Linear Algebra and the C Language/a04l

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

At node A:

      Input = Output
    I1 + I2 = I3

At node B :

      Input = Output
         I3 = I1 + I2

Then:

a)      I1 + I2 - I3 = 0

Left part of the circuit:

       I2 R2 - I1 R1 = 0

b)   - I1 R1 + I2 R2 = 0

Right part of the circuit:

 120 - I2 R2 - I3 R3 = 0

c)     I2 R2 + I3 R3 = 120

External part of the circuit:

 120 - I3 R3 - I1 R1 = 0

d)     I1 R1 + I3 R3 = 120

Then:

a)  I1    + I2    - I3     = 0    
b) -I1 R1 + I2 R2          = 0
c)          I2 R2 + I3 R3  = 120 
d)  I1 R1         + I3 R3  = 120

With R1 = 60, R2 = 30, R3 = 20

a)  I1    + I2    - I3     = 0    
b) -I1 60 + I2 30          = 0
c)          I2 30 + I3 20  = 120 
d)  I1 60         + I3 20  = 120

Let's rectify the system:

   I1    I2    I3
   +1    +1    -1    = 0    
  -60   +30    +0    = 0
   +0   +30   +20    = 120 
  +60    +0   +20    = 120

The code in C language:

double ab[RA*(CA+Cb)]={
// I1    I2    I3    
   +1,   +1,   -1,   +0,    +0,
  -60,  +30,   +0,   +0,    +0,  
   +0,  +30,  +20,   +0,    +120,
  +60,   +0,  +20,   +0,    +120
};

The solution is given by solving the system:

  I1    I2    I3  
  +1    +0    +0    +0    +1 
  +0    +1    +0    +0    +2 
  -0    -0    +1    -0    +3 
  +0    +0    +0    +0    +0 

 I1 = 1A; I2 = 2A; I3 = 3A;