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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 38
disp("CHAPTER 1");
disp("EXAMPLE 38");
//VARIABLE INITIALIZATION
I=2; //current source in Amperes
r1=2; //in Ohms
r2=1; //in Ohms
r3=1; //in Ohms
r4=2; //in Ohms
//SOLUTION
//Thevenin Equivalent circuit
I1=1; //since there is equal resistance of 3Ω, hence, current=1A
vth=(I1*r2)+(-I1*r4);
req1=r1+r2;
req2=r3+r4;
rth=(req1*req2)/(req1+req2);
disp("THEVENIN EQUIVALENT CIRCUIT IS-");
disp(sprintf(" Thevenin voltage= %d V",vth));
disp(sprintf(" Thevenin resistance= %f Ω",rth));
//Norton Equivalent circuit
v1=I/((1/r2)+(1/r4));
v2=-I/((1/r3)+(1/r1));
req1=r1+r2;
req2=r3+r4;
rn=(req1*req2)/(req1+req2);
Isc=(v1/r4)+v2;
disp("NORTON EQUIVALENT CIRCUIT IS-");
disp(sprintf(" Norton current= %f A",Isc));
disp(sprintf(" Norton resistance= %f Ω",rn));
//END
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