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Question
Two resistors when connected in parallel give the resultant resistance of 2 ohms, but when connected in series the effective resistance becomes 9 ohms. Calculate the value of each resistance.
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Solution
Resultant resistance of parallel combination RP = 2 Ω
Resultant resistance of series combination RS = 9 Ω
`1/"R"_p` = `1/"R"_1`+`1/"R"_2`
`1/2` = `1/"R"_1`+`1/"R"_2`
`1/2` = `("R"_1+"R"_2)/("R"_1"R"_2)`
2(R1 + R2) = R1R2 ….. (1)
RS = R1 + R2
⇒ 9 = R1 + R2
⇒ R1 = 9 – R2 ….. (2)
Substitute equation (2) in equation (1)
2(9 – R2 + R2) = (9 – R2) R2
⇒ \(18=9 \mathrm{R}_{2}-\mathrm{R}_{2}^{2}\)
⇒ \(\mathrm{R}_{2}^{2}-9 \mathrm{R}_{2}+18=0\)
⇒ (R2 – 3) (R2 – 6) = 0
⇒ R2 = 3, 6
(i) If R2 = 3; R1 = 9 – R2 = 9 – 3 = 6 Ω
(ii) If R2 = 6; R1 = 9 – R2 = 9 – 6 = 3 Ω.
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