Question

A series connection of two resistors results in an equivalent resistance of 20 Ω. When they are connected in parallel, the effective resistance is 4.8 Ω. Calculate the value of individual resistances.

Answer 1

Given Data:

Equivalent resistance in series (R_s) = 20 Ω

Effective resistance in parallel (R_p) = 4.8 Ω

Let the two individual resistors be R₁ and R₂.

From the series combination,

R₁ + R₂ = 20 Ω

From the parallel combination,

`(R_1R_2)/(R_1 + R_2) = 4.8`

`(R_1R_2)/20 = 4.8`

R₁R₂ = 4.8 × 20

R₁R₂ = 96

Form the quadratic equation:

x² − 20x + 96 = 0

(x − 12)(x − 8) = 0

R₁ = 12 Ω, R₂ = 8 Ω

The values of the two resistances are 12 Ω and 8 Ω.