Question

When two resistors of resistances R1 and R2 are connected in parallel, the net resistance is 3 W . When connected in series, its value is 16 W . Calculate the values of R1 and R2 .

Answer 1

Answer:

Explanation:

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 :

Let the resistors be R and r

When the resistors are connected in series,

$\sf \: \: R + r = 16 \: \Omega - - - - - - - - -(1)$

When the resistors are connected in parallel,

$\sf \: \dfrac{1}{R} + \dfrac{1}{r} = \dfrac{1}{3} \: \Omega \: \\ \\ \longrightarrow \: \sf \: \dfrac{R + r}{Rr} = \dfrac{1}{3} \\ \\ \longrightarrow \: \sf \: Rr = 48 \Omega \: \: \: \: \: \: |from \: (1)|$

We know that,

$\sf \: (a - b) {}^{2} = (a + b) {}^{2} - 4ab$

Thus,

$\sf \: (R - r) {}^{2} = (R + r) {}^{2} - 4Rr \\ \\ \implies \sf (R - r) {}^{2} = (16) {}^{2} - 4 \times 48 \\ \\ \implies \sf \: (R - r) {}^{2} = 256 - 192 \\ \\ \implies \ \sf \: (R - r) {}^{2} = 64 \\ \\ \implies \: \sf \: R - r = 8 \Omega - - - - - - - - - (2)$

Adding equations (1) and (2),

$\longrightarrow \sf \: 2R = 24 \\ \\ \longrightarrow \: \boxed{ \sf R = 12 \: \Omega}$

Putting value of R equation in equation (2),

$\longrightarrow \sf \: 12 - r = 8 \\ \\ \longrightarrow \: \boxed{ \sf r = 4 \Omega}$

Thus,both the resistors are 12Ω and 4Ω