Explanation

$R_{1} = (6 \pm 0.3)k\Omega,\ R_{2} = (10 \pm 0.2)k\Omega$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R_{parallel} = \frac{R_{1}R_{2}}{\left( R_{1} + R_{2} \right)}$ Let $\left( R_{1} + R_{2} \right) = x$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R_{P} = \frac{R_{1}R_{2}}{x}$ Taking log of both sides $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ln\ R_{P} = ln\ R_{1} + \ln R_{2} - \ln x$ Differentiating, $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\mathrm{\Delta}R_{P}}{R_{P}} = \frac{\mathrm{\Delta}R_{1}}{R_{1}} + \frac{\mathrm{\Delta}R_{2}}{R_{2}} + \left( - \frac{\mathrm{\Delta}x}{x} \right)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}x_{mean} = \frac{0.3 + 0.2}{2} = 0.25\ \Omega$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R_{mean} = \frac{6 + 10}{2} = 8\ \Omega$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{6 + 10}{2} = 8\ \Omega$ $\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\ \mathrm{\Delta}x}{x} = \frac{0.25}{8}$ $\therefore Total\ error = \frac{0.3}{6} + \frac{0.2}{10} + \frac{0.25}{8}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.05 + 0.02 + 0.03125 = 0.10125$ $\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{\mathrm{\Delta}R_{P}}{R_{P}} = 10.125\%$