A. Find vc(0-), which is the voltage across the capacitor just before the switch moves. Vs = 3.0V , R1 = 10.0\Omega , R2 = 10.0\Omega , R3 = 3.0\Omega , R4 = 2.0\Omega , C = 5.0F , \alpha = 50.0V/A .
B. Find vc(0 ), which is the voltage across the capacitor immediately after the switch moves.
C. Calculate the Thevenin equivalent values for the part of the circuit to the right of C and D (which is also the part within the box).
D. Write an expression for the capacitor voltage, v(t), for t \geq 0.
Vs = 3.0V , R1 = 10.0\Omega , R2 = 10.0\Omega , R3 = 3.0\Omega , R4 = 2.0\Omega , C = 5.0F , \alpha = 50.0V/A.
E. Assume that the capacitor breaks down when the terminal voltage reaches Vx= 50.0 V, how many seconds elapse before the capacitor breaks down.
Vs = 3.0V , R1 = 10.0\Omega , R2 = 10.0\Omega , R3 = 3.0\Omega , R4 = 2.0\Omega , C = 5.0F , \alpha = 50.0V/A.

answer:

okay

explanation:

answer: i want to be brainest and get free points

explanation:

answer:

// here is code in C++.

#include <bits/stdc++.h>

using namespace std;

//main function

int main()

{

  // variable

 int n;

 cout<<"enter a number:";

 // read the input

 cin>>n;

 // find the square root of the absolute value

 float square_root=sqrt(abs(n));

 // print the square_root

 cout<<"square root of "<<n<<" is: "<<square_root<<endl;

  return 0;

}

Explanation:

Read a number from user and assign it to "n".Then find its absolute value with abs() function.Then calculate the square root of the absolute value with sqrt() function.Print the square root of the number.

Output:

enter a number:-17

square root of -17 is: 4.12311

                                                         This is easy

                                                     sinserly Natalie 7 grader (this is not that hard)