[Drazzminius]

I'm trying to get a calculation to work for a nested loop, but I'm having a problem with the initial calculation.

The question reads:

Quote

Write a program to compute the power loss in a transmission line with a
resistance of 0.05 ohms/mile. Compute the power loss if 500 kW of power is
transmitted from a power generating station to cities at distances of 20, 30,
50, 50, ..., 100 miles at 100V and 200V.

power transmitted in watts
i = ------------------------------
volts transmitted

The total resistance R is computed from the equation:

R = r x miles

where r is the resistance per mile.

The power loss is computed from the equation:

Power loss = i^2 x R.

Here is my code:

#include <stdio.h>
#include <iostream>


int main(void) 
{

int v=1;
int dist=1;
int i = (500 / 100);
float r = 0.05;
int val = 20;
float R = 0.05 * val;
float pwr_loss = (val * r) * (i * i);

  
	printf("\t  Voltage  \t  Distance  \t  Power Loss\n");
  	printf("\t=============================================\n");
 
for (dist=1; dist<=9; dist++) 

{
    printf("\t %7d \t %8d  \t %8d\n", v*100, val, pwr_loss); 
    val = 10 + val;
	pwr_loss = (0.05 * val)*(i*i);	
  }
	printf("\n\n");
	system("PAUSE");
  return 0;

}

I don't want exact answers necessarily, I'm trying to figure out why I can't get the code to run. I keep getting zero's for the power loss.

Any nudges in the right direction are appreciated.

[jjl]

The error resides in your print statement, make sure you use the correct format when printing floating point values (%f)

printf("\t %7d \t %8d  \t %8f\n", v*100, val, pwr_loss); //changed %8d to %8f

[Drazzminius]

Awesome! Thank you for the quick reply!!!

[mtrouard29]

how do you add the nested loop for voltage of 200?

[Adak]

It wouldn't be a nested loop, it would be a second for loop - very similar to the first one, but using the higher voltage, and using changed variables (like the ones listed above the for loop, here). Change them appropriately for the higher voltage, etc., and then add the second for loop for these new equations.