Problem 1560

I just solved a problem on TJU online judge. Just an easy problem. So here is the problem

Problem 1560 – Bode Plot

Problem

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 v₁ = V_Scos t
and v₂ = V_Rcos(t + ) where V_S is the voltage of the source, is the frequency (in radians per second), and t is time. V_R is the magnitude of the voltage drop across the resistor, and is its phase.

You are to write a program to determine V_R for different values
of . You will need two laws of electricity to solve this problem. The first
is Ohm’s Law, which states v₂ = iR where i is the current in the circuit, oriented
clockwise. The second is i = C d/dt(v₁ – v₂) which relates the current to the
voltage on either side of the capacitor. “d/dt” indicates the derivative with
respect to t.

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, .

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
100.0

Sample Output

0.010
0.032
0.100
0.302
0.707
0.953
0.995
1.000
1.000

Well, if you are an Electrical Engineer, you don’t need the second law given in the problem i = C d/dt(v₁ – v₂) to solve this problem. All you need is phasor, then you can just transform the capacitor value to a “resistor” with value and the problem will be very easy enough to solve with a simple loop to get V_R.

After solving a rather simple math equation, I get the value of V₂ is this equation:

We are asked to find V_R and not V₂, in the equation above V_R is the amplitude of V₂. So we just ignore the inverse tangent part, FYI, that part mean the phase, . So we get:

Just translate it into a C code:

#include 
#include 

int main() {
	double Vs,R,C,omega;
	int i,n;
	scanf("%lf %lf %lf %d",&Vs,&R,&C,&n);
	for (i=0;i<n;i++) {
		scanf("%lf",&omega);
		printf("%.3lf\n",Vs*R/sqrt(R*R+1/(omega*omega*C*C)));
	}
	return 0;
}

And I got accepted!

FYI, this problem only solved twice this year, by gxlmoon and me. Computer Scientist don’t know how to use phasor maybe?