If two resistors are connected in series, the total resistance is 45 Ω and if the same resistors are connected in parallel, the total resistance becomes 10 Ω . Find the values of individual resistors. Solution: Data : Rs = 45Ω , Rp = 10Ω , R1 = ? , R2 = ?

In Series combination: R_s = R₁ + R₂ = 45Ω  ........ eq. no. (1)

∴ R₂ = 45 – R₁ ......... eq. no. (2)

In Parallel combination : 1/R_p = 1/R₁ + 1/R₂ = 1/10Ω

∴ (R₁ + R₂)/R₁R₂ = 1/10Ω

∴ R₁R₂/(R₁ + R₂) = 10Ω

∴ R₁ (45 – R₁) / 45 = 10   [From eq. no. (1) & (2) ]

∴ 45R₁ – R₁² = 450

∴ R₁² – 45R₁  + 450 = 0

∴ R₁² – 30R₁ – 15R₁ + 450 = 0

∴ R₁ (R₁ – 30) – 15 (R₁ – 30) = 0

∴ (R₁ – 30) (R₁ – 15) = 0

∴ R₁ – 30 = 0    OR   R₁ – 15 = 0

∴ R₁ = 30   OR   R₁ = 15

∴ The values of individual resistors are 30Ω and 15Ω