Two resistors are measured as R 1 = ( 5.0 ± 0.2 ) Ω and R 2 = ( 10.0 ± 0.1 ) Ω . Find the maximum percentage error in their series resistance.

Δ R s = 0.2 + 0.1 = 0.3 Ω . Percentage error = 15.0 0.3 × 100 = 2% .

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Two resistors are measured as $R_{1} = (5.0 \pm 0.2) Ω$ and $R_{2} = (10.0 \pm 0.1) Ω$ . Find the maximum percentage error in their series resistance.

Answer:

Solution:

$Δ R_{s} = 0.2 + 0.1 = 0.3 Ω$ .

Percentage error $= \frac{0.3}{15.0} \times 100 = 2%$ .

Stream:JEESubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

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