Two resistances are given as R _1=(10 pm 0.5) | vedclass

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Question

$\frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R _2}$

Differentiating both sides, we get

$\frac{\Delta R }{ R ^2}=\frac{\Delta R _1}{ R _1^2}+\frac{\D

Vedclass · Accepted Answer

$4.33$

Vedclass · Answer

$\frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R _2}$

Differentiating both sides, we get

$\frac{\Delta R }{ R ^2}=\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}\left[ R =\frac{ R _1 R _2}{ R _1+ R _2}=\frac{10 	imes 15}{10+15}=6ight]$

$\Rightarrow \frac{\Delta R }{ R }=\left(\frac{\Delta R _1}{ R _1^2}+\frac{\Delta R _2}{ R _2^2}ight) R$

$=\left(\frac{0.5}{100}+\frac{0.5}{225}ight) 6$

$=\left(\frac{6 	imes 0.5}{25}ight)\left(\frac{1}{4}+\frac{1}{9}ight)=\frac{13}{300}$

$\frac{\Delta R }{ R } 	imes 100=\frac{13}{3}=4.33 \%$

Two resistances are given as $R _1=(10 \pm 0.5)\,\Omega$ and $R_2=(15 \pm 0.5)\, \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

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