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
\[\frac{-\Delta
{{R}_{p}}}{R_{p}^{2}}=\frac{\Delta {{R}_{1}}}{R_{1}^{2}}-\frac{\Delta
{{R}_{2}}}{R_{2}^{2}}\]
\[\Delta {{\text{R}}_{\text{p}}}=\Delta
{{R}_{1}}{{\left( \frac{{{R}_{p}}}{{{R}_{1}}} \right)}^{2}}+\Delta
{{R}_{2}}{{\left( \frac{{{R}_{p}}}{{{R}_{2}}} \right)}^{2}}=0.3{{\left(
\frac{12}{16} \right)}^{2}}+0.5{{\left( \frac{12}{48} \right)}^{2}}\]
\[=\text{
}0.\text{16875 }+\text{ }0.0\text{3125 }=\text{ }0.\text{2}0\text{ ohm}\]
\[\frac{\Delta {{R}_{p}}}{{{R}_{p}}}\times \text{
1}00\times \frac{0.20}{12}\times \text{ 1}00\text{ }=\text{ 1}.\text{6
}\!\!%\!\!\text{ }\]
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