Question

For two resistors R₁ and R₂, connected in parallel, the relative error in their equivalent resistance is (where R₁ = (10.0 ± 0.1) Ω and R₂ = (20.0 ± 0.4) Ω) 

(a) 0.08 

(b) 0.05 

(c) 0.01 

(d) 0.04

Answer 1

Correct option is (c) 0.01

Given,

\(R_1 = (10.0 \pm0.1)\Omega\)

\(R_2 = (20.0 \pm 0.4 )\Omega\)

\(\frac1{R_eq} = \frac1{R_1} + \frac1{R_2}\)    .....(1)

Differentiate equation (1)

\(+ \frac1{{R_e}^2} \Delta R_{eq} = + \frac1{{R_1}^2} \Delta R_1 + \frac1{{R_2}^2} \Delta R_2\)   ....(2)

Both side multiply Re, so

\(\frac{\Delta R_{eq}}{R_{eq}} = R_{eg} \left[\frac{\Delta R_1}{{R_1}}^2 + \frac{\Delta R_2}{{R_2}^2}\right]\)    .....(3)

 ∵ \(R_{eq} = \frac{R_1 + R_2}{R_1 + R_2}= \frac{10 \times 20}{30} = \frac{200}{30}\)

\(R_{eq} = \frac{20}3\)

Put all value in equation (3)

\(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 \left[\frac{0.1}{(10)^2} + \frac{0.4}{(20)^2}\right]\)

\(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 \left[\frac{0.1}{100} + \frac{0.4}{400}\right]\)

\(\frac{\Delta R_{eq}}{R_{eq}} = \frac{20}3 [0.001 + 0.001]\)

\(\frac{\Delta R_{eq}}{R_{eq}} = 0.01\)