Why DC series motors are not started with out load

If we start DC series motor with out load the speed increases which in turn  increases back EMF(proportional to speed) but due to increase in back EMF armature current decreases. Since field winding is in series the flux also decreases due to decrease in the armature current. We know that flux is inversely proportional to speed, so decrease in flux causes increase in speed. This process continues.

Theoretically speed reaches to infinity as flux tends to zero.

Practically due to high speeds the bearings,armature shaft,etc get damaged.

So keep in mind never start a DC series motor at no load.

C program for sum of powers of numbers

#include<stdio.h>
int main()
{
int a,s;
scanf("%d",&a);
s=((6*a*a*a*a*a)+(15*a*a*a*a)+(10*a*a*a)-a)/30;
printf("%d",s);
return 0;
}

C program for sum of adjacent pairs

#include<stdio.h>
int main()
{
int a,b,s;
scanf("%d %d",&a,&b);
s=0;
while(!(b==-1))
{
s=a+b;
a=b;
scanf("%d",&b);
printf("%d",s);
if(b!=-1)
printf(" ");
if(b==-1)
printf("\n");
}
return 0;
}

C program for inverted right triangle

#include<stdio.h>
int main()
{
int n,i,j,t,m;
scanf("%d",&n);
m=n;
for(i=0;i<m;i++)
  {
  t=i;
for(j=0;j<n;j++)
  {
  i=i+1;
   if(i<=9)
     printf("%d",i);
   if(i>=10)
     printf("%d",i%10);
  }
  n=n-1;
   printf(" \n ");
  i=t;
  }
return 0;
}

C program to find second largest element

#include<stdio.h>
int main()
{
int a,b,temp,t;
scanf("%d%d",&a,&b);
if(a<b)
{
 t=a;
 a=b;
 b=t;
}
scanf("%d",&temp);
while(temp!=-1)
{
 if(temp<b)
 scanf("%d",&temp);
 else if(temp>b)
 {
 b=temp;
 scanf("%d",&temp);
 }
 if(a<b)
 {
 t=a;
 a=b;
 b=t;
 }
}
printf("%d",b);
return 0;
}

Pythagorean triples

#include<stdio.h>
int main()
{
int a,b,c;
scanf("%d %d %d",&a,&b,&c);
if((a*a)==(b*b)+(c*c))
printf("yes");
else if((b*b)==(c*c)+(a*a))
printf("yes");
else if((c*c)==(a*a)+(b*b))
printf("yes");
else
 printf("no");
return 0;
}

How three phase system reduces the copper loss

Consider a 5KW load.If a single phase system it then we require 2 wires,one carrying the current towards the load and other carrying current away from the load. Assume that the applied voltage is 100V  then the current is 30A.if we see the copper losses in it(if resistance of each wire is 1 ohm) the Total copper loss is 1800 watts (900+900).
Now consider a three phase balanced supply.It has three wires each carrying 10 amps current then calculate the losses which are less when compared to that of single phase.