Two resistances are measured in ohm and are given as: R _ 1 = 3Ω ± 1% R _ 2 = 6Ω ± 2% When they are connected in parallel, the percentage error in equivalent resistance is

Correct answer: D. 1.33% (1)/(Req) = (1)/(R1) + (1)/(R2) d R _ eq R2 = d R _ 1 R _ 1 ^ 2 + d R _ 2 R _ 2 ^ 2 ⇒ d R _ eq Req = (d R _ 1 R _ 1 ^ 2 + d R _ 2 R _ 2 ^ 2 ) × R _ eq ⇒ ((1)/(3) + (2)/(6)) × 2 ⇒ ∴ % error = (4)/(3)%

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Physics
Chapter: Units and Dimensions

NTA Abhyas 2020 · Physics · Units and Dimensions

Two resistances are measured in ohm and are given as: R _ 1 = 3Ω ±...

Two resistances are measured in ohm and are given as:

{R}_{1} = 3Ω \pm 1%

{R}_{2} = 6Ω \pm 2%

When they are connected in parallel, the percentage error in equivalent resistance is

Options

A $3%$
B $4.5%$
C $0.67%$
D $1.33%$

Correct answer

D. $1.33%$

Step-by-step solution

\dfrac{1}{Req} = \dfrac{1}{R1} + \dfrac{1}{R2}

\dfrac{d{R}_{eq}}{R2} = d{R}_{1}{R}_{1}^{2} + d{R}_{2}{R}_{2}^{2} ⇒ \dfrac{d{R}_{eq}}{Req} = \left(d{R}_{1}{R}_{1}^{2} + d{R}_{2}{R}_{2}^{2}\right) \times {R}_{eq}

⇒ \left(\dfrac{1}{3} + \dfrac{2}{6}\right) \times 2

⇒ ∴ %

error

= \dfrac{4}{3}%

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