1.

The potential difference across the ends of a wire is found to be `(100pm2)` volt and the current flowing in the wire is found to be `(10pm0.1)`A. What is the maximum percentage error in the measurement of resistance?A. `2%`B. `3%`C. `4%`D. `5%`

Answer» Correct Answer - B
`R=(V)/(I),V=100V,deltaV=2V,I=10A,delta1=0.1A`
`therefore` Max. percentage error in R is
`((deltaR)/(R))_(max)xx100=((deltaV)/(V)xx100)+((deltaI)/(I)xx100)`
`=(2)/(100)xx100+(0.1)/(10)xx100=2%+1%=3%`
Note : Errors are to be added.


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