Physics, asked by ritusmita2005, 8 months ago

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 .

Answers

Answered by hamzak4086
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

Answered by Anonymous
11

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

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