//Chapter 4,Ex4.1,Pg 4.8

funcprot(0)

function [polar] = rect2polar(x,y) //Function to convert rectangular coordinates to polar coordinates

polar=ones(1,2)

polar(1)=sqrt((x^2)+(y^2))

polar(2)=atan(y/x)

polar(2)=(polar(2)*180)/%pi

endfunction

clc;

//Refer to the data given in the question

Vl=400 //Supply voltage

Il=34.65 //Current in Amperes

P=14.4*10^3 //Active power in volts

Vph=Vl

Iph=Il/sqrt(3)

printf("\n Iph=%.0f A \n",Iph)

Zph=Vph/Iph

printf("\n Zph=%.0f ohms \n",Zph)

//We can calculate the power factor easily from the given data

pf=P/(sqrt(3)*400*34.65)

printf("\n Power factor=%.1f \n",pf)

phi=acos(pf)

phi=phi*180/%pi

printf("\n phi=%.2f degrees \n",phi)

Rph=Zph*pf

printf("\n Rph=%.0f ohms \n",Rph)

Xph=Zph*sqrt(1-pf^2)

printf("\n Xph=%.0f ohms \n",Xph)

Life is essentially a probability based (statistical) understanding including expectations. Randomness also may need to be included. It is a good problem, however. And don't you think both the dielectric quality and the contact surfaces may have to be taken separately and may be other quantities too and by adequate or appropriate weight-age, you may be able to arrive at a life cycle evaluation?

Life depends on number of operations, and value of fault current to handled, by breaker....