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Two carbon resistances are given by R(1)...

Two carbon resistances are given by `R_(1)=(4pm0.4)` ohm and `R_(2)=(12pm0.6)Omega`. What is their net resistance with percentage error, if they are connected in series?

A

`16Omegapm5%`

B

`16Omegapm6.25%`

C

`16Omegapm8%`

D

`16Omegapm12%`

Text Solution

Verified by Experts

The correct Answer is:
b
`R_(1)=4Omega,DeltaR_(1)=pm0.4,R_(2)=12OmegaandDeltaR_(2)=0.6Omega`
When they are connected in series, their combined resistance is `R_(s)=R_(1)+R_(2)=4+12=16Omega`
and `DeltaR_(s)=DeltaR_(1)+DeltaR_(2)=0.4+0.6=1Omega`
`therefore (DeltaR_(s))/(R_(s))xx100=(1)/(16)xx100=6.25%`
`therefore R_(s)=16Omegapm06.25%`
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