若有三个电阻R1、R2、R3并联,则U1=U2=U3=U,每个电阻上的电流分别为:I1=U1/R1;I2=U2/R2;I3=U3/R3;流过这三个电阻的总电流I=I1+I2+I3=U1/R1+U2/R2+U3/R3
=U(1/R1+1/R2+1/R3);从三个电阻并联的端口来看,相当于产生了一个新的电阻R=U/I=1(1/R1+1/R2+1/R3);即1/R并=1/R1+1/R2+1/R3
#include <stdio.h>#include <stdlib.h>
int main()
{
float r1, r2, r3, i, u
printf("please input R1: ")
scanf("%f", &r1)
printf("please input R2: ")
scanf("%f", &r2)
printf("please input R3: ")
scanf("%f", &r3)
printf("please input voltage V: ")
scanf("%f", &u)
if (r2 == 0 || r3 == 0)
{
printf("input error, cannot calculator.\n")
system("pause")
return 0
}
i = u/(r1+1.0/(1.0/r2+1.0/r3)) // R=R1+1/(1/R2+1/R3), I=U/R
printf("Current is: %.2f", i)
system("pause")
return 0
}