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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 37
disp("CHAPTER 1");
disp("EXAMPLE 37");
//VARIABLE INITIALIZATION
v1=90; //voltage source in Volts
r1=8; //in Ohms
r2=6; //in Ohms
r3=5; //in Ohms
r4=4; //in Ohms
r5=8; //diagonal resistance in Ohms
r6=8; //in Ohms
//SOLUTION
//using Thevenin's Theorem
//(3)v1+(-2)v2=90...........eq (1)
//(-2)v1+(4)v2=-90..........eq (2)
A=[3 -2;-2 4];
b=[90;-90];
x=inv(A)*b;
v1=x(1,:);
v2=x(2,:);
vth=v1;
req1=(r1*r5)/(r1+r5);
req2=req1+r4;
req3=(req2*r6)/(req2+r6);
rth=req3+r2;
vab1=(vth*r3)/(rth+r3);
disp(sprintf("By Thevenin Theorem, the value of V_ab is %f V",vab1));
//using Norton's Theorem
//(13)v1+(-7)v2=270.........eq (1)
//(7)v1+(-13)v2=0...........eq (2)
A=[13 -7;7 -13];
b=[270;0];
x=inv(A)*b;
v1=x(1,:);
v2=x(2,:);
req1=(r1*r5)/(r1+r5);
req2=req1+r4;
req3=(req2*r6)/(req2+r6);
rn=req3+r2;
if(v1>v2) then
In=(v1-v2)/r2;
else
In=(v2-v1)/r2;
end;
vab2=(r3*In)*(rn/(rth+r3));
disp(sprintf("By Norton Theorem, the value of V_ab is %f V",vab2));
//END
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