Question

Two resistance R1 and R2 have effective resistance Rs when connected in series and Rp , when connected in parallal . If Rs Rp =16 and R1/R2 =4,calculate the value ofR1 and R2 (in units of resistance)

Answer 1

Acc to me,

Formula for parallel

1/Rp=1/R1+1/R2

1/Rp=(R1+R2)/R1R2

Rp=R1*R2/R1+R2

Formula for series

Rs=R1+R2

given

RsRp=16

put Rs and Rp from above

(R1+R2)*R1*R2/R1+R2=16

R1+R2 Will cancel

R1*R2=16-----EQN1

R1/R2=4-----EQN2

R1=4R2

Put R1 in eqn 1

4R2*R2=16

R2*R2=4

R2=Root4

R2=2 ohm

Put R2 in eqn 2

R1/R2=4

R1/2=4

R1=8 ohm

I hope it help

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