Answer:
(a) the inductance of each conductor due to internal flux linkages only is 0.05 mH/km per conductor
(b) the inductance of each conductor due to both internal and external flux linkage is 0.8894 mH/km
(c) the total inductance of the line is 1.78 mH/km per circuit
Explanation:
(a) the inductance of each conductor due to internal flux linkages only
per conductor.
(b) the inductance of each conductor due to both internal and external flux linkage is given by
but
where D is the spacing between conductors = 0.5 m
d = diameter=1.5 cm= 0.015 m
per conductor
(c) the total inductance of the line L
L≅ 1.780 mH/km per circuit
Answer:
import java.util.Scanner;
public class CountByAnything
{
public static void main (String args[])
{
int beginning = 5;
int ending = 200;
Scanner s = new Scanner(System.in);
System.out.println("Enter the value to count by:");
int valueToCount = s.nextInt();
int counter = 0;
for(int i = beginning; i <= ending; i += valueToCount)
{
System.out.print(i + " ");
if((counter+1) % 10 == 0)
System.out.println();
counter++;
}
}
}
Explanation:
- Initialize the beginning and ending.
- Loop through the value of ending variable.
- Check if numbers per line is equal to ten and Increment the counter
.