Description
Consider the AC circuit below. We will assume that the circuit is in steady-state. Thus, the voltage at nodes 1 and 2 are given by v 1 = V S cos wt and v 2 = V Rcos ( wt + q ) where V S is the voltage of the source, w is the frequency (in radians per second), and t is time. V R is the magnitude of the voltage drop across the resistor, and q is its phase. You are to write a program to determine V R for different values of w. You will need two laws of electricity to solve this problem. The first is Ohm's Law, which states v 2 = iR where i is the current in the circuit, oriented clockwise. The second is i = C d/dt (v 1-v 2) which relates the current to the voltage on either side of the capacitor. "d/dt"indicates the derivative with respect to t.Input
The input will consist of one or more lines. The first line contains three real numbers and a non-negative integer. The real numbers are V S, R, and C, in that order. The integer, n, is the number of test cases. The following n lines of the input will have one real number per line. Each of these numbers is the angular frequency, w.Output
For each angular frequency in the input you are to output its corresponding V R on a single line. Each V R value output should be rounded to three digits after the decimal point.Sample Input
1.0 1.0 1.0 9 0.01 0.031623 0.1 0.31623 1.0 3.1623 10.0 31.623 100.0Sample Output
0.010 0.032 0.100 0.302 0.707 0.953 0.995 1.000 1.000 //这道题纯属公式题，推导出计算VR的公式即可。利用iR=CR d/dt (VScos(wt)-VRcos (wt + q ))=VRcos (wt + q ) 得到CRW(VRsin(wt+q)-VSsin(wt))=VRcos (wt + q ) .之后分别令t=0和wt+q=0产生两个式子，合并之后得VR=CRW*Vs/（sqrt(CRW*CRW)+1） #include <iostream> #include <cstdio> #include <cmath> using namespace std; int main(int argc, char const *argv[]) { int n; double a,b,c; scanf("%lf %lf %lf %d",&a,&b,&c,&n); while(n--) { double w,value; scanf("%lf",&w); value=a*b*c*w/sqrt(1+pow(b*c*w,2)); printf("%.3f\n",value); } return 0; }