Three resistors, out of which two are identical, when connected in series have an effective resistance of 30 ohms. When they are connected in parallel, the effective resistance is 3 ohms. Find the resistance of each resistors.
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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
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