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PhysicsErrorMiscellaneousEasy2 minPYQ_2020

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The heat generated in a circuit is given by $Q = I^{2} Rt,$ where I is current, $R$ is resistance and tis time. If the percentage errors in measuring I. $R$ and $t$ are $2%, 1%$ and $1%$ respectively, then the maximum error in measuring heat will be

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Solution:

$\frac{Δ Q}{Q} \times 100 = \frac{2Δ I}{I} \times 100 + \frac{Δ R}{R} \times 100 + \frac{Δ t}{t} \times 100 = 2 \times 2% + 1% + 1% = 6%$

Stream:BITSATSubject:PhysicsTopic:ErrorSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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