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A resistor of 6kOmega with tolerance 10%...

A resistor of `6kOmega` with tolerance `10%` and another of `4kOmega` with tolerance `10%` are connected in series. The tolerance of combination is about :

A

`5%`

B

`10%`

C

`12%`

D

`40%`

Text Solution

Verified by Experts

The correct Answer is:
B
`(DeltaR_(1))/(R_(1))xx100=10impliesR_(1)=(6xx10)/(100)=0*6kOmega`
Similarly `(DeltaR_(2))/(R_(2))xx100=10`
`impliesDeltaR_(2)=(4xx10)/(100)=0*4kOmega`
In series, `R_(s)=R_(1)+R_(2)=6+4=10kOmega`
`DeltaR_(s)=DeltaR_(1)+DeltaR_(2)=0*6+0*4`
`=1kOmega`
`:.(DeltaR_(s))/(R_(s))xx100=(1)/(10)xx100=10%`
Hence correct choice is `(b)`.
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