In Ohm's law experiment , the potential drop acros a resistance was as `V = 5.0 V` and the current was measured as `I = 2.00 A`. Find the maximum permissible error in resistance.
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` R = (V)/( i) = V xx i^(-1)` `((d R)/( R))_(max) = ( Delta V) / (V) + ( Delta i)/( i)` `V = 5.0 V rarr Delta V = 0.1 V` `I = 2.00 A rarr Delta I = 0.01 A` `% ((d R)/( R ))_(max) = (( 0.1)/( 5.0) + ( 0.01)/( 2.00)) xx 100% = 2.5%` `= Value of R` From the observation, `R = ( V)/( i) = ( 5.0)/( 2.00) = 2.5 Omega`. So , we can write `R = ( 2.5 +- 2.5%) Omega`
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