Question

when two resistors of resistance R1 and R2 are connected in parallel the net resistance is 3 ohm when connected in series its value is 60 ohm calculate the value of R1 and R2​

Answer 1

Answer:

R₁ and R₂ are 57 Ω or 3 Ω and 3 Ω or 57 Ω respectively.

Explanation:

There are two resistors, R₁ and R₂. They are connected parallel which means equivalent resistance will be Rₚ which is equal to 3 Ω.

When they are connected in series then equivalent resistance Rₛ is equal to 60 Ω.

• 1/3 = 1/R₁ + 1/R₂
• 60 = R₁ + R₂

→ 1/3 = (R₁ + R₂)/R₁R₂

→ 1/3 = 60/R₁R₂

→ R₁R₂ = 180

→ R₂ = 180/R₁

→ 60 = R₁ + 180/R₁

→ 60 = (R₁² + 180)/R₁

→ R₁² - 60R₁ + 180 = 0

→ D = b² - 4ac

→ D = 3600 - 720

→ D = 2880

→ R₁ = (60 ± √2880)/2

→ R₁ = (60 ± 54)/2

→ R₁ = 57 Ω or 3 Ω

→ R₂ = 60 - 57

→ R₂ = 3 Ω or 57 Ω

Answer 2

Equivalent resistance in series = R1 +R2

Equivalent Resistance in parallel = 1/R1 +1/R2

Here we take √2880 as 54 for calculation convenience.

The final two resistances are thus 57 ohm and 3 ohm.