0.00  -0.00  +0.00  -0.00  -0.00  +2.00 
+0.00  +1.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +1.00 
+0.00  +0.00  +1.00  +0.00  +0.00  +0.00  +0.00  +0.00  +1.00 
+0.00  +0.00  +0.00  +1.00  +0.00  +0.00  +0.00  +0.00  +2.00 
+0.00  +0.00  +0.00  +0.00  +1.00  +0.00  +0.00  +0.00  +1.00 
+0.00  +0.00  +0.00  +0.00  +0.00  +1.00  +0.00  +0.00  +1.00 
+0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00 
+0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00  +0.00 

I1 = 2,    I2 = 1,   I3 = 1,   I4 = 2,   I5 = 1,   I6 = 1,