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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 14
clc;
disp("CHAPTER 1");
disp("EXAMPLE 14");
//VARIABLE INITIALIZATION
v=10; //voltage source in Volts
I3=-5; //current source in Amperes
r1=2; //in Ohms
r2=2; //in Ohms
r3=4; //in Ohms
//SOLUTION
//by loop analysis
//(1)I1+(-1)I2=0.........eq (1)
//(4)I1+(-2)I2=10........eq (2)
//solving the equations by matrix method
A=[1 -1;4 -2];
b=[0;10];
x=inv(A)*b;
I1=x(1,:); //to access 1st element of 2X1 matrix
I2=x(2,:); //to access 2nd element of 2X1 matrix
In=I2-I3;
rn=(r1*r2)/(r1+r2);
I=(rn*In)/(rn+r3);
disp(sprintf("By Norton Theorem, the current through resistor R3 is %d A",I));
//END
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