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  1. #include <stdio.h>
  2.  
  3. int main(void) {
  4. 	// your code goes here
  5. 	#include <stdio.h>
  6. #include <math.h>
  7.  
  8. int main(void) {
  9.     double E = 5, R = 1000, C = 0.1e-6, t, I;
  10.  
  11.     printf("t\tI\n");
  12.     for (t = 0.0000; t < 0.001; t = t + 0.0001) {
  13.         I = E / R * exp(-t / (C * R));
  14.         printf("%f\t%f\n", t, I);
  15.     }
  16.  
  17.     return 0;
  18. }
  19.  
  20. 	return 0;
  21. }
  22.  
Success #stdin #stdout 0s 5276KB
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https://ideone.com/6XCqRP
language:
C (gcc 8.3)
created:
4 days ago
visibility:
public

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