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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Units and Measurement
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Two resistances are expressed as $R_1=(4\pm0.5) \Omega$ and $R_2=(12 \pm 0.5) \Omega$. what is the net resistance when they are connected 1) series 2) in parallel, with percentage error ?

\[(a)\;16 \Omega \pm 23 \%  ,3 \Omega \pm 6.25 \% \] \[(b)\;3 \Omega \pm 2.3 \%,3 \Omega \pm 6.25 \% \]\[ (c)\;3 \Omega \pm 23 \%,16 \Omega \pm 6.25 \% \] \[ (d)\;16 \Omega \pm 6.25 \%,3 \Omega \pm 23 \%\]

Can you answer this question?
 
 

1 Answer

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$R_S=R_1+R_2=16 \Omega$
$R_P=\large\frac{R_1R_2}{R_1+R_2}$$=3 \Omega$
$\large\frac{\Delta R_S}{R_S} $$\times 100$ is $\%$ error in $R_S$
$=\large\frac{1}{16}$$ \times 100=6.25 \%$
$=>R_S=16 \Omega \pm 6.25 \%$
$\large\frac{\Delta R_P}{R_P} $$\times 100$ is $\%$ error in $R_P$
$\bigg(\large\frac{\Delta R_1}{R_1} +\frac{\Delta R_2}{R_2}+\frac{\Delta R_S}{R_S}\bigg)$$ \times 100$
$=\bigg(\large\frac{.5}{4}+\frac{.5}{12}+\frac{1}{16}\bigg) $$ \times 100=23 \%$
$=>R_P=3 \Omega \pm 23 \%$
Hence d is the correct answer. 

 

answered Jun 17, 2013 by meena.p
edited Jan 9, 2014 by meena.p
 

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