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The values of two resistors are R(1)=(6p...

The values of two resistors are `R_(1)=(6pm0.3)kOmegaandR_(2)=(10pm0.2)kOmega`. The percentage error in the equivalent resistance when they are connected in parallel is

A

A)`5.125%`

B

B)`2%`

C

C)`3.125%`

D

D)`10.125%`

Text Solution

Verified by Experts

The correct Answer is:
D
`R_(1)=(6pm0.3)kOmega,R_(2)=(10pm0.2)kOmega`
`R_("parallel")=(R_(1)R_(2))/((R_(1)+R_(2)))`
Let `(R_(1)+R_(2))=x`
`implies R_(p)=(R_(1)R_(2))/(x)`
`lnR_(p)=lnR_(1)+lnR_(2)-lnx`
Differentiating,
`(DeltaR_(p))/(R_(p))=(DeltaR_(1))/(R_(1))+(DeltaR_(2))/(R_(2))+(-(Deltax)/(x))`
`Deltax_("mean")=(0.3+0.2)/(2)=0.25Omega`
`R_("mean")=(6+10)/(2)=8Omega`
`therefore x=(6+10)/(2)=8Omega`
`implies (Deltax)/(x)=(0.25)/(8)`
`therefore` Total error `=(0.3)/(6)+(0.2)/(10)+(0.25)/(8)`
`=0.05+0.02+0.03125`
`=0.10125`
`therefore (DeltaR_(p))/(R_(p))=10.125%`
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