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Given Resistance R(1) = (8 +- 0.4) Omega...

Given Resistance `R_(1) = (8 +- 0.4) Omega`and Resistence, `R_(2) = (8 +- 0.6) Omega` What is the net resistence when `R_(1)` and `R_(2)` are connected in series?

A

`(16 +- 0.4 Omega)`

B

`(3.45 +- 0.3) Omega`

C

`(3.45 +- 0.4) Omega`

D

`(3.45 +- 0.5) Omega`

Text Solution

Verified by Experts

The correct Answer is:
C
When resistances are connected in series, then net resistence `R = R_(1) + R_(2)`
`R = (8 + 8) +- (0.4 + 0.6)Omega = (16+- 1.0)Omega`
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