A) 5.125%
B) 2%
C) 3.125%
D) 7%
E) 10.125%
Correct Answer: E
Solution :
\[{{R}_{1}}=(6\pm 0.3)k\Omega ,{{R}_{2}}=(10\pm 0.2)k\Omega \] \[{{R}_{parallel}}=\frac{{{R}_{1}}{{R}_{2}}}{({{R}_{1}}+{{R}_{2}})}\] Let \[({{R}_{1}}+{{R}_{2}})=x\] \[\Rightarrow \] \[{{R}_{P}}=\frac{{{R}_{1}}{{R}_{2}}}{x}\] \[\ln \,{{R}_{p}}=\ln \,{{R}_{1}}\,+\ln \,{{R}_{2}}\,-\ln \,x\] Differentiating, \[\frac{\Delta {{R}_{p}}}{{{R}_{p}}}=\frac{\Delta {{R}_{1}}}{{{R}_{1}}}+\frac{\Delta {{R}_{2}}}{{{R}_{2}}}+\left( -\frac{\Delta x}{x} \right)\] \[\Delta {{x}_{mean}}=\frac{0.3+0.2}{2}=0.25\,\Omega \] \[{{R}_{mean}}=\frac{6+10}{2}=8\,\Omega \] \[\therefore \] \[x=\frac{6+10}{2}=8\,\Omega \] \[\Rightarrow \] \[\frac{\Delta x}{x}=\frac{0.25}{8}\] \[\therefore \]Total error\[=\frac{0.3}{6}+\frac{0.2}{10}+\frac{0.25}{8}\] \[=0.05+0.02+0.03125=0.10125\] \[\therefore \] \[\frac{\Delta {{R}_{p}}}{{{R}_{p}}}=10.125%\]
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