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Question
Three resistances, of which two are equal when connected in series have an effective resistance of 30 ohms. When the three resistances are connected in parallel, the effective resistances is 3 ohms. Find the values of the individual resistances.
Solution
When connected in series, R = R1 + R1 + R2 (two are equal)
R = 2R1 + R2
∴ 2R1 + R2 = 30 ....(R = 30 Ω)
∴ 2R1 = 30 - R2
∴ R1 = `(30 - "R")/2` .....(i)
When connected in parallel,
`1/"R" = 1/"R"_1 + 1/"R"_1 + 1/"R"_2`
Or `1/"R" = 1/((30 - "R"_2)/2) + 1/((30 - "R"_2)/2) + 1/"R"_2` ....(From equation (i))
∴ `1/"R" = 2/(30 - "R"_2) + 2/(30 - "R"_2) + 1/"R"_2`
`1/3 = 4/(30 - "R"_2) + 1/"R"_2` ...(as R = 3Ω)
∴ `1/3 = (4"R"_2 + 30 - "R"_2)/("R"_2
(30 - "R"_2)) = (3"R"_2 + 30)/("R"_2(30 - "R"_2))`
∴ R2 `(30 - "R"_2)/3 = 9 "R"_2 + 90`
`30"R"_2 - "R"_2^2 = 9"R"_2 + 90`
`"R"_2^2 - 21 "R"_2 + 90 = 0`
∴ R2 - 6R2 - 15R2 + 90 = 0
∴ R2 = 6 or 15
∴ R1 = `(30 - 6)/2` or `(30 - 15)/2`
R1 = 12 Ω or 7.5 Ω
The three resistances can have values 12 Ω, 12 Ω and 6 Ω or 15 Ω, 7.5 Ω and 7.5 Ω.
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