question_answer

Two resistances \[{{\text{R}}_{\text{1}}}=\left( \text{16}\pm 0.\text{3} \right)\]ohm and \[{{\text{R}}_{\text{2}}}=\left( \text{48 }\pm 0.\text{5} \right)\]ohm are connected in parallel. Find the total resistance of the combination and maximum percentage error.

Answer:

                Here,                                                                           \[{{\text{R}}_{\text{1}}}=\text{ 16 ohm}\] ,                                                          \[\Delta {{\text{R}}_{\text{1}}}=\text{ }0.\text{3}\Omega\]                                                                            \[{{\text{R}}_{\text{2}}}=\text{48 ohm}\] ,                                                          \[\Delta {{\text{R}}_{\text{2}}}=\text{ }0.\text{5}\Omega\] ,                                                                                 \[{{\text{R}}_{\text{p}}}=\text{ }?\] \[\frac{1}{{{R}_{p}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}=\frac{1}{16}+\frac{1}{48}=\frac{3+1}{48}=\frac{4}{48}=\frac{1}{12}\]                                                                           \[{{\text{R}}_{\text{p}}}=\text{ 12 ohm}\] On differenting,                                                        \[\frac{1}{{{R}_{p}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\]  we get          \[\frac{-\Delta {{R}_{p}}}{R_{p}^{2}}=\frac{\Delta {{R}_{1}}}{R_{1}^{2}}-\frac{\Delta {{R}_{2}}}{R_{2}^{2}}\] \[\Delta {{\text{R}}_{\text{p}}}=\Delta {{R}_{1}}{{\left( \frac{{{R}_{p}}}{{{R}_{1}}} \right)}^{2}}+\Delta {{R}_{2}}{{\left( \frac{{{R}_{p}}}{{{R}_{2}}} \right)}^{2}}=0.3{{\left( \frac{12}{16} \right)}^{2}}+0.5{{\left( \frac{12}{48} \right)}^{2}}\]                                   \[=\text{ }0.\text{16875 }+\text{ }0.0\text{3125 }=\text{ }0.\text{2}0\text{ ohm}\] \[\frac{\Delta {{R}_{p}}}{{{R}_{p}}}\times \text{ 1}00\times \frac{0.20}{12}\times \text{ 1}00\text{ }=\text{ 1}.\text{6 }\!\!%\!\!\text{ }\]