A) \[5.125%\]
B) \[2%\]
C) \[3.125%\]
D) \[10.125%\]
Correct Answer: D
Solution :
\[{{R}_{1}}=(6\underline{+}\,3)\] k \[\Omega \]and \[{{R}_{2}}=(10\,\underline{+}\,0.2)\,k\Omega .\] Then equivalent resistance, \[{{R}_{parallel}}=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\] Let \[({{R}_{1}}+{{R}_{2}})=x\Rightarrow {{R}_{p}}=\frac{{{R}_{1}}{{R}_{2}}}{x}\] Taking log of both sides, In \[{{R}_{p}}\]= In \[{{R}_{1}}+\] In \[{{R}_{2}}-\] In x Differentiating \[\frac{\Delta \,{{R}_{p}}}{{{R}_{p}}}=\frac{\Delta \,{{R}_{1}}}{{{R}_{1}}}+\frac{\Delta {{R}_{2}}}{{{R}_{2}}}+\left( -\frac{\Delta \,x}{x} \right)\] for error \[\frac{\Delta \,{{R}_{p}}}{{{R}_{p}}}=\frac{\Delta \,{{R}_{1}}}{{{R}_{1}}}+\frac{\Delta {{R}_{2}}}{{{R}_{2}}}+\frac{\Delta \,x}{x}\]positive \[\frac{\Delta \,x}{x}\] is used because error increases in each step of calculation. \[\Delta x=0.3+0.2=0.5\Omega \] \[x=6+10=16\Omega \] \[\Rightarrow \,\frac{\Delta \,x}{x}=\frac{0.5}{16}\] \[\therefore \] Total error \[=\frac{0.3}{6}+\frac{0.2}{10}+\frac{0.5}{16}\] \[=0.5+0.02+0.03125=0.10125\] \[\therefore \]Percentage error = 10.125%
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