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In ohm's law experiment potential drop a...

In ohm's law experiment potential drop across a resistance was measured as `v = 5.0` volt and current was measured as `I = 2.0amp` If least count of the volmeter and ammeter are `0.1V` and `0.01A` respectively then find the maximum permissible error in resistance .

Text Solution

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`R = (v)/(i) = v x i^(-1)`
`((DeltaR)/(R))_(max)=(Deltav)/(v)+(Deltai)/(i)`
`Deltav = 0.1` volt (least count0
`Deltai = 0.01` amp (least count)
`%((DeltaR)/(R))_(max)=((0.1)/(5.0)+(0.01)/(2.00))xx100%=2.5%` .
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