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Given R(1)=5.0+-0.2Omega, R(2)=10*0+-0.1...

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

A

`15Omega+-2%`

B

`3.3Omega+-7%`

C

`15Omega+-7%`

D

`3.3Omega2%`

Text Solution

Verified by Experts

The correct Answer is:
B
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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