Question

The program has to be written in C

OPTIONAL TAKE-HOME QUIZ WORTH 10 POINTS OUT OF A GRAND TOTAL OF 110 FOR THE COURSE A DC electrical circuit consists of an ideal battery with user-suppled terminal voltage VB which drives a resistive load comprised of a user-supplied number m of branches connected in parallel across each other, with each branch consisting of a variable user-supplied number n of resistors and user-supplied values of the resistors. NOTE: the number of resistors in a branch and their values, which are user supplied, vary in general from branch to branch. of call by reference which, when Design a program using either the method of call by value or the method run, will do the following in the indicated sequence: Ask the user to type on the keyboard the value of the battery voltage VB Ask the user to type on the keyboard the value of the number m of the resistive branches connected in parallel. Ask the user to type on the keyboard the number n of resistors in branch O followed by the values of these resistances expressed as a one-dimensional array. The program will then, using a prototype function called "series", compute the equivalent resistance to all.of the resistors connected in series in branch 0 Ask the user to repeat step 3 for branch 1, then branch 2, and so on until the last branch m-1 is reached Having reduced each branch of resistors to a single equivalent resistance, the program will next compute the equivalent resistance req to the m parallel-connected single resistances computed in steps 3 and 4 using a prototype function called "parallel The program will, using a prototype function called SumentaNpowe supplied by the battery, a one-dimensional array whose elements are the currents Io, I, I-i in the m branches, and a one-dimensional array whose elements are the powers dissipated in the m resistive branche:s The program will finally compute the power supplied by the battery to the resistive load. 1. 2. 3. 4. 5. 6. next compute the power PB 7. NOTE: For sample output, use the following values for the circuit elements: m-3 (three resistor branches) For the m-0 branch, n-4, with r10-10.0, r[1]-20.0, r[2]-30.0 and r[3]#40.0 For the m-1 branch, n-3, with rfo]-50.0, r[1]-40.0, r[2-10.0 For the m-2 branch, n-2, with o]-20.0 and 1]-30.0 Vo=10V All resistor values in ohms The output screen on the console corresponding to the above inputted quantities is attached for your convenience

The expected output:

Please make sure to use two-dimensional arrays.

Answer 1

#include <stdio.h>
int main(int argc, char const *argv[])
{
int branch,resistor;
double VB, r[50][50], req[50]={0}, total_eq=0,eq;
printf("Please enter value for battery voltage VB in volts: ");
scanf("%lf",&VB);
printf("Please enter value for number m of branches: ");
scanf("%d",&branch);
printf("\n");
for (int i = 0; i < branch; ++i)
{
printf("Please enter the number of resistors n in brach %d : ", i);
scanf("%d",&resistor);
for (int j = 0; j < resistor ; ++j)
{
printf("Please enter a value of resistor r[%d][%d] in ohms: ",i,j);
scanf("%lf", &(r[i][j]));
req[i]+=r[i][j];
printf("Equivalent resistance #%d in branch #%d is r[%d] = ", i,j,j);
printf("%lf ohms \n", r[i][j]);
}
printf("Result: Equiv. resistance to resistors in brach #%d is =%f ohms \n \n", i, req[i]);
total_eq+=req[i];
}
eq=total_eq/VB;
printf("\n Equiv. resistance seen by battery req= %lf ohms \n", eq);
printf("Current & power supplied by battery are: IB = %lf A, PB= %lf W\n", VB/eq , (VB/eq)*VB);
for (int i = 0; i < branch; ++i)
printf("Current in and power dissipated in branch %d are I[%d]=%lf A, P[%d]= %lf W \n", i, i, VB/req[i], (VB/req[i])*VB);
return 0;
}

Output:

dps@machine:~/Downloads$ gcc main.c-o kewl dps@machine:~/Downloads$./kewl Please enter value for battery voltage VB in volts: 10 Please enter value for number m of branches 3 Please enter the number of resistors n in brach 0 : 4 Please enter a value of resistor r[0][0] in ohms: 10 Equivalent resistance #Θ in branch #Θ is r[0] = 10.00ΘΘ00 ohms Please enter a value of resistor r[0][1] in ohms: 2Θ Equivalent resistance #θ in branch #1 is r[1] = 2θ.θθθθθθ ohms Please enter a value of resistor r[0] [2] in ohms: 3Θ Equivalent resistance #θ in branch #2 is r[2] -38.θθθθθθ ohms Please enter a value of resistor r[0][3] in ohms: 4Θ Equivalent resistance #0 in branch #3 is r[3] -40.000000 ohms Result: Equiv. resistance to resistors in brach #0 1s =100.000000 ohms Please enter the number of resistors n in brach 13 Please enter a value of resistor r[1] [0] in ohms: 5Θ Equivalent resistance #1 in branch #θ is r[0] = 5θ.θθθθθθ ohms Please enter a value of resistor r[1][1] in ohms: 40 Equivalent resistance #1 in branch #1 is r[1] -48.θθθθθθ ohms Please enter a value of resistor r[1] [2] in ohms: 10 Equivalent resistance #1 in branch #2 is r[2] = 10.θθθθθθ ohms Result: Equiv. resistance to resistors in brach #1 is =100.000000 ohms Please enter the number of resistors n in brach 22 Please enter a value of resistor r[2] [0] in ohms: 2Θ Equivalent resistance #2 in branch #Θ is r[0] = 20.00ΘΘ00 ohms Please enter a value of resistor r[2] [1] in ohms: 3Θ Equivalent resistance #2 in branch #1 is r[1] -38.θθθθθθ ohms Result: Equiv. resistance to resistors in brach #2 is -50.000000 ohms Equiv. resistance seen by battery req 25.000000 ohms Current & powe r supplied by battery are : IB = θ.4ΘΘΘΘΘ A, PB= 4. OΘΘΘΘΘ w Current in and power dissipated in branch θ are 1[0]-0.100000 A, P[0]= 1.000000 Current in and powe r dissipated in branch 1 are 1[1]-0.100000 A, P[1]= 1.000000 Current in and powe r dissipated in branch 2 are 1[2]-0.290000 A, P[2]= 2.000000