Linear Algebra and the C Language/a04l

__NOTOC__ Analysis of an electrical circuit thumb| 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 }; thumb| 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;