Two resistance are expressed as R1=(2.0±0.1)Ω and R2=(3.0±0.1)Ω. The equivalent resistance when they are connected in series with percentage error is

R=R1+R2=2.0+3.0=5.0Ω△RR×100=△R1R1×100+△R2R2×100=0.12.0×100+0.13.0×100=5+3.33=8% after rounding off

Two resistance are expressed as R1=(2.0±0.1)Ω and R2=(3.0±0.1)Ω. The equivalent resistance when they are connected in series with percentage error is

R=R1+R2=2.0+3.0=5.0Ω△RR×100=△R1R1×100+△R2R2×100=0.12.0×100+0.13.0×100=5+3.33=8% after rounding off

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Two resistance are expressed as $R_{1} = (2.0 \pm 0.1) Ω$ and $R_{2} = (3.0 \pm 0.1) Ω$ . The equivalent resistance when they are connected in series with percentage error is

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$(5.0 \pm 10 %) Ω$

$(5.0 \pm 8 %) Ω$

$(5.0 \pm 13 %) Ω$

$(5.0 \pm 4 %) Ω$

answer is B.

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Detailed Solution

$R = R_{1} + R_{2} = 2.0 + 3.0 = 5.0 Ω$
$\frac{△ R}{R} \times 100 = (\frac{△ R_{1}}{R_{1}} \times 100) + (\frac{△ R_{2}}{R_{2}} \times 100)$
$= (\frac{0.1}{2.0} \times 100) + (\frac{0.1}{3.0} \times 100)$
$= 5 + 3.33 = 8 %$ after rounding off

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Two resistance are expressed as R1=(2.0±0.1)Ω and R2=(3.0±0.1)Ω. The equivalent resistance when they are connected in series with percentage error is

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Two resistance are expressed as $R_{1} = (2.0 \pm 0.1) Ω$ and $R_{2} = (3.0 \pm 0.1) Ω$ . The equivalent resistance when they are connected in series with percentage error is