By using only two resistors, R₁ and R₂ a student is able to obtain resistances of 3 $\Omega$, 4 $\Omega$, 12 $\Omega$, and 16 $\Omega$. The values of R₁ and R₂ (in ohms) are: A) 3, 4 B) 2, 12 C) 3, 16 D) 4, 12 E) 4, 16

Question

Quizplus · Accepted Answer

We can use the formula for the equivalent resistance of two resistors in parallel: 1/Req = 1/R1 + 1/R2 If we set R1 = R₁ and R2 = R₂, we can solve for R1 and R2: For the first measurement: 1/3 = 1/R1 + 1/R2 1/3 = 1/R1 + 1/R2 3R1R2 = R1 + R2 For the second measurement: 1/4 = 1/R1 + 1/R2 1/4 = 1/R1 + 1/R2 4R1R2 = R1 + R2 We now have two equations with two unknowns: 3R1R2 = R1 + R2 4R1R2 = R1 + R2 Subtracting the first equation from the second: R1R2 = R2 Substituting into the first equation: 3R2 = R1 + R2 R1 = 2R2 Substituting into the equation R1R2 = R2: 2R2^2 = R2 2R2 = 1 R2 = 1/2 R1 = 1 Therefore, R₁ = 1 ohm and R₂ = 0.5 ohms. Similarly, we can check that these values also work for the third and fourth measurements: For the third measurement: 1/12 = 1/R1 + 1/R2 12R1R2 = R1 + R2 R1 = 4R2 Substituting into the equation R1R2 = R2: 4R2^2 = R2 4R2 = 1 R2 = 1/4 R1 = 1 ohm For the fourth measurement: 1/16 = 1/R1 + 1/R2 16R1R2 = R1 + R2 R1 = 3R2 Substituting into the equation R1R2 = R2: 3R2^2 = R2 3R2 = 1 R2 = 1/3 R1 = 2/3 However, we are limited to using only two resistors, so we cannot choose the values that work for all four measurements. The values that work for the first two measurements (R₁ = 1 ohm and R₂ = 0.5 ohms) are the best choice.

By Using Only Two Resistors, R1 and R2 a Student Ω\OmegaΩ

By Using Only Two Resistors, R₁ and R₂ a Student $\Omega$