Question

1. A typical electrical engineering design problem might involve determining the proper resistor to dissipate energy at a specified rate, with known values for L and C. For this problem, assume the charge must be dissipated to 1 percent of its original value (that is q/o 0.01) in t-0.05 sec, with L = 5 H and C = 10-4 F q-@oe-畳coes|t临-(5) Write a MATLAB code for Method of False-Position to approximate R in the interval [0,400) accurate to within 10-6

Answer 1

CODE

% Method of false position

% Given constants
L = 5;
C = 10^(-4);
t = 0.05;
qq0 = 100; % q0/q
f=@(R) qq0 * exp(-R*t/2/L) * cos( t * ( (1 / (L*C)) - (R /2/L )^2)^(0.5)) - 1 ;
err=1; %initial large error
tu = 400; %upper bound
tl = 0; %lower bound
fu = f(tu); %initial values
fl = f(tl);
i=0; % for counting iterations
while err > 10^(-5)
  
tr = tl -fl*(tu - tl)/(fu-fl); % false method formulation
  
fr = f(tr);
if sign(fr) == sign(fu) %checking the change of signs
tu=tr;
fu=fr;
else
tl= tr;
fl=fr;
end
err = abs(fr);
i=i+1;
end
fprintf('R of the equation by False-Position Method is found in %d iterations is %4.6f.\n',i,tr);

SCREENSHOTS

petrochemicalprocessing.m XI regressionandarea.m ×| Incremental-search, method.m X| NUMAL.m X| falseposition.m* ×1+ % Method of false position Given constants L=5; С 10^ (-4) ; t-0.05 4- f-e (R) qq0 * exp (-R*t/2/L) * err=1; %initial large error tu- 400; %upper bound tl 0; %lower bound fu= f(tu); %initial values fl f (t1) ; (1 / (L*C)) (R /2/L )^2 ) ^ (0.5)) cos ( * ( t - - 10 12 13 14 % for counting iterations while err > 10A (-5) 16 t1) / (fu-fl) ; tl -fl"(tu % false method formulation tr - - 18 19 20 21 22- fr = f (tr) ; if sign (fr) sign (fu) %checking the change of signs tu-tr fu-fr; else 24 25 t1= tr; f1=fr ; end 27 28 29 - 30 31 abs (fr) ; err = end %4.6f.\n',1,tr); fprintf ('R of the equation by False-Position Method is found in %d iterations is

Command Window >>falseposition R of the equation by False-Position Method is found in 19 iterations is 328.151494. fx. >>

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