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Two resistance 60.36 Omega and 30.09 Ome...

Two resistance `60.36 Omega and 30.09 Omega` are connected in parallel. The equivalent resistance is

A

`20 pm 0.08 Omega`

B

`20 pm 0.06 Omega`

C

`20 pm 0.03 Omega`

D

`20 pm 0.10 Omega`

Text Solution

Verified by Experts

The correct Answer is:
A
Equivalent resistance is
`R_("parallel") = (R_(1)R_(2))/(R_(1)+R_(2))=(60xx30)/(60+30)=20 Omega` and tolerance value is
`rArr Delta R_(P) =R_(P) {(Delta R_(1))/(R_(1))+(Delta R_(2))/(R_(2))-(Delta R_(1)+Delta R_(2))/(R_(1)+R_(2))}`
`=20 {(0.36)/(60)+(0.09)/(30) -(0.36+0.09)/(90)}`
`=20{0.006+0.003-0.005}=0.085 Omega` So, resistance in parallel is `R_(p) =20 pm 0.08`
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