Question

Two resistors when connected in parallel give resultant value of $2\Omega$, when connected in series the value becomes $8\Omega$ , the resistance of each resistor is

• A. 5Ω, 3Ω
• B. 2Ω, 6Ω
• C. 4Ω, 4Ω
• D. 7Ω, 1Ω

Answer 1

The correct option is C

4Ω, 4Ω

Let the resistances of the two resistors be: R₁ and R_2.
In parallel combination:
$\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{2}$
Or $\frac{R_1\times R_2}{R_1+R_2}=2$
In series combination, $R_1+R_2=8\Omega$
Substituting $R_1+R_2=8\Omega$ in the above equation.
16 = R₁. R₂
16 = (8 - R₂) R₂
R₂² - 8 R₂ + 16 = 0
(R₂ - 4) (R₂ - 4) = 0
R₂ = 4Ω , R₁ = 4 Ω