clc
// Fundamental of Electric Circuit 
// Charles K. Alexander and  Matthew N.O Sadiku  
// Mc Graw Hill of New York 
// 5th Edition 

// Part 2     :  AC Circuits 
// Chapter 11 :  AC power Analysis 
// Example 11 - 4

clear; clc; close; // Clear the work space and console.
//
// Given data
I_mag_1   =  4.0000;                     // I_mag_1    =  Magnitude of Source Current 4 Ampere for Mesh 1
I_angle_1 =  0.0000;                     // I_angle_1  =  Angle of Source Current 0 Degree for Mesh 1
I_mag_2   = 10.5800;                     // I_mag_2    =  Magnitude of Source Current 10.58 Ampere for Mesh 2
I_angle_2 = 79.1000;                     // I_angle_2  =  Angle of Source Current 79.10 Degree for Mesh 2
V_mag_v   = 60.0000;                     // V_mag_v    =  Magnitude of Source Voltage 60 Volt
V_angle_v = 30.0000;                     // V_angle_v  =  Angle of Source Voltage 30 Degree 
V_mag_i   = 184.9840;                    // V_mag_i    =  Magnitude of Source Voltage 184.9840 Volt
V_angle_i =   6.2100;                    // V_angle_i  =  Angle of Source Current 6.2100 Degree 
R2        =  20.0000;                    // R2         =  Resistance of Resistor 20 Ohm
XL        =  10.0000;                    // XL         =  Reactance of Inductor 10 Ohm
XC        =   5.0000;                    // XC         =  Reactance of Capasitor 5 Ohm
//
// Calculations Average Power Generated by The Source Voltage
P_5   = 0.5000 * V_mag_v* I_mag_2 * cosd(V_angle_v - I_angle_2);          // P_5  = Average Power Generated by The Source Voltage
// Calculations Average Power Generated by The Source Current 
P_1   = -0.5000 * V_mag_i* I_mag_1 * cosd(V_angle_i - I_angle_1);         // P_1  = Average Power Generated by The Source Current
// Calculations Power Absorbed by Resistor 
V_20   = R2 * I_mag_1;                                                    // V_20     = Voltage of Resistor 20 Ohm 
P_2 = 0.5000 * V_20 * I_mag_1;                                            // P_2   = Power Absorbed by Resistor 20 Ohm
// Calculations Power Absorbed by Inductor 
I_mag_1_real  = I_mag_1*cosd(I_angle_1);                                  // I_mag_1_real     = Real Part of Current for Mesh 1
I_mag_1_imag  = I_mag_1*sind(I_angle_1);                                  // I_mag_1_imag     = Imaginary Part of Current for Mesh 1
I_mag_2_real  = I_mag_2*cosd(I_angle_2);                                  // I_mag_2_real     = Real Part of Current for Mesh 2
I_mag_2_imag  = I_mag_2*sind(I_angle_2);                                  // I_mag_2_imag     = Imaginary Part of Current for Mesh 2
I_L_10_mag_real  = I_mag_1_real - I_mag_2_real;                           // I_L_10_mag_real  = Real Part of Current Through Inductor
I_L_10_mag_imag  = I_mag_1_imag - I_mag_2_imag;                           // I_L_10_mag_imag  = Imaginary Part of Current Through Inductor 
I_L_10_mag = norm(complex(I_L_10_mag_real,I_L_10_mag_imag));              // V_L_10_mag       = Magnitude of Current Through Inductor
V_L_10_mag   = XL*I_L_10_mag;                                             // P_3              = Power Absorbed by Inductor  
P_3 = 0.5000 * V_L_10_mag * I_L_10_mag * cosd(90.0000)      
// Calculations Power Absorbed by Capasitor 
V_C_5_mag = norm(complex(I_mag_2_real,I_mag_2_imag))*XC;                  // V_C_5_mag       = Magnitude of Current Through Capasitor
P_4 = 0.5000 *V_C_5_mag*norm(complex(I_mag_2_real,I_mag_2_imag))*cosd(90.0000); // P_4       = Power Absorbed by Capasitor
//
disp("Example 11-4 Solution : ");
printf(" \n P_1  = Average Power Generated by Source Current   = %.3f Watt",P_1)
printf(" \n P_2  = Average Power Absorbed  by Resistor 20 Ohm  = %.3f Watt",P_2)
printf(" \n P_3  = Average Power Absorbed  by Inductor 10 Ohm  = %.3f Watt",P_3)
printf(" \n P_4  = Average Power Absorbed  by Capasitor 5 Ohm  = %.3f Watt",P_4)
printf(" \n P_5  = Average Power Generated by Source Voltage   = %.3f Watt",P_5)