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.

Answer 1

Resultant resistance of parallel combination R_P = 2 Ω
Resultant resistance of series combination R_S = 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)
R_S = 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.