/* ovaj program racuna ekvivalentnu otpornost tri otpornika prema vrsti veze, koju bira korisnik*/
#include <stdio.h>
void main(void)
{
int izbor;
double R1, R2, R3; //deklaracija promenljivih
printf("Moguce kombinacije: \n");
printf("-------------------------------\n\n");
printf(" 1. R1-R2-R3 \n");
printf(" 2. R1||R2||R3\n");
printf(" 3. R1-(R2||R3)\n");
printf(" 4. (R1||R2)-R3\n");
printf(" 5. (R1-R2)||R3\n");
printf(" 6. R1||(R2-R3)\n\n");
printf("-------------------------------\n");
printf("Vas izbor je:");
scanf("%d", &izbor);
switch (izbor) //ubacuje se izbor i testira se svaka od mogucnosti
{
/* unos podataka za svih sest slucajeva*/
case 1: case 2: case 3: case 4: case 5: case 6:
printf(“R1=”);
scanf(“%lf”,&R1);
printf(“R2=”);
scanf(“%lf”,&R2);
printf(“R3=”);
scanf(“%lf”,&R3);
switch (izbor);
{
case 1: //ukoliko se izabere kombinacija R1+R2+R3 (1.)
Re1=R1+R2+R3;
printf("Ekvivalentna otpornost je:”);
printf(“Re=%.2f\n",Re1);
break;
case 2:
Re2=(R1*R2*R3)/(R2*R3+R1*R3+R1*R2);
printf("Ekvivalentna otpornost je:”);
printf("Re=%.2f\n",Re2);
break;
case 3:
Re3=(R1*R2+R1*R3+R2*R3)/(R2*R3);
printf("Ekvivalentna otpornost je:”);
printf("Re=%.2f\n",Re3);
break;
case 4:
Re4=(R1*R2+R1*R3+R2*R3)/(R1+R2);
printf("Ekvivalentna otpornost je:”);
printf(“Re=%.2f\n”,Re4);
break;
case 5:
Re5=((R1+R2)*R3)/(R1+R2+R3);
printf("Ekvivalentna otpornost je:”);
printf(“Re=%.2f\n”,Re5);
break;
case 6:
Re6=(R1*(R2+R3))/(R1+R2+R3);
printf("Ekvivalentna otpornost je:”);
printf(“Re=%.2f\n”,Re6);
break;
}
}
}