Question

when two resistors of resistance R1 and R2 they are connected in parallel the net resistance is 3 ohm when they are connected in series the value 16 ohm calculate R1 and R2

Answer 1

hi
I will answer this
 Resistance equivalent in parallel is given by R1 *R2 / (R1+R2)
Resistance equivalent in series is R1 + R2 
Given R1 + R2 = 16
and R eq in parallel as 3
sustitute the above values in the first equation of parallel equivalence 
we get R1*R2 = 3*16 = 48
The values of R1 and R2 are 12,4
By using basic calculations of (R1-R2)^2 = (R1 +R2)^2 - 4R1R2 
R1 - R2 = sqrt(256-192) = sqrt(64) = 8
solving R1 and R2 we get r1 =12 and r2 =4

Answer 2

Answer:

R1 = 12 ohms and R2 = 4 ohms

Explanation:

Given :- Rp = 3 ohms and RS =16 ohms

When resistors are connected in parallel:-

1/Rp= 1/R1+1/R2

1/Rp=R2+R1/R1×R2

1/Rp=R1+R2/R1×R2

Now,

By cross multiplication

Rp=R1×R2/R1+R2

R1×R2=Rp(R1+R2).....................(1)

Now, resistors are connected in series:-

RS=R1+R2

16=R1+R2......................(2)

Use equation 1:-

R1×R2=3×16=48..................(3)

Now,

By the help of basic mathematics calculation:-

(R1-R2)²=(R1+R2)²-4R1×R2

(R1-R2)²=(16)²-4×48 (from eq. 2)

(R1-R2)²=256-192

(R1-R2)²=256-192

(R1-R2)=√64

(R1-R2)=8.....................(4)

From equation 2 and 4

R1+R2=16

R1-R2=8

2R1=24

R1=12

Put the value of R1=12 ohm in equation 2.

16=R1+R2

16=12+R2

R2=16-12

R2=4 ohms

So, the value of R1=12 ohms and R2=4ohms