Question

Resistors R(1) = (200 + or - 6) ohm and R(2) = (100 + or - 4) ohm are connected in a) series b) parallel. Find equivalent resistances of the a ) series combination b ) parallel combination. Give your errors in terms of errors expressed in percentage.

Answer 1

$$\begin{gathered} R_1=200\pm 6ohm \\ {R}_2=100\pm 4ohm \\ equiva\ln etresisa\tan ceinseriescombination \\ {R}_s=R_1+R_2 \\ {R}_s=200+100=300ohm \\ ∆R_s=∆R_1+∆R_2 \\ ∆R_s=\frac{6}{200}\times 100+\frac{4}{100}\times 100=3\%+4\%=7\% \\ equivalentresis\tan ceinparallelcommbination: \\ {R}_p=\frac{R_1{R}_2}{R_1+R_2} \\ {R}_1+R_2=R_s \\ {R}_p=\frac{R_1{R}_2}{R_s}=\frac{200\times 100}{300}=\frac{200}{3}ohm \\ \frac{∆R_1}{R_1}+\frac{∆R_2}{R_2}+\frac{∆R_s}{R_s} \\ \frac{∆R_p}{{R_p}^2}=\frac{∆R_1}{{R_1}^2}+\frac{∆R_2}{{R_2}^2}=\frac{3\%}{{200}^2}+\frac{4\%}{{100}^2} \\ ∆R_p={\left(\frac{200}{3}\right)}^2\left(\frac{3\%}{{200}^2}+\frac{4\%}{{100}^2}\right) \\ ∆R_p=\frac{200\times 200}{9}\times \frac{1}{100\times 100}\left(\frac{3\%}{4}+4\%\right) \\ ∆R_p=\frac{4}{9}\times \frac{19}{4}\%=2.11\% \end{gathered}$$