two resistance R1=(4+-0.5)&R2=(12+-0.5). What is the …

{tex}\eqalign{ & {\text{Here, }}{{\text{R}}_1} = 4 \pm 0.5 \cr & {\text{ }}{{\text{R}}_2} = 12 \pm 0.5 \cr & {\text{Case - I: When }}{{\text{R}}_{{\text{1 }}}}{\text{\& }}{{\text{R}}_{\text{2}}}{\text{ are connected in series:}} \cr & {\text{we know, in series:}} \cr & {{\text{R}}_{net}} = {R_1} + {R_2}{\text{ }} \cr & \Rightarrow {R_{net}} = (4 \pm 0.5) + (12 \pm 0.5) = 16 \pm 1 \cr & \Rightarrow {\text{% Error in }}{{\text{R}}_{net}} = 16 \pm \frac{1}{{16}} \times 100 = 16 \pm 6.25\% \cr & \cr & {\text{Case - II: When }}{{\text{R}}_{{\text{1 }}}}{\text{& }}{{\text{R}}_{\text{2}}}{\text{ are connected in parallel:}} \cr & {{\text{R}}_{{\text{net}}}} = \frac{{{{\text{R}}_{{\text{1 }}}}{{\text{R}}_{\text{2}}}}}{{{{\text{R}}_{{\text{1 }}}} + {{\text{R}}_{\text{2}}}}}{\text{ - (1)}} \cr & {\text{Now,}} \cr & {\text{% Error in }}{{\text{R}}_1} = \frac{{{\text{absolute error in }}{{\text{R}}_1}}}{{{\text{True value of }}{{\text{R}}_1}}} \times 100 \cr & \Rightarrow {\text{% Error in }}{{\text{R}}_1} = \frac{{0.5}}{4} \times 100 = 12.5\% \cr & {\text{Similarly, }} \cr & {\text{% Error in }}{{\text{R}}_2} = \frac{{0.5}}{{12}} \times 100 = 4.16\% \cr & \therefore {\text{ }}{{\text{R}}_1} = 4 \pm 12.5\% \cr & {\text{and }}{R_2} = 12 \pm 4.16\% \cr & \therefore {\text{ }}{{\text{R}}_1}{R_2} = (4{\text{ohm}} \pm 12.5\% )(12{\text{ohm}} \pm 4.16\% ) \cr & {\text{ = (48 oh}}{{\text{m}}^2} \pm 16.66\% ) \cr & and{\text{ }}{R_1} + {R_2}{\text{ }} = 16{\text{ ohm}} \pm 1\% \cr & {\text{Putting these values in equation (1):}} \cr & \Rightarrow {R_{net}} = \frac{{{\text{(48 oh}}{{\text{m}}^2} \pm 16.66\% )}}{{16{\text{ ohm}} \pm 1\% }} = 3{\text{ ohm}} \pm {\text{16}}{\text{.66\% }} \cr} {/tex}