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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS
//Example 42
disp("CHAPTER 1");
disp("EXAMPLE 42");
//VARIABLE INITIALIZATION
v=10; //in Volts
I=0.5; //in Amperes
r1=4; //top LHS resistance in Ohms
r2=2; //top RHS resistance in Ohms
r3=2; //first resistance in Ohms
r4=2; //second resistance in Ohms
//SOLUTION
//using Thevenin theorem
rth=(r1*r3)/(r1+r3);
vth=v*(r3/(r1+r3));
//solving for R directly,
R=(40-(56*I))/(24*I);
disp(sprintf("(i) By Thevenin Theorem, the value of R is %d Ω",R));
//using nodal analysis
//solving the quadratic equation formed by comparing eq(1) and eq(2)
//(3)R^2+(-3)R+(0)=0
a=3;
b=-3;
c=0;
D=(b^2)-(4*a*c); //discriminant
R1=(-b+sqrt(D))/(2*a);
R2=(-b-sqrt(D))/(2*a);
if(R1==1) then
disp(sprintf("(ii) By Nodal analysis, the value of R is %d Ω",R1));
else
disp(sprintf("(ii) By Nodal analysis, the value of R is %d Ω",R1));
end;
//END
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