# 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 \%$ $(d)\;16 \Omega \pm 6.25 \%,3 \Omega \pm 23 \%$

$R_S=R_1+R_2=16 \Omega$
$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.

edited Jan 9, 2014 by meena.p