Question

.In an experiment the values of two resistances were measured to be as given below. R1 = 5.0 ± 0.2 ohm and R2 = 10.0 ± 0.1 ohms. Find their combined resistance in series.​

Answer 1

Answer:

Combined resistance in series is 15±0.3Ω

Explanation:

Given that,

In an experiment the values of two resistances were measured to be as given below.

$R_{1}=5.0±0.2 \: Ω \\ </p><p>R_{2}=10.0±0.1 \: Ω$

Combined resistance in series is the sum of individual resistances.Therefore the resistance in series is

$R_{s}=R_{1}+R_{2}$

Substituting values of given resistances we get

$R_{s}=(5.0 ±0.2Ω)+(10.0±0.1Ω) \\ = (5 + 10)±(0.2 + 0.1) \\ = 15±0.3Ω$

Therefore,Combined resistance in series is 15±0.3Ω

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