Question

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answer 1

Answer:

Δ$R_{e}$ = 84   Ω,     $R_{e}$ = (40 ± 8) 10¹   Ω

Explanation:

The formula for parallel equivalent resistance is

          1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

           R₁ = 500 ± 50  Ω

          R₂ = 2000 ± 5%

This percentage equals

        0.05 = ΔR₂ / R₂

        ΔR₂ = 0.05 R₂

        ΔR₂ = 0.05 2000 = 100   Ω

We write the resistance

        R₂ = 2000 ± 100    Ω

We apply the initial formula

        1 / $R_{e}$ = 1 / R₁ + 1 / R₂

        1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

        $R_{e}$  = 400    Ω

Let's look for the error  (uncertainly) of Re

      $R_{e}$ = R₁R₂ / (R₁ + R₂)

       R’= R₁ + R₂

       $R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

      Δ$R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

      ΔR’= ΔR₁ + ΔR₂

We substitute the values

     Δ$R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

     Δ$R_{e}$ / 400 = 0.1 + 0.05 + 0.06

     Δ$R_{e}$ = 0.21 400

     Δ$R_{e}$ = 84   Ω

Let's write the resistance value with the correct significant figures

    $R_{e}$ = (40 ± 8) 10¹   Ω