Question

The values of two resistors are $ {{R}_{1}}=(6\pm 0.3)k\Omega $ and $ {{R}_{2}}=(10\pm 0.2)k\Omega $ . The percentage error in the equivalent resistance when they are connected in parallel is

• A. 5.125%
• B. 2%
• C. 3.125%
• D. 7%
• E. 10.125%

Answer 1

Answer: (E)

$ {{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 $ %