Question

The equivalent resistance of a series and a parallel connection of two resistance are given by the

formula :

[10]

R1 = r1 + r2 (series)

R2 = r1 x r2

r1 + r2

(parallel)

Write a program to input the value of r1& r2.



Guys please answer to this question for God sake

Don't give rubbish answer to me ​

Answer 1

Explanation:

When resistances are connected in parallel, the mathematical formula for calculating it is given by,

1/Rp=[(1/R1)+(1/R2)],

which is also written as,

Rp=(R1*R2)/(R1+R2)

irrespective of whether which is smaller.

But if thought logically,

when two resistances are connected in parallel and one of them is smaller(R1) and one is larger(R2), then the equivalent resistance of the parallel combination will always be smaller than the smallest resistance(R1).

For example,

R1=1k and R2=2k,

then the equivalent resistance when connected in parallel will be 0.667k

which is less than R1.

In other case if R1=1k and R2=10k.

the equivalent resistance when connected in parallel will be 0.909

which is slightly less than R1.

Hope it was helpful!

Answer 2

import java.util.*;

public class Resistances

{

public static void main (String args [])

{

Scanner in = new Scanner (System.in);

int n;

double r1, r2, rs, rp;

System.out.println(" enter 1 for series resistance, enter 2 for parallel resistance");

System.out.println("enter your choice:");

n = in.nextInt();

switch (n)

{

case 1:

System.out.println(" enter the value of r1 and r2:");

r1 = in.nextInt();

r2 = in.nextInt();

rs = r1+r2;

System.out.println("Resistance in series =" +rs);

break;

case 2:

System.out.println(" enter the value of r1 and r2:");

r1 = in.nextInt();

r2 = in.nextInt();

rp = (r1*r2)/(r1+r2);

System.out.println(" Resistance in parallel =" +rp);

break;

default:

System.out.println(" Wrong Choice !!!");

}

}

}