1.

Given R_(1)=5.0+-0.2Omega, R_(2)=10*0+-0.1Omega. What is total resistance in parallel with possible % error.?

Answer»

`15Omega+-2%`
`3.3OMEGA+-7%`
`15Omega+-7%`
`3.3Omega2%`

Solution :In parallel, `R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(5*0xx10*0)/(5*0+10*0)`
`=(50)/(15)=3*3Omega`
ALSO `(DeltaR_(p))/(R_(p))XX100=(DeltaR_(1))/(R_(1))xx100+(DeltaR^(2))/(R^(2))xx100`
`+(DELTA(R_(1)+R_(2)))/(R_(1)+R_(2))xx100`
`=(0*2)/(5*0)xx100+(0*1)/(10*0)xx100+(0*3)/(15)xx100=7%`
`:.R_(p)=3*3Omega+-7%`
Hence correct answer is `(b)`.


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