Using Taylor s Series method taking algorithm of order 3. solve th.pdf

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Using source combination to first simplify the circuit, determine the power dissipated by the 5.8 k Ohm resistor in the circuit shown. Express your answers to four significant figures. Solution Using Super postion theorem we can solve this problem current through 5.8Kohm due to 1mA current source alone: I5.8kohm_1mA=1mA*(4.7K/(4.7k+5.8k) =1mA*4.7/10.5 I5.8kohm_1mA =0.447mA current through 5.8Kohm due to 2mA current source alone: I5.8kohm_2mA=2mA*(4.7K/(4.7k+5.8k) =2mA*4.7/10.5 I5.8kohm_2mA=0.895mA current through 5.8Kohm I5.8kohm=0.89mA+0.447mA=1.342mA Power due to 5.8kohm resitance P5.8kohm=(1.342mA)2 *5.8kohm P5.8kohm=10.44mW.

Formation

Using Taylor ' s Series method taking algorithm of order 3. solve the initial value problem y'
=1 + y ; y(0)=0 h = 0.5 At X=1
Solution
%% % taylororder 3 rule
%%
clear all
fork=1:2
h=0.5/k ;
tmax=1;
n=tmax /h;
x(1)= 0;
y(1)= 0;
tout=x(1);
yout=y(1).';
true(1) = 0;
ero(1) =0;
trout=true(1).';
erout=ero(1);
ero(1) =0;
disp(' x actual Taylor(order3)error ')
disp('~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~')
fori=1:(n + 1 )
x(i+1) = x(1)+ i*h;
T3 = (1+y(i)+(h/2)*(1+y(i)+x(i) )+h^2*(1+x(i)+y(i)) );
y(i+1) = y(i) + h*T3;
%
true(i+1) = exp(x(i)) -1;
ero(i+1) =abs( y(i) -true(i+1) );
%
tout=[tout;x(i)];
yout=[yout,y(i).'];
trout=[trout,true(i).'];
erout=[erout,ero(i).'];
%

ifrem(i-1 ,k) ==0% x(i)<= 0.1 %rem(i-1,10) == 0
fprintf('%5.2f%12.7f%12.7f%15.7f ' ,x(i),true(i+1),y(i), ero(i+1) );
end% if
end% i

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