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In the circuit shown, voltmeter is ideal...

In the circuit shown, voltmeter is ideal and its least count is 0.1 V. The least count of ammeter is 1 mA. Reading of the voltmeter be 30.0 V and the reading of ammeter is 0.020 A. We shell calculate the value of resistance R within error limits.

Text Solution

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`V =30.0,I=0.020A,R=(V)/(I)=(30.0)/(0.020)=1.50kOmega`
Error : As `R=(V)/(I):.(DeltaR)/(R)=(DeltaV)/(V)+(DeltaI)/(I)rArr DeltaR=R((DeltaV)/(V)+(DeltaI)/(I))`
`=1.50xx10^(3)((0.1)/(30.0)+(0.001)/(0.020))=0.080kOmega`
So, resistance is `(1.2 pm 0.08 k Omega)`
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