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

1 R e q = 1 R 1 + 1 R 2 d R e q R 2 = d R 1 R 1 2 + d R 2 R 2 2 ⇒ d R e q R e q = d R 1 R 1 2 + d R 2 R 2 2 × R e q ⇒ 1 3 + 2 6 × 2 ⇒ ∴ % error = 4 3 %

Physics - Error Question with Solution | TestHub

PhysicsErrorMiscellaneousEasy2 minPYQ_2020

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Two resistances are measured in ohm and are given as:
$R_{1} = 3 Ω \pm 1 %$
$R_{2} = 6 Ω \pm 2 %$
When they are connected in parallel, the percentage error in equivalent resistance is

Options:

Answer:

Solution:

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{d R_{e q}}{R^{2}} = \frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}} \Rightarrow \frac{d R_{e q}}{R_{e q}} = (\frac{d R_{1}}{R_{1}^{2}} + \frac{d R_{2}}{R_{2}^{2}}) \times R_{e q}$
$\Rightarrow (\frac{1}{3} + \frac{2}{6}) \times 2$
$\Rightarrow ∴ %$ error $= \frac{4}{3} %$

Stream:NTA_ABHYASSubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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