我正在学习C ++,并在尝试使用公式来计算电流时遇到了这个问题。
我得到:0.628818
答案应该是:
f = 200赫兹
R = 15欧姆
C = 0.0001(100µF)
L = 0.01476(14.76mH)
E = 15伏
答案:I = 0.816918A(计算得出)
下面是我的代码:
#include <iostream>
#include <cmath>
int main()
{
const double PI = 3.14159;
double r = 15;
double f = 200;
double c = 0.0001;
double l = 0.01476;
double e = 15;
double ans = e / std::sqrt(std::pow(r, 2) + (std::pow(2 * PI*f*l - (1.0 / 2.0 * PI*f*c), 2)));
std::cout << "I = " << ans << "A" << std::endl;
}
我已经阅读了有关截断错误的信息,并尝试使用1.0 / 2.0,但似乎也不起作用。
最佳答案
截断误差是指仅使用无限序列的前N个项来估计值。因此,您的问题的答案是“否”。您可能会发现以下内容值得关注...。
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
template<typename T>
T fsqr(T x) { return x * x; }
// Numerically stable and non-blowuppy way to calculate
// sqrt(a*a+b*b)
template<typename T>
T pythag(T a, T b) {
T absA = fabs(a);
T absB = fabs(b);
if (absA > absB)
{
return absA*sqrt(1.0 + fsqr(absB / absA));
} else if (0 == absB) {
return 0;
} else {
return absB*sqrt(1.0 + fsqr(absA / absB));
}
}
int main () {
double e, r, f, l, c, ans;
const double PI = 3.14159265358972384626433832795028841971693993751058209749445923078164062862089986280348253421170;
cout << "Insert value for resistance: " << endl;
cin >> r ;
cout << "Insert value for frequency: " << endl;
cin >> f;
cout << "Insert value for capacitance: " << endl;
cin >> c;
cout << "Insert value for inductance: " << endl;
cin >> l;
cout << "Insert value for electromotive force (voltage): " << endl;
cin >> e;
ans = e / pythag(r, 2*PI*f*l - (1/(2*PI*f*c)) );
cout << "I = " << ans << "A" << endl;
system("pause");
return 0;
}
只是开个玩笑。