Question

When two resistors are connected in series, their effective resistance is 80 Ω. When they are connected in parallel, their effective resistance is 20 Ω. What are the values of the two resistances? 

Answer 1

Let the two resistances be R₁ and R₂. Therefore,
R₁ + R₂ = 80 Ω
or, R₁ = 80 −R₂-R₂   .....(i)
and 
1/R₁+1/R₂=1/20

⇒R₁R₂=20(R₁+R₂)

As, R₁=80−R₂

⇒(80−R₂)R₂=20(80 −R₂+R₂)

80R₂−R₂²=1600

R₂²−80R₂+1600=0

(R₂−40)(R₂−40)=0

⇒R₂=40 Ω

From (i), we get R₁=80−40=40 Ω

R₁=40 Ω and R₂=40 Ω