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If voltage `V=(100pm5)` V and current `I=(10pm0.2)` A, the percentage error in resistance `R` isA. `5.2`%B. 0.25C. 0.07D. 0.1 |
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Answer» Correct Answer - C Given, voltage, `V=(100pm5)`V Current, `I=(10pm0.2)`A `therefore` Resistance, `R=(V)/(I)` Maximum percentage error in resistance `((DeltaR)/(R)xx100)=((DeltaV)/(V)xx100)+((DeltaI)/(I)xx100)` =`((5)/(100)xx100)+((0.2)/(10)xx100)` = 5+2=7% |
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