When the voltage and current in a conductor are measured as 100 ± 4 V and 5 ± 0.2 A , then the percentage of error in the calculation of resistance is

Given, V ± ∆ V = 100 ± 4 V I ± ∆ I = 5 ± 0.2 A ⇒ ∆ R R = ∆ V V + ∆ I I = 4 100 + 0.2 5 = 0.08 ⇒ ∆ R R % = 0.08 × 100 = 8 %

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Physics - Unit & Dimension Question with Solution | TestHub

PhysicsUnit & DimensionMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

When the voltage and current in a conductor are measured as $(100 \pm 4) V$ and $(5 \pm 0.2) A$ , then the percentage of error in the calculation of resistance is

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Answer:

Solution:

Given, $V \pm ∆ V = (100 \pm 4) V$

$I \pm ∆ I = (5 \pm 0.2) A$

$\Rightarrow \frac{∆ R}{R} = \frac{∆ V}{V} + \frac{∆ I}{I} = \frac{4}{100} + \frac{0.2}{5} = 0.08$

$\Rightarrow (\frac{∆ R}{R}) % = 0.08 \times 100 = 8 %$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Unit & DimensionSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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