Question

Three resistances of which two are equal when connected in a series have an effective resistance of 30 ohm when these three resistance are connected in parallel the effective resistance is 3 ohm find the value of individual resistance

Answer 1

Answer:

When connected in series, R = R1 + R1 + R2 (two are equal)

R = 2R1 + R2

∴ 2R1 + R2 = 30 (R = 30Ω)

∴ 2R1 = 30 – R2

∴ R1 = 30 – R/2 …(i)

When connected in parallel, 1/R = 1/R1 + 1/R1 + 1/R2

Or 1/R = 1/(30 – R2)/2 + 1/(30 – R2)/2 + 1/R2 [from equation (i)]

∴ 1/R = 2/30 – R2 + 2/30 – R2 + 1/R2

1/3 = 4/(30 – R2) + 1/R2

∴ 1/3 = 4R2 + 30 – R2/R2(30 – R2) = 3R2 + 30/R2(30 – R2)

∴ R2(30 – R2)/3 = 9R2 + 90

30R2 – R22 = 9R2 + 90

R22 – 21R2 + 90 = 0

∴ R2 – 21R2 + 90 = 0

R2 – 6R2 – 15R2 + 90 = 0

∴ R2 = 6 or 15

∴ R1 = 30 – 6/2 or 30 – 15/2

R1 = 12Ω or 7.5Ω

The three resistance can have values 12Ω, 12Ω and 6Ω or 15Ω, 7.5Ω and 7.5Ω