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When resistors 1 and 2 are connected in series, the equivalent resistance is 15.0 Ω. When they are connected in parallel, the equivalent resistance is 2.67 Ω. What are (a) the smaller resistance and (b) the larger resistance of these two resistors

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MathPhys

Answer:

a) 3.5 Ω

b) 11.5 Ω

Explanation:

In series:

R₁ + R₂ = 15.0

In parallel:

1/R₁ + 1/R₂ = 1/2.67

Multiply both sides by R₁ R₂:

R₂ + R₁ = R₁ R₂ / 2.67

Solve the system of equations with substitution:

15.0 = R₁ (15.0 − R₁) / 2.67

40.1 = 15.0 R₁ − R₁²

R₁² − 15.0 R₁ + 40.1 = 0

R₁ = [ 15 ± √(225 − 4(1)(40.1)) ] / 2

R₁ = 11.5 Ω

R₂ = 15.0 − R₁

R₂ = 3.5 Ω

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