лабораторная работа / мэмс / 6lw

6 lABORATORY WORK # 6

TOPIC: Modeling of steady state processes in nonlinear electric circuits of direct current by numerical method in MATLAB environment. Part 1.

PURPOSE OF THE WORK: Problem statement, development of the program of calculation of electrical state in the simple nonlinear circuit of a direct current by the Newton method.

Mathematical model

Figure 6.1 – Given electrical circuit

EMF – E= 10 V;

Linear resistance – R=0.5 Ohm;

The given current-voltage characteristic:

I=2.5U³

Nonlinear resistance Rn is given analytically accord to VACH:

U_n=a·I^⅓

where U_n – voltage across of nonlinear resistor, a=0.4^{}.

It is require the determination of the current through the given circuit.

Then accord to second Kirchhoff law the equation is as following:

R1·I+ U_n =E

Let’s substitute U_n into the above represented equation and the nonlinear algebraic equation concerning the current I is obtained:

R1·I+a·I^⅓=E

Accord to the Newton algorithm k+1 approach could be found as:

.

Expression for determination of f'(I) could be obtained from (6.4):

f'(I)=R1+

Determination of zero approximation

Let zero approximation of the current through the circuit is I₀=0, then at the first iteration step difference between initial and following approximation will be:

I₁ - I₀= f (I₀)/ f'(I₀)= -10/∞ =0,

i.e, accord to given approximation process is been divergent.

Let zero approximation I₀=1, then:

I₁ - I₀= f (I₀)/ f'(I₀)= -8.7632/ 0.7456=-11.7532.

Hence, I₀=1 could be taken as a zero approximation

Block-diagram

Program of calculation in MATLAB environment

%newton method

%initial data

E=10;

R=0.5;

a=0.4^(1/3);

k=1;

N=40;

I(k)=1;

U(1)=-4;

Ii(1)=U(1)^3*2.5;

for k=2:N

fun=R*I(k-1)+a*I(k-1)^(1/3)-E;

dfun=R+a*(1/3)*I(k-1)^(-2/3);

I(k)=I(k-1)-0.5*fun/dfun;

%current curve

U(k)=U(k-1)+0.2;

Ii(k)=U(k)^3*2.5;

end

%voltage curve

Iii=-160;

Uii(1)=-4;

for n=2:8*N;

Iii(n)=Iii(n-1)+1;

Uii(n)=sign(Iii(n))*(abs(Iii(n))/2.5)^(1/3);

end;

%and others

subplot(3,1,1), plot(I);

grid on; title ('Approaching of the varistor current')

xlabel('iterations');

ylabel('Current');

legend('I1',-1);

subplot(3,1,2), plot(U,Ii);

grid on;title ('Voltage')

xlabel('voltage');

ylabel('current');

legend('U',-1);

subplot(3,1,3),

plot(Iii,Uii);grid on;

title ('current')

xlabel('voltage');

ylabel('current');legend('I',-1);

Results of calculation

In the figure 6.2 graphical dependence of calculated current through the given circuit (fig. 6.1.) accord to the every iteration step is represented.

The above represented curve I(n) demonstrates that practically it is enough of 10 iteration steps for satisfactory accuracy of solution.

Figure 6.2 – Graphical dependence I(n), I(U), U(I)

Conclusion: The current via varistor was found. The graph of VAD is presented. The graph of differential as absent as there is no complicate sign in the given function.