Question

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0. If two resistors are connected in series, the equivalent resistance
becomes 16 S2. If the same are connected in parallel, the equivalent
resistance is 3 12. Find the individual resistances.​

Answer 1

Answer:

Individual resistances are 4 Ω and 12 Ω

Explanation:

Given,

Equivalent resistance joined in series is 16 ohms.

R₁ + R₂ = 16    ----- (1)

Equivalent resistance joined in parallel is 3 ohms.

(R₁ + R₂)/R₁R₂ = 3     ------ (2)

Therefore,

R₁R₂ = 3 * 16 = 48.

Convert this into a quadratic equation

Let R₁ be α & R₂ be β.

α β = 48, α + β = 16

General form of quadratic equation =  x² - x(α +β) + α β

Substituting values, we get,

=> x² - 16x + 48

=> x² - 12x- 4x + 48

=> x(x - 12) - 4(x - 12)

=> (x - 4)(x - 12) are factors.

Thus, zeroes are 4 & 12.

Thus, α = R₁ = 4 ohms, β = R₂ = 12 ohms.

Therefore,

Individual resistances are 4 Ω and 12 Ω

Hope it helps!